Algebra - Completing the Square Review Questions

Solutions

Question 1

a) $\begin{align}x^2+6x+7&=[x^2+6x]+7\,\,\,\,\,\,\text{half 6 and square it to get 9}\\&=[x^2+6x+9-9]+7\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=(x^2+6x+9)-9+7\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x+3)^2-9+7\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x+3)^2-2\end{align}$

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b) $\begin{align}x^2+8x+20&=[x^2+8x]+20\,\,\,\,\,\,\text{half 8 and square it to get 16}\\&=[x^2+8x+16-16]+20\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=(x^2+8x+16)-16+20\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x+4)^2-16+20\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x+4)^2+4\end{align}$

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c) $\begin{align}x^2+12x+25&=[x^2+12x]+25\,\,\,\,\,\,\text{half 12 and square it to get 36}\\&=[x^2+12x+36-36]+25\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=(x^2+12x+36)-36+25\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x+6)^2-36+25\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x+6)^2-11\end{align}$

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d) $\begin{align}x^2+2x+9&=[x^2+2x]+9\,\,\,\,\,\,\text{half 2 and square it to get 1}\\&=[x^2+2x+1-1]+9\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=(x^2+2x+1)-1+9\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x+1)^2-1+9\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x+1)^2+8\end{align}$

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e) $\begin{align}x^2-10x+15&=[x^2-10x]+15\,\,\,\,\,\,\text{half -10 and square it to get 25}\\&=[x^2-10x+25-25]+15\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=(x^2-10x+25)-25+15\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x-5)^2-25+15\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x-5)^2-10\end{align}$

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f) $\begin{align}x^2-14x+1&=[x^2-14x]+1\,\,\,\,\,\,\text{half -14 and square it to get 49}\\&=[x^2-14x+49-49]+1\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=(x^2-14x+49)-49+1\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x-7)^2-49+1\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=(x-7)^2-48\end{align}$

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Question 2

a) $\begin{align}x^2+10x+21&=0\\{\bf x^2+10x}+21&=0\,\,\,\,\,\,\,\,\,\text{half the 10 and square it to get 25}\\{\bf x^2+10x+25-25}+21&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+10x+25}-25+21&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{(x+5)^2}-25+21&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\(x+5)^2-4&=0\\(x+5)^2&=4\\x+5&=\pm\sqrt4\\x+5=2&\text{ or }x+5=-2\\x=-3&\text{ or }x=-7\end{align}$

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b) $\begin{align}x^2+16x+48&=0\\{\bf x^2+16x}+48&=0\,\,\,\,\,\,\,\,\,\text{half the 16 and square it to get 64}\\{\bf x^2+16x+64-64}+48&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+16x+64}-64+48&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{(x+8)^2}-64+48&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\(x+8)^2-16&=0\\(x+8)^2&=16\\x+8&=\pm\sqrt{16}\\x+8=4&\text{ or }x+8=-4\\x=-4&\text{ or }x=-12\end{align}$

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c) $\begin{align}x^2+22x+57&=0\\{\bf x^2+22x}+57&=0\,\,\,\,\,\,\,\,\,\text{half the 22 and square it to get 121}\\{\bf x^2+22x+121-121}+57&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+22x+121}-121+57&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{(x+11)^2}-121+57&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\(x+11)^2-64&=0\\(x+11)^2&=64\\x+11&=\pm\sqrt{64}\\x+11=8&\text{ or }x+11=-8\\x=-3&\text{ or }x=-19\end{align}$

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d) $\begin{align}x^2+4x&=0\\{\bf x^2+4x}&=0\,\,\,\,\,\,\,\,\,\text{half the 4 and square it to get 4}\\{\bf x^2+4x+4-4}&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+4x+4}-4&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{(x+2)^2}-4&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\(x+2)^2-4&=0\\(x+2)^2&=4\\x+2&=\pm\sqrt{4}\\x+2=2&\text{ or }x+2=-2\\x=0&\text{ or }x=-4\end{align}$

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e) $\begin{align}x^2-18x-19&=0\\{\bf x^2-18x}-19&=0\,\,\,\,\,\,\,\,\,\text{half the -18 and square it to get 81}\\{\bf x^2-18x+81-81}-19&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2-18x+81}-81-19&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{(x-9)^2}-81-19&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\(x-9)^2-100&=0\\(x-9)^2&=100\\x-9&=\pm\sqrt{100}\\x-9=10&\text{ or }x-9=-10\\x=19&\text{ or }x=-1\end{align}$

