Question

Solve the following quadratic equations by completing the square method.
(1) x² + x – 20 = 0
(2) x² + 2x – 5 = 0
(3) m² – 5m = –3
(4) 9y² – 12y + 2 = 0
(5) 2y² + 9y +10 = 0
(6) 5x² = 4x + 7

Answer 1

(1) x² + x – 20 = 0
$$\begin{gathered} x^2+x+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2-20=0 \\ \Rightarrow {\left(x+\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^2+20 \\ \Rightarrow {\left(x+\frac{1}{2}\right)}^2=\frac{1}{4}+20 \\ \Rightarrow {\left(x+\frac{1}{2}\right)}^2=\frac{81}{4} \\ \Rightarrow x+\frac{1}{2}=\frac{9}{2}\mathrm{or}x+\frac{1}{2}=-\frac{9}{2} \\ \Rightarrow x=4\mathrm{or}x=-5 \end{gathered}$$
(2) x² + 2x – 5 = 0
$$\begin{gathered} \Rightarrow x^2+2x+{\left(\frac{2}{2}\right)}^2-{\left(\frac{2}{2}\right)}^2-5=0 \\ \Rightarrow \left(x^2+2x+1\right)-1-5=0 \\ \Rightarrow {\left(x+1\right)}^2-6=0 \\ \Rightarrow {\left(x+1\right)}^2=6 \\ \Rightarrow {\left(x+1\right)}^2={\left(\sqrt{6}\right)}^2 \\ \Rightarrow x+1=\sqrt{6}\mathrm{or}x+1=-\sqrt{6} \\ \Rightarrow x=\sqrt{6}-1\mathrm{or}x=-\sqrt{6}-1 \end{gathered}$$

(3) m² – 5m = –3
$$\begin{gathered} \Rightarrow m^2-5m+{\left(\frac{-5}{2}\right)}^2-{\left(\frac{-5}{2}\right)}^2=-3 \\ \Rightarrow \left(m^2-5m+\frac{25}{4}\right)-\frac{25}{4}=-3 \\ \Rightarrow {\left(m-\frac{5}{2}\right)}^2=-3+\frac{25}{4} \\ \Rightarrow {\left(m-\frac{5}{2}\right)}^2=\frac{13}{4} \\ \Rightarrow {\left(m-\frac{5}{2}\right)}^2={\left(\frac{\sqrt{13}}{2}\right)}^2 \\ \Rightarrow m-\frac{5}{2}=\frac{\sqrt{13}}{2}\mathrm{or}m-\frac{5}{2}=-\frac{\sqrt{13}}{2} \\ \Rightarrow m=\frac{\sqrt{13}}{2}+\frac{5}{2}\mathrm{or}m=-\frac{\sqrt{13}}{2}+\frac{5}{2} \\ \Rightarrow m=\frac{\sqrt{13}+5}{2}\mathrm{or}m=\frac{5-\sqrt{13}}{2} \end{gathered}$$

(4) 9y² – 12y + 2 = 0
$$\begin{gathered} \Rightarrow y^2-\frac{12}{9}y+\frac{2}{9}=0 \\ \Rightarrow y^2-\frac{4}{3}y+\frac{2}{9}=0 \\ \Rightarrow y^2-\frac{4}{3}y+{\left(\frac{{\displaystyle \frac{4}{3}}}{2}\right)}^2-{\left(\frac{{\displaystyle \frac{4}{3}}}{2}\right)}^2+\frac{2}{9}=0 \\ \Rightarrow y^2-\frac{4}{3}y+{\left(\frac{2}{3}\right)}^2-{\left(\frac{2}{3}\right)}^2+\frac{2}{9}=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left[y^2-\frac{4}{3}y+\left(\frac{4}{9}\right)\right]-\left(\frac{4}{9}\right)+\frac{2}{9}=0 \\ \Rightarrow {\left(y-\frac{2}{3}\right)}^2-\frac{2}{9}=0 \\ \Rightarrow {\left(y-\frac{2}{3}\right)}^2=\frac{2}{9} \\ \Rightarrow {\left(y-\frac{2}{3}\right)}^2={\left(\frac{\sqrt{2}}{3}\right)}^2 \end{gathered}$$
$$\begin{gathered} \Rightarrow y-\frac{2}{3}=\frac{\sqrt{2}}{3}\mathrm{or}y-\frac{2}{3}=-\frac{\sqrt{2}}{3} \\ \Rightarrow y=\frac{\sqrt{2}}{3}+\frac{2}{3}\mathrm{or}y=-\frac{\sqrt{2}}{3}+\frac{2}{3} \\ \Rightarrow y=\frac{\sqrt{2}+2}{3}\mathrm{or}y=\frac{-\sqrt{2}+2}{3} \end{gathered}$$

(5) 2y² + 9y +10 = 0
$$\begin{gathered} \Rightarrow y^2+\frac{9}{2}y+5=0 \\ \Rightarrow y^2+\frac{9}{2}y+{\left(\frac{{\displaystyle \frac{9}{2}}}{2}\right)}^2-{\left(\frac{{\displaystyle \frac{9}{2}}}{2}\right)}^2+5=0 \\ \Rightarrow \left[y^2+\frac{9}{2}y+{\left(\frac{9}{4}\right)}^2\right]-{\left(\frac{9}{4}\right)}^2+5=0 \\ \Rightarrow {\left(y+\frac{9}{4}\right)}^2=\frac{81}{16}-5 \\ \Rightarrow {\left(y+\frac{9}{4}\right)}^2=\frac{1}{16} \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(y+\frac{9}{4}\right)}^2={\left(\frac{1}{4}\right)}^2 \\ \Rightarrow y+\frac{9}{4}=\frac{1}{4}\mathrm{or}y+\frac{9}{4}=-\frac{1}{4} \\ \Rightarrow y=\frac{1}{4}-\frac{9}{4}=\frac{-8}{4}=-2\mathrm{or}y=-\frac{1}{4}-\frac{9}{4}=\frac{-10}{4}=-\frac{5}{2} \\ \Rightarrow y=-2,-\frac{5}{2} \end{gathered}$$
(6) 5x² = 4x + 7
$$\begin{gathered} \Rightarrow 5x^2-4x-7=0 \\ \Rightarrow x^2-\frac{4}{5}x-\frac{7}{5}=0 \\ \Rightarrow x^2-\frac{4}{5}x+{\left(\frac{{\displaystyle \frac{4}{5}}}{2}\right)}^2-{\left(\frac{{\displaystyle \frac{4}{5}}}{2}\right)}^2-\frac{7}{5}=0 \\ \Rightarrow \left[x^2-\frac{4}{5}x+{\left(\frac{2}{5}\right)}^2\right]-{\left(\frac{2}{5}\right)}^2-\frac{7}{5}=0 \\ \Rightarrow {\left(x-\frac{2}{5}\right)}^2=\frac{7}{5}+\frac{4}{25}=\frac{35+4}{25}=\frac{39}{25} \\ \Rightarrow {\left(x-\frac{2}{5}\right)}^2=\frac{39}{25} \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(x-\frac{2}{5}\right)}^2={\left(\frac{\sqrt{39}}{5}\right)}^2 \\ \Rightarrow x-\frac{2}{5}=\frac{\sqrt{39}}{5}\mathrm{or}x-\frac{2}{5}=-\frac{\sqrt{39}}{5} \\ \Rightarrow x=\frac{2+\sqrt{39}}{5}\mathrm{or}x=\frac{2-\sqrt{39}}{5} \end{gathered}$$