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

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Answer 1 · 2016-07-23T13:35:33

$-1.118 " or " x = -2.882$

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

Compare with $5x^2 = 20, " "rArr x^2 = 4 " "rArrx= +-2$

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

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

$x + 2 = +-sqrt(7/9)$

$x = +sqrt7/3-2 " or " x = -sqrt7/3 -2$

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

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