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f) $\begin{align}x^2-2x-8&=0\\{\bf x^2-2x}-8&=0\,\,\,\,\,\,\,\,\,\text{half the -2 and square it to get 1}\\{\bf x^2-2x+1-1}-8&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2-2x+1}-1-8&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{(x-1)^2}-1-8&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\(x-1)^2-9&=0\\(x-1)^2&=9\\x-1&=\pm\sqrt{9}\\x-1=3&\text{ or }x-1=-3\\x=4&\text{ or }x=-2\end{align}$

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Question 3

a) $\begin{align}x^2+3x+1&=[x^2+3x]+1\,\,\,\,\,\,\text{half 3 and square it to get }\frac94\\&=\left[x^2+3x+\frac94-\frac94\right]+1\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=\left(x^2+3x+\frac94\right)-\frac94+1\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x+\frac32\right)^2-\frac94+1\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x+\frac32\right)^2-\frac54\end{align}$

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b) $\begin{align}x^2+5x-1&=[x^2+5x]-1\,\,\,\,\,\,\text{half 5 and square it to get }\frac{25}{4}\\&=\left[x^2+5x+\frac{25}{4}-\frac{25}{4}\right]-1\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=\left(x^2+5x+\frac{25}{4}\right)-\frac{25}{4}-1\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x+\frac52\right)^2-\frac{25}{4}-1\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x+\frac52\right)^2-\frac{29}{4}\end{align}$

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c) $\begin{align}x^2+9x+\frac14&=[x^2+9x]+\frac14\,\,\,\,\,\,\text{half 9 and square it to get }\frac{81}{4}\\&=\left[x^2+9x+\frac{81}{4}-\frac{81}{4}\right]+\frac14\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=\left(x^2+9x+\frac{81}{4}\right)-\frac{81}{4}+\frac14\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x+\frac92\right)^2-\frac{81}{4}+\frac14\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x+\frac92\right)^2-\frac{80}{4}\\&=\left(x+\frac92\right)^2-20\end{align}$

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d) $\begin{align}x^2+x+2\frac14&=[x^2+x]+2\frac14\,\,\,\,\,\,\text{half 1 and square it to get }\frac14\\&=\left[x^2+x+\frac14-\frac14\right]+2\frac14\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=\left(x^2+x+\frac14\right)-\frac14+2\frac14\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x+\frac12\right)^2-\frac14+2\frac14\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x+\frac12\right)^2+2\end{align}$

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e) $\begin{align}x^2-7x-5&=[x^2+7x]-5\,\,\,\,\,\,\text{half -7 and square it to get }\frac{49}{4}\\&=\left[x^2+7x+\frac{49}{4}-\frac{49}{4}\right]-5\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=\left(x^2-7x+\frac{49}{4}\right)-\frac{49}{4}-5\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x-\frac72\right)^2-\frac{49}{4}-5\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x-\frac72\right)^2-\frac{69}{4}\end{align}$

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f) $\begin{align}x^2-x+\frac{71}{4}&=[x^2-x]+\frac{71}{4}\,\,\,\,\,\,\text{half -1 and square it to get }\frac14\\&=\left[x^2-x+\frac14-\frac14\right]+\frac{71}{4}\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=\left(x^2-7x+\frac14\right)-\frac14+\frac{71}{4}\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x-\frac12\right)^2-\frac14+\frac{71}{4}\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=\left(x-\frac12\right)^2+\frac{35}{2}\end{align}$

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Question 4

a) $\begin{align}x^2+3x+2&=0\\{\bf x^2+3x}+2&=0\,\,\,\,\,\,\,\,\,\text{half the 3 and square it to get }\frac94\\{\bf x^2+3x+\frac94-\frac94}+2&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+3x+\frac94}-\frac94+2&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+\frac32\right)^2}-\frac94+2&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+\frac32\right)^2-\frac14&=0\\\left(x+\frac32\right)^2&=\frac14\\x+\frac32&=\pm\sqrt{\frac14}\\x+\frac32=\frac12&\text{ or }x+\frac32=-\frac12\\x=-1&\text{ or }x=-2\end{align}$

