How do you solve the equation and express in simplest radical form for ${(4 x + 3)}^{2} = 7$ $(4x+3)^2=7$ ?

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Answer 1 · 2018-06-20T08:14:31

$x = \frac{- 3 \pm \sqrt{7}}{4}$ $color(green)(x = (-3 +- sqrt7)/4$

${(4 x + 3)}^{2} = 7$ $(4x + 3)^2 = 7$

${(4 x + 3)}^{2} = {(\sqrt{7})}^{2}$ $(4x + 3)^2 = (sqrt7)^2$

$4 x + 3 = \pm \sqrt{7}$ $4x + 3 = +- sqrt7$

$4 x = - 3 \pm \sqrt{7}$ $4x = -3 +- sqrt7$

$x = \frac{- 3 \pm \sqrt{7}}{4}$ $color(green)(x = (-3 +- sqrt7)/4$

How do you solve the equation and express in simplest radical form for (4x+3)2=7(4x+3)^2=7?