91Ó°ÊÓ

Complete the square to solve the quadratic equations by isolating $x^2$ andbx-terms.

$$\begin{gathered} 2x^2−36=−6x \\ 2x^2+6x−36=0\left[\text{ â¶Ä‰S³Ù²¹²Ô»å²¹°ù»å â¶Ä‰F´Ç°ù³¾â€‰â¶Ä‰}\right] \\ \frac{2x^2+6x−66}{2}=\frac{0}{2}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\left[\text{ â¶Ä‰D¾±±¹¾±»å±ð â¶Ä‰b²â â¶Ä‰2 â¶Ä‰}\right] \\ {x}^2+3x−18=0 \\ {x}^2+3x−18+\frac{9}{4}=0+\frac{9}{4}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\left[\text{ S¾±²Ô³¦±ð â¶Ä‰}{\left(\frac{3}{2}\right)}^2=\frac{9}{4}\text{ â¶Ä‰a»å»å â¶Ä‰t´Ç â¶Ä‰e²¹³¦³ó â¶Ä‰s¾±»å±ð }\right] \\ {x}^2+3x+\frac{9}{4}−18+18=\frac{9}{4}+18\text{ â¶Ä‰â¶Ä‰}\left[\text{‼î»å»å â¶Ä‰6 â¶Ä‰t´Ç â¶Ä‰e²¹³¦³ó â¶Ä‰s¾±»å±ð â¶Ä‰}\right] \\ {x}^2+3x+\frac{9}{4}=\frac{81}{4} \end{gathered}$$

First factor $x^2+3x+\frac{9}{4}$, then take square root of each side.

$$\begin{gathered} {\left(x+\frac{3}{2}\right)}^2=\frac{81}{4} \\ \sqrt{{\left(x+\frac{3}{2}\right)}^2}=Â\pm \sqrt{\frac{81}{4}} \\ x+\frac{3}{2}=Â\pm \frac{9}{2} \end{gathered}$$

Subtract $\frac{3}{2}$from both sides and separate the solutions.

$$\begin{gathered} x+\frac{3}{2}−\frac{3}{2}=−\frac{3}{2}Â\pm \frac{9}{2} \\ x=−\frac{3}{2}−\frac{9}{2}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰or â¶Ä‰â¶Ä‰}x=−\frac{3}{2}+\frac{9}{2} \\ x=−6\text{ â¶Ä‰â¶Ä‰â€‰or  }x=3 \end{gathered}$$

Therefore, the solutions of the equation are $-6$and 3.