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b) $\begin{align}x^2+7x+12&=0\\{\bf x^2+7x}+12&=0\,\,\,\,\,\,\,\,\,\text{half the 7 and square it to get }\frac{49}{4}\\{\bf x^2+7x+\frac{49}{4}-\frac{49}{4}}+12&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+7x+\frac{49}{4}}-\frac{49}{4}+12&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+\frac72\right)^2}-\frac{49}{4}+12&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+\frac72\right)^2-\frac14&=0\\\left(x+\frac72\right)^2&=\frac14\\x+\frac72&=\pm\sqrt{\frac14}\\x+\frac72=\frac12&\text{ or }x+\frac72=-\frac12\\x=-3&\text{ or }x=-4\end{align}$

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c) $\begin{align}x^2+11x+28&=0\\{\bf x^2+11x}+28&=0\,\,\,\,\,\,\,\,\,\text{half the 11 and square it to get }\frac{121}{4}\\{\bf x^2+11x+\frac{121}{4}-\frac{121}{4}}+28&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+11x+\frac{121}{4}}-\frac{121}{4}+28&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+\frac{11}{2}\right)^2}-\frac{121}{4}+28&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+\frac{11}{2}\right)^2-\frac94&=0\\\left(x+\frac{11}{2}\right)^2&=\frac94\\x+\frac{11}{2}&=\pm\sqrt{\frac94}\\x+\frac{11}{2}=\frac32&\text{ or }x+\frac{11}{2}=-\frac32\\x=-4&\text{ or }x=-7\end{align}$

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d) $\begin{align}x^2-5x-6&=0\\{\bf x^2-5x}-6&=0\,\,\,\,\,\,\,\,\,\text{half the -5 and square it to get }\frac{25}{4}\\{\bf x^2-5x+\frac{25}{4}-\frac{25}{4}}-6&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2-5x+\frac{25}{4}}-\frac{25}{4}-6&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x-\frac52\right)^2}-\frac{25}{4}-6&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x-\frac52\right)^2-\frac{49}{4}&=0\\\left(x-\frac52\right)^2&=\frac{49}{4}\\x-\frac52&=\pm\sqrt{\frac{49}{4}}\\x-\frac52=\frac72&\text{ or }x-\frac52=-\frac72\\x=6&\text{ or }x=-1\end{align}$

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e) $\begin{align}x^2-x-42&=0\\{\bf x^2-x}-42&=0\,\,\,\,\,\,\,\,\,\text{half the -1 and square it to get }\frac14\\{\bf x^2-x+\frac14-\frac14}-42&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2-x+\frac14}-\frac14-42&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x-\frac12\right)^2}-\frac14-42&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x-\frac12\right)^2-\frac{169}{4}&=0\\\left(x-\frac12\right)^2&=\frac{169}{4}\\x-\frac12&=\pm\sqrt{\frac{169}{4}}\\x-\frac12=\frac{13}{2}&\text{ or }x-\frac12=-\frac{13}{2}\\x=7&\text{ or }x=-6\end{align}$

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f) $\begin{align}x^2-3x+\frac54&=0\\{\bf x^2-3x}+\frac54&=0\,\,\,\,\,\,\,\,\,\text{half the -3 and square it to get }\frac94\\{\bf x^2-3x+\frac94-\frac94}+\frac54&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2-3x+\frac94}-\frac94+\frac54&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x-\frac32\right)^2}-\frac94+\frac54&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x-\frac32\right)^2-1&=0\\\left(x-\frac32\right)^2&=1\\x-\frac32&=\pm\sqrt{1}\\x-\frac32=1&\text{ or }x-\frac32=-1\\x=\frac52&\text{ or }x=\frac12\end{align}$

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Question 5

a) $\begin{align}2x^2+12x+5&=2[x^2+6x]+5\,\,\,\,\,\,\text{half 6 and square it to get 9}\\&=2[x^2+6x+9-9]+5\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=2(x^2+6x+9)-18+5\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=2(x+3)^2-18+5\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=2(x+3)^2-13\end{align}$

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b) $\begin{align}3x^2+12x-8&=3[x^2+4x]-8\,\,\,\,\,\,\text{half 4 and square it to get 4}\\&=3[x^2+4x+4-4]-8\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=3(x^2+4x+4)-12-8\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=3(x+2)^2-12-8\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=3(x+2)^2-20\end{align}$

