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) $3x^2-2\sqrt{6}x+2=0$
(9) $2m\left(m-24\right)=50$
(10) $25m^2=9$
(11) $7m^2=21m$
(12) $m^2-11=0$

Answer 1

(1) x² – 15x + 54 = 0
$$\begin{gathered} \Rightarrow x^2-9x-6x+54=0 \\ \Rightarrow x\left(x-9\right)-6\left(x-9\right)=0 \\ \Rightarrow \left(x-6\right)\left(x-9\right)=0 \\ \Rightarrow \left(x-6\right)=0\mathrm{or}\left(x-9\right)=0 \\ \Rightarrow x=6\mathrm{or}x=9 \end{gathered}$$
So, 6 and 9 are the roots of the given quadratic equation.

(2) x² + x – 20 = 0
$$\begin{gathered} \Rightarrow x^2+5x-4x-20=0 \\ \Rightarrow x\left(x+5\right)-4\left(x+5\right)=0 \\ \Rightarrow \left(x-4\right)\left(x+5\right)=0 \\ \Rightarrow \left(x-4\right)=0\mathrm{or}\left(x+5\right)=0 \\ \Rightarrow x=4\mathrm{or}x=-5 \end{gathered}$$
So, 4 and $-$5 are the roots of the given quadratic equation.

(3) 2y² + 27y + 13 = 0
$$\begin{gathered} 2y^2+26y+y+13=0 \\ \Rightarrow 2y\left(y+13\right)+\left(y+13\right)=0 \\ \Rightarrow \left(2y+1\right)\left(y+13\right)=0 \\ \Rightarrow \left(2y+1\right)=0\mathrm{or}\left(y+13\right)=0 \\ \Rightarrow y=\frac{-1}{2}\mathrm{or}y=-13 \end{gathered}$$
So, $\frac{-1}{2}\mathrm{and}-13$ are the roots of the given quadratic equation.

(4) 5m² = 22m + 15
$$\begin{gathered} \Rightarrow 5m^2-22m-15=0 \\ \Rightarrow 5m^2-25m+3m-15=0 \\ \Rightarrow 5m\left(m-5\right)+3\left(m-5\right)=0 \\ \Rightarrow \left(5m+3\right)\left(m-5\right)=0 \\ \Rightarrow \left(5m+3\right)=0\mathrm{or}\left(m-5\right)=0 \\ \Rightarrow m=\frac{-3}{5}\mathrm{or}m=5 \end{gathered}$$
So, $\frac{-3}{5}\mathrm{and}5$ are the roots of the given quadratic equation.

(5) 2x² – 2x + $\frac{1}{2}$ = 0
$$\begin{gathered} \Rightarrow 4x^2-4x+1=0 \\ \Rightarrow {\left(2x\right)}^2-2\times 1\times 2x+1=0 \\ U\sin gtheidentity, \\ {\left(a-b\right)}^2=a^2+b^2-2ab \\ {\left(2x\right)}^2-2\times 1\times 2x+1=0 \\ \Rightarrow {\left(2x-1\right)}^2=0 \\ \Rightarrow x=\frac{1}{2},\frac{1}{2} \end{gathered}$$
So, $\frac{1}{2}\mathrm{and}\frac{1}{2}$ are the roots of the given quadratic equation.

(6) $6x-\frac{2}{x}=1$
$$\begin{gathered} \Rightarrow 6x^2-2=1x \\ \Rightarrow 6x^2-1x-2=0 \\ \Rightarrow 6x^2-4x+3x-2=0 \\ \Rightarrow 2x\left(3x-2\right)+1\left(3x-2\right)=0 \\ \Rightarrow \left(3x-2\right)\left(2x+1\right)=0 \\ \Rightarrow \left(3x-2\right)=0\mathrm{or}\left(2x+1\right)=0 \\ \Rightarrow x=\frac{2}{3}\mathrm{or}x=\frac{-1}{2} \end{gathered}$$
$\frac{2}{3}\mathrm{and}\frac{-1}{2}$ are the roots of the given quadratic equation.

(7) $\sqrt{2}{x}^2+7x+5\sqrt{2}=0$
$$\begin{gathered} \sqrt{2}{x}^2+\boxed{5x}+\boxed{2x}+5\sqrt{2}=0 \\ x\left(\sqrt{2}x+5\right)+\sqrt{2}\left(\sqrt{2}x+5\right)=0 \\ \left(\sqrt{2}x+5\right)\left(x+\sqrt{2}\right)=0 \\ \left(\sqrt{2}x+5\right)=0\mathrm{or}\left(x+\sqrt{2}\right)=0 \\ \therefore x=\boxed{\frac{-5}{\sqrt{2}}}\mathrm{or}x=-\sqrt{2} \\ \therefore \boxed{\frac{-5}{\sqrt{2}}}\mathrm{and}-\sqrt{2}\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}. \end{gathered}$$

(8) $3x^2-2\sqrt{6}x+2=0$
$$\begin{gathered} \Rightarrow 3x^2-\sqrt{6}x-\sqrt{6}x+2=0 \\ \Rightarrow \sqrt{3}x\left(\sqrt{3}x-\sqrt{2}\right)-\sqrt{2}\left(\sqrt{3}x-\sqrt{2}\right)=0 \\ \Rightarrow \left(\sqrt{3}x-\sqrt{2}\right)\left(\sqrt{3}x-\sqrt{2}\right)=0 \\ \Rightarrow x=\frac{\sqrt{2}}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}} \end{gathered}$$

(9) $2m\left(m-24\right)=50$
$$\begin{gathered} \Rightarrow m\left(m-24\right)=25 \\ \Rightarrow m^2-24m=25 \\ \Rightarrow m^2-24m-25=0 \\ \Rightarrow m^2-25m+m-25=0 \\ \Rightarrow m\left(m-25\right)+1\left(m-25\right)=0 \\ \Rightarrow \left(m+1\right)\left(m-25\right)=0 \\ \Rightarrow m=-1,25 \end{gathered}$$

(10) $25m^2=9$
$$\begin{gathered} \Rightarrow 25m^2-9=0 \\ \Rightarrow {\left(5m\right)}^2-3^2=0 \\ \Rightarrow \left(5m-3\right)\left(5m+3\right)=0\left(\because \left(x-a\right)\left(x+a\right)=x^2-a^2\right) \\ \Rightarrow \left(5m-3\right)=0or\left(5m+3\right)=0 \\ \Rightarrow m=\frac{3}{5},\frac{-3}{5} \end{gathered}$$

(11) $7m^2=21m$
$$\begin{gathered} \Rightarrow 7m^2-21m=0 \\ \Rightarrow 7m\left(m-3\right)=0 \\ \Rightarrow 7m=0\mathrm{or}m-3=0 \\ \Rightarrow m=0\mathrm{or}m=3 \end{gathered}$$

(12) $m^2-11=0$
$$\begin{gathered} \Rightarrow m^2-{\left(\sqrt{11}\right)}^2=0 \\ \Rightarrow \left(m-\sqrt{11}\right)\left(m+\sqrt{11}\right)=0 \\ \Rightarrow m=\sqrt{11},-\sqrt{11} \end{gathered}$$