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Question

# Solve the following quadratic equations by factorisation. (1) x2 – 15x + 54 = 0 (2) x2 + x – 20 = 0 (3) 2y2 + 27y + 13 = 0 (4) 5m2 = 22m + 15 (5) 2x2 – 2x + $\frac{1}{2}$ = 0 (6) $6x-\frac{2}{x}=1$ (7) $\sqrt{2}{x}^{2}+7x+5\sqrt{2}=0$ to solve this quadratic equation by factorisation, complete the following activity. (8) $3{x}^{2}-2\sqrt{6}x+2=0$ (9) $2m\left(m-24\right)=50$ (10) $25{m}^{2}=9$ (11) $7{m}^{2}=21m$ (12) ${m}^{2}-11=0$

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Solution

## (1) x2 – 15x + 54 = 0 $⇒{x}^{2}-9x-6x+54=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-9\right)-6\left(x-9\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-6\right)\left(x-9\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-6\right)=0\mathrm{or}\left(x-9\right)=0\phantom{\rule{0ex}{0ex}}⇒x=6\mathrm{or}x=9$ So, 6 and 9 are the roots of the given quadratic equation. (2) x2 + x – 20 = 0 $⇒{x}^{2}+5x-4x-20=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+5\right)-4\left(x+5\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-4\right)\left(x+5\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-4\right)=0\mathrm{or}\left(x+5\right)=0\phantom{\rule{0ex}{0ex}}⇒x=4\mathrm{or}x=-5$ So, 4 and $-$5 are the roots of the given quadratic equation. (3) 2y2 + 27y + 13 = 0 $2{y}^{2}+26y+y+13=0\phantom{\rule{0ex}{0ex}}⇒2y\left(y+13\right)+\left(y+13\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(2y+1\right)\left(y+13\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(2y+1\right)=0\mathrm{or}\left(y+13\right)=0\phantom{\rule{0ex}{0ex}}⇒y=\frac{-1}{2}\mathrm{or}y=-13$ So, $\frac{-1}{2}\mathrm{and}-13$ are the roots of the given quadratic equation. (4) 5m2 = 22m + 15 $⇒5{m}^{2}-22m-15=0\phantom{\rule{0ex}{0ex}}⇒5{m}^{2}-25m+3m-15=0\phantom{\rule{0ex}{0ex}}⇒5m\left(m-5\right)+3\left(m-5\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(5m+3\right)\left(m-5\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(5m+3\right)=0\mathrm{or}\left(m-5\right)=0\phantom{\rule{0ex}{0ex}}⇒m=\frac{-3}{5}\mathrm{or}m=5$ So, $\frac{-3}{5}\mathrm{and}5$ are the roots of the given quadratic equation. (5) 2x2 – 2x + $\frac{1}{2}$ = 0 $⇒4{x}^{2}-4x+1=0\phantom{\rule{0ex}{0ex}}⇒{\left(2x\right)}^{2}-2×1×2x+1=0\phantom{\rule{0ex}{0ex}}U\mathrm{sin}gtheidentity,\phantom{\rule{0ex}{0ex}}{\left(a-b\right)}^{2}={a}^{2}+{b}^{2}-2ab\phantom{\rule{0ex}{0ex}}{\left(2x\right)}^{2}-2×1×2x+1=0\phantom{\rule{0ex}{0ex}}⇒{\left(2x-1\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{2},\frac{1}{2}$ So, $\frac{1}{2}\mathrm{and}\frac{1}{2}$ are the roots of the given quadratic equation. (6) $6x-\frac{2}{x}=1$ $⇒6{x}^{2}-2=1x\phantom{\rule{0ex}{0ex}}⇒6{x}^{2}-1x-2=0\phantom{\rule{0ex}{0ex}}⇒6{x}^{2}-4x+3x-2=0\phantom{\rule{0ex}{0ex}}⇒2x\left(3x-2\right)+1\left(3x-2\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(3x-2\right)\left(2x+1\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(3x-2\right)=0\mathrm{or}\left(2x+1\right)=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{2}{3}\mathrm{or}x=\frac{-1}{2}\phantom{\rule{0ex}{0ex}}$ $\frac{2}{3}\mathrm{and}\frac{-1}{2}$ are the roots of the given quadratic equation. (7) $\sqrt{2}{x}^{2}+7x+5\sqrt{2}=0$ $\sqrt{2}{x}^{2}+\overline{)5x}+\overline{)2x}+5\sqrt{2}=0\phantom{\rule{0ex}{0ex}}x\left(\sqrt{2}x+5\right)+\sqrt{2}\left(\sqrt{2}x+5\right)=0\phantom{\rule{0ex}{0ex}}\left(\sqrt{2}x+5\right)\left(x+\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}\left(\sqrt{2}x+5\right)=0\mathrm{or}\left(x+\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}\therefore x=\overline{)\frac{-5}{\sqrt{2}}}\mathrm{or}x=-\sqrt{2}\phantom{\rule{0ex}{0ex}}\therefore \overline{)\frac{-5}{\sqrt{2}}}\mathrm{and}-\sqrt{2}\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}.$ (8) $3{x}^{2}-2\sqrt{6}x+2=0$ $⇒3{x}^{2}-\sqrt{6}x-\sqrt{6}x+2=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x\left(\sqrt{3}x-\sqrt{2}\right)-\sqrt{2}\left(\sqrt{3}x-\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(\sqrt{3}x-\sqrt{2}\right)\left(\sqrt{3}x-\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{\sqrt{2}}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}$ (9) $2m\left(m-24\right)=50$ $⇒m\left(m-24\right)=25\phantom{\rule{0ex}{0ex}}⇒{m}^{2}-24m=25\phantom{\rule{0ex}{0ex}}⇒{m}^{2}-24m-25=0\phantom{\rule{0ex}{0ex}}⇒{m}^{2}-25m+m-25=0\phantom{\rule{0ex}{0ex}}⇒m\left(m-25\right)+1\left(m-25\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(m+1\right)\left(m-25\right)=0\phantom{\rule{0ex}{0ex}}⇒m=-1,25$ (10) $25{m}^{2}=9$ $⇒25{m}^{2}-9=0\phantom{\rule{0ex}{0ex}}⇒{\left(5m\right)}^{2}-{3}^{2}=0\phantom{\rule{0ex}{0ex}}⇒\left(5m-3\right)\left(5m+3\right)=0\left(\because \left(x-a\right)\left(x+a\right)={x}^{2}-{a}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒\left(5m-3\right)=0or\left(5m+3\right)=0\phantom{\rule{0ex}{0ex}}⇒m=\frac{3}{5},\frac{-3}{5}$ (11) $7{m}^{2}=21m$ $⇒7{m}^{2}-21m=0\phantom{\rule{0ex}{0ex}}⇒7m\left(m-3\right)=0\phantom{\rule{0ex}{0ex}}⇒7m=0\mathrm{or}m-3=0\phantom{\rule{0ex}{0ex}}⇒m=0\mathrm{or}m=3$ (12) ${m}^{2}-11=0$ $⇒{m}^{2}-{\left(\sqrt{11}\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒\left(m-\sqrt{11}\right)\left(m+\sqrt{11}\right)=0\phantom{\rule{0ex}{0ex}}⇒m=\sqrt{11},-\sqrt{11}$

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