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c) $\begin{align}5x^2+10x-1&=5[x^2+2x]-1\,\,\,\,\,\,\text{half 2 and square it to get 1}\\&=5[x^2+2x+1-1]-1\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=5(x^2+2x+1)-5-1\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=5(x+1)^2-5-1\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=5(x+1)^2-6\end{align}$

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d) $\begin{align}4x^2-24x+19&=4[x^2-6x]+19\,\,\,\,\,\,\text{half -6 and square it to get 9}\\&=4[x^2-6x+9-9]+19\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=4(x^2-6x+9)-36+19\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=4(x+3)^2-36+19\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=4(x+3)^2-17\end{align}$

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e) $\begin{align}7x^2-56x+13&=7[x^2-8x]+13\,\,\,\,\,\,\text{half -8 and square it to get 16}\\&=7[x^2-8x+16-16]+13\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=7(x^2-8x+16)-112+13\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=7(x-4)^2-112+13\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=7(x-4)^2-99\end{align}$

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f) $\begin{align}11x^2-22x-15&=11[x^2-2x]-15\,\,\,\,\,\,\text{half -2 and square it to get 1}\\&=11[x^2-2x+1-1]-15\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=11(x^2-2x+1)-11-15\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=11(x-1)^2-11-15\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=11(x-1)^2-26\end{align}$

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Question 6

a) $\begin{align}2x^2+8x+6&=0\,\,\,\,\,\,\,\,\,\text{divide the equation by 2}\\x^2+4x+3&=0\\{\bf x^2+4x}+3&=0\,\,\,\,\,\,\,\,\,\text{half the 4 and square it to get }4\\{\bf x^2+4x+4-4}+3&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+4x+4}-4+3&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+2\right)^2}-4+3&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+2\right)^2-1&=0\\\left(x+2\right)^2&=1\\x+2&=\pm\sqrt{1}\\x+2=1&\text{ or }x+2=-1\\x=-1&\text{ or }x=-3\end{align}$

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b) $\begin{align}3x^2+24x+45&=0\,\,\,\,\,\,\,\,\,\text{divide the equation by 3}\\x^2+8x+15&=0\\{\bf x^2+8x}+15&=0\,\,\,\,\,\,\,\,\,\text{half the 8 and square it to get }16\\{\bf x^2+8x+16-16}+15&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+8x+16}-16+15&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+4\right)^2}-16+15&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+4\right)^2-1&=0\\\left(x+4\right)^2&=1\\x+4&=\pm\sqrt{1}\\x+4=1&\text{ or }x+4=-1\\x=-3&\text{ or }x=-5\end{align}$

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c) $\begin{align}4x^2+8x+3&=0\,\,\,\,\,\,\,\,\,\text{divide the equation by 4}\\x^2+2x+\frac34&=0\\{\bf x^2+2x}+\frac34&=0\,\,\,\,\,\,\,\,\,\text{half the }2\text{ and square it to get }1\\{\bf x^2+2x+1-1}+\frac34&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+2x+1}-1+\frac34&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+1\right)^2}-1+\frac34&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+1\right)^2-\frac14&=0\\\left(x+1\right)^2&=\frac14\\x+1&=\pm\sqrt{\frac14}\\x+1=\frac12&\text{ or }x+1=-\frac12\\x=-\frac12&\text{ or }x=-\frac32\end{align}$

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d) $\begin{align}9x^2+18x-7&=0\,\,\,\,\,\,\,\,\,\text{divide the equation by 9}\\x^2+2x-\frac79&=0\\{\bf x^2+2x}-\frac79&=0\,\,\,\,\,\,\,\,\,\text{half the }2\text{ and square it to get }1\\{\bf x^2+2x+1-1}-\frac79&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+2x+1}-1-\frac79&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+1\right)^2}-1-\frac79&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+1\right)^2-\frac{16}{9}&=0\\\left(x+1\right)^2&=\frac{16}{9}\\x+1&=\pm\sqrt{\frac{16}{9}}\\x+1=\frac43&\text{ or }x+1=-\frac43\\x=\frac13&\text{ or }x=-\frac73\end{align}$

