How do you solve $9 {(x + 2)}^{2} - 7 = 0$ $9(x + 2)^2 - 7 = 0$ ?

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Answer 1 · 2016-07-23T13:35:33

$- 1.118 or x = - 2.882$ $-1.118 " or " x = -2.882$

While it is possible to simply and change this into the form $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ , it can be used as it is.

Compare with $5 x^{2} = 20, \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$ $5x^2 = 20, " "rArr x^2 = 4 " "rArrx= +-2$

$9 {(x + 2)}^{2} - 7 = 0 move the constant and divide by 9$ $9(x + 2)^2 - 7 = 0 " move the constant and divide by 9"$

${(x + 2)}^{2} = \frac{7}{9} square root both sides$ $(x + 2)^2 = 7/9" square root both sides"$

$x + 2 = \pm \sqrt{\frac{7}{9}}$ $x + 2 = +-sqrt(7/9)$

$x = + \frac{\sqrt{7}}{3} - 2 or x = - \frac{\sqrt{7}}{3} - 2$ $x = +sqrt7/3-2 " or " x = -sqrt7/3 -2$

= $- 1.118 or x = - 2.882$ $-1.118 " or " x = -2.882$

How do you solve 9(x+2)2−7=09(x + 2)^2 - 7 = 0?