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e) $\begin{align}2x^2+9x+4&=0\,\,\,\,\,\,\,\,\,\text{divide the equation by 2}\\x^2+\frac92x+2&=0\\{\bf x^2+\frac92x}+2&=0\,\,\,\,\,\,\,\,\,\text{half the }\frac92\text{ and square it to get }\frac{81}{16}\\{\bf x^2+\frac92x+\frac{81}{16}-\frac{81}{16}}+2&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+\frac92x+\frac{81}{16}}-\frac{81}{16}+2&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+\frac94\right)^2}-\frac{81}{16}+2&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+\frac94\right)^2-\frac{49}{16}&=0\\\left(x+\frac94\right)^2&=\frac{49}{16}\\x+\frac94&=\pm\sqrt{\frac{49}{16}}\\x+\frac94=\frac74&\text{ or }x+\frac94=-\frac74\\x=-\frac12&\text{ or }x=-4\end{align}$

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f) $\begin{align}6x^2-7x-20&=0\,\,\,\,\,\,\,\,\,\text{divide the equation by 6}\\x^2-\frac76x-\frac{10}{3}&=0\\{\bf x^2-\frac76x}-\frac{10}{3}&=0\,\,\,\,\,\,\,\,\,\text{half the }-\frac76\text{ and square it to get }\frac{49}{144}\\{\bf x^2-\frac76x+\frac{49}{144}-\frac{49}{144}}-\frac{10}{3}&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2-\frac76x+\frac{49}{144}}-\frac{49}{144}-\frac{10}{3}&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x-\frac{7}{12}\right)^2}-\frac{49}{144}-\frac{10}{3}&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x-\frac{7}{12}\right)^2-\frac{529}{144}&=0\\\left(x-\frac{7}{12}\right)^2&=\frac{529}{144}\\x-\frac{7}{12}&=\pm\sqrt{\frac{529}{144}}\\x-\frac{7}{12}=\frac{23}{12}&\text{ or }x-\frac{7}{12}=-\frac{23}{12}\\x=\frac52&\text{ or }x=-\frac43\end{align}$

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Question 7

a) $\begin{align}5x^2+20x+\frac{5}{4}&=5[x^2+4x]+\frac54\,\,\,\,\,\,\text{half }4\text{ and square it to get }4\\&=5[x^2+4x+4-4]+\frac54\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=5(x^2+4x+4)-20+\frac54\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=5(x+2)^2-20+\frac54\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=5(x+2)^2-\frac{75}{4}\end{align}$

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b) $\begin{align}8x^2+24x+\frac{3}{2}&=8\left[x^2+3x\right]+\frac32\,\,\,\,\,\,\text{half }3\text{ and square it to get }\frac94\\&=8\left[x^2+3x+\frac94-\frac94\right]+\frac32\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=8\left(x^2+3x+\frac94\right)-18+\frac32\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=8\left(x+\frac32\right)^2-18+\frac32\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=8\left(x+\frac32\right)^2-\frac{33}{2}\end{align}$

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c) $\begin{align}6x^2+5x+\frac{2}{3}&=6[x^2+\frac56x]+\frac23\,\,\,\,\,\,\text{half }\frac56\text{ and square it to get }\frac{25}{144}\\&=6[x^2+\frac56x+\frac{25}{144}-\frac{25}{144}]+\frac23\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=6(x^2+\frac56x+\frac{25}{144})-\frac{25}{24}+\frac23\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=6(x+\frac{5}{12})^2-\frac{25}{24}+\frac23\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=6(x+\frac{5}{12})^2-\frac{3}{8}\end{align}$

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d) $\begin{align}2x^2+2x+\frac{1}{2}&=2[x^2+x]+\frac12\,\,\,\,\,\,\text{half 1 and square it to get }\frac14\\&=2[x^2+x+\frac14-\frac14]+\frac12\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=2(x^2+x+\frac14)-\frac12+\frac12\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=2(x+\frac12)^2-\frac12+\frac12\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=2(x+\frac12)^2\end{align}$

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e) $\begin{align}7x^2+4x+1&=7\left[x^2+\frac47x\right]+1\,\,\,\,\,\,\text{half }\frac47\text{ and square it to get }\frac{4}{49}\text{ (simplified)}\\&=7\left[x^2+\frac47x+\frac{4}{49}-\frac{4}{49}\right]+1\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=7\left(x^2+\frac47x+\frac{4}{49}\right)-\frac47+1\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=7\left(x+\frac27\right)^2-\frac47+1\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=7\left(x+\frac27\right)^2+\frac{3}{7}\end{align}$

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f) $\begin{align}11x^2-2x-\frac{5}{11}&=11\left[x^2-\frac{2}{11}x\right]-\frac{5}{11}\,\,\,\,\,\,\text{half }\frac{2}{11}\text{ and square it to get }\frac{1}{121}\text{ (simplified)}\\&=11\left[x^2-\frac{2}{11}x+\frac{1}{121}-\frac{1}{121}\right]-\frac{5}{11}\,\,\,\,\,\,\text{the expression in brackets are the same}\\&=11\left(x^2-\frac{2}{11}x+\frac{1}{121}\right)-\frac{1}{11}-\frac{5}{11}\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=11\left(x-\frac{1}{11}\right)^2-\frac{1}{11}-\frac{5}{11}\,\,\,\,\,\,\text{the expression in the brackets is a perfect square}\\&=11\left(x-\frac{1}{11}\right)^2-\frac{6}{11}\end{align}$

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Question 8

a) $\begin{align}3x^2-\frac{x}{2}-1&=0\,\,\,\,\,\,\,\,\,\text{divide the equation by 3}\\x^2-\frac16x-\frac13&=0\\{\bf x^2-\frac16x}-\frac13&=0\,\,\,\,\,\,\,\,\,\text{half the }\frac{1}{6}\text{ and square it to get }\frac{1}{144}\\{\bf x^2-\frac16x+\frac{1}{144}-\frac{1}{144}}-\frac13&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2-\frac16x+\frac{1}{144}}-\frac{1}{144}-\frac13&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x-\frac{1}{12}\right)^2}-\frac{1}{144}-\frac13&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x-\frac{1}{12}\right)^2-\frac{49}{144}&=0\\\left(x-\frac{1}{12}\right)^2&=\frac{49}{144}\\x-\frac{1}{12}&=\pm\sqrt{\frac{49}{144}}\\x-\frac{1}{12}=\frac{7}{12}&\text{ or }x-\frac{1}{12}=-\frac{7}{12}\\x=\frac23&\text{ or }x=-\frac12\end{align}$

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b) $\begin{align}6x^2+13x+6&=0\,\,\,\,\,\,\,\,\,\text{divide the equation by 6}\\x^2+\frac{13}{6}x+1&=0\\{\bf x^2+\frac{13}{6}x}+1&=0\,\,\,\,\,\,\,\,\,\text{half the }\frac{13}{6}\text{ and square it to get }\frac{169}{144}\\{\bf x^2+\frac{13}{6}x+\frac{169}{144}-\frac{169}{144}}+1&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+\frac{13}{6}x+\frac{169}{144}}-\frac{169}{144}+1&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+\frac{13}{12}\right)^2}-\frac{169}{144}+1&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+\frac{13}{12}\right)^2-\frac{25}{144}&=0\\\left(x+\frac{13}{12}\right)^2&=\frac{25}{144}\\x+\frac{13}{12}&=\pm\sqrt{\frac{25}{144}}\\x+\frac{13}{12}=\frac{5}{12}&\text{ or }x+\frac{13}{12}=-\frac{5}{12}\\x=-\frac23&\text{ or }x=-\frac32\end{align}$

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c) $\begin{align}20x^2+\frac{1}{2}x-\frac{1}{4}&=0\,\,\,\,\,\,\,\,\,\text{divide the equation by 20}\\x^2+\frac{1}{40}x-\frac{1}{80}&=0\\{\bf x^2+\frac{1}{40}x}-\frac{1}{80}&=0\,\,\,\,\,\,\,\,\,\text{half the }\frac{1}{40}\text{ and square it to get }\frac{1}{6400}\\{\bf x^2+\frac{1}{40}x+\frac{1}{6400}-\frac{1}{6400}}-\frac{1}{80}&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+\frac{1}{40}x+\frac{1}{6400}}-\frac{1}{6400}-\frac{1}{80}&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+\frac{1}{80}\right)^2}-\frac{1}{6400}-\frac{1}{80}&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+\frac{1}{80}\right)^2-\frac{81}{6400}&=0\\\left(x+\frac{1}{80}\right)^2&=\frac{81}{6400}\\x+\frac{1}{80}&=\pm\sqrt{\frac{81}{6400}}\\x+\frac{1}{80}=\frac{9}{80}&\text{ or }x+\frac{1}{80}=-\frac{9}{80}\\x=\frac{1}{10}&\text{ or }x=-\frac18\end{align}$

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d) $\begin{align}8x^2+\frac{18x}{5}-\frac{9}{5}&=0\,\,\,\,\,\,\,\,\,\text{divide the equation by 8}\\x^2+\frac{9}{20}x-\frac{9}{40}&=0\\{\bf x^2+\frac{9}{20}x}-\frac{9}{40}&=0\,\,\,\,\,\,\,\,\,\text{half the }\frac{9}{20}\text{ and square it to get }\frac{81}{1600}\\{\bf x^2+\frac{9}{20}x+\frac{81}{1600}-\frac{81}{1600}}-\frac{9}{40}&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+\frac{9}{20}x+\frac{81}{1600}}-\frac{81}{1600}-\frac{9}{40}&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+\frac{9}{40}\right)^2}-\frac{81}{1600}-\frac{9}{40}&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+\frac{9}{40}\right)^2-\frac{441}{1600}&=0\\\left(x+\frac{9}{40}\right)^2&=\frac{441}{1600}\\x+\frac{9}{40}&=\pm\sqrt{\frac{441}{1600}}\\x+\frac{9}{40}=\frac{21}{40}&\text{ or }x+\frac{9}{40}=-\frac{21}{40}\\x=\frac{3}{10}&\text{ or }x=-\frac34\end{align}$

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e) $\begin{align}10x^2-3x+\frac{2}{9}&=0\,\,\,\,\,\,\,\,\,\text{divide the equation by 10}\\x^2-\frac{3}{10}x+\frac{2}{90}&=0\\{\bf x^2-\frac{3}{10}x}+\frac{2}{90}&=0\,\,\,\,\,\,\,\,\,\text{half the }-\frac{3}{10}\text{ and square it to get }\frac{9}{400}\\{\bf x^2-\frac{3}{10}x+\frac{9}{400}-\frac{9}{400}}+\frac{2}{90}&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2-\frac{3}{10}x+\frac{9}{400}}-\frac{9}{400}+\frac{2}{90}&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x-\frac{3}{20}\right)^2}-\frac{9}{400}+\frac{2}{90}&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x-\frac{3}{20}\right)^2-\frac{1}{3600}&=0\\\left(x-\frac{3}{20}\right)^2&=\frac{1}{3600}\\x-\frac{3}{20}&=\pm\sqrt{\frac{1}{3600}}\\x-\frac{3}{20}=\frac{1}{60}&\text{ or }x-\frac{3}{20}=-\frac{1}{60}\\x=\frac{1}{6}&\text{ or }x=\frac{2}{15}\end{align}$

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f) $\begin{align}\frac{x^2}{2}+\frac{8x}{3}-2&=0\,\,\,\,\,\,\,\,\,\text{multiply the equation by 2}\\x^2+\frac{16}{3}x-4&=0\\{\bf x^2+\frac{16}{3}x}-4&=0\,\,\,\,\,\,\,\,\,\text{half the }\frac{16}{3}\text{ and square it to get }\frac{64}{9}\\{\bf x^2+\frac{16}{3}x+\frac{64}{9}-\frac{64}{9}}-4&=0\,\,\,\,\,\,\,\,\,\text{the expressions in bold are the same}\\\underline{x^2+\frac{16}{3}x+\frac{64}{9}}-\frac{64}{9}-4&=0\,\,\,\,\,\,\,\,\,\text{the expression underlined is a perfect square}\\\underline{\left(x+\frac83\right)^2}-\frac{64}{9}-4&=0\,\,\,\,\,\,\,\,\,\text{the expressions underlined are the same}\\\left(x+\frac83\right)^2-\frac{100}{9}&=0\\\left(x+\frac83\right)^2&=\frac{100}{9}\\x+\frac83&=\pm\sqrt{\frac{100}{9}}\\x+\frac83=\frac{10}{3}&\text{ or }x+\frac83=-\frac{10}{3}\\x=\frac23&\text{ or }x=-6\end{align}$

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