How do you solve ${(8 x - 9)}^{2} = 6$ $(8x-9)^2=6$ ?

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Answer 1 · 2017-02-27T18:23:59

$x = 1.431 and x = 0.819$ $x = 1.431" and "x =0.819$

If you had an equation such as $x^{2} = 25,$ $x^2 = 25,$

you could solve it easily by taking the square root of both sides:

$x = \pm \sqrt{25} = \pm 5$ $x = +-sqrt25 = +-5$

The given equation is in exactly the same form:

${(8 x - 9)}^{2} = 6 \leftarrow$ $(8x-9)^2 =6" "larr$ take the square root of both sides

$8 x - 9 = \pm \sqrt{6} \leftarrow$ $8x-9 = +-sqrt6" "larr$ remember there are two roots

$8 x = \pm \sqrt{6} + 9$ $8x = +-sqrt6 +9$

This will give us two answers:

$x = \frac{+ \sqrt{6} + 9}{8} and x = \frac{- \sqrt{6} + 9}{8}$ $x = (+sqrt6+9)/8" and "x = (-sqrt6+9)/8$

$x = 1.431 and x = 0.819$ $x = 1.431" and "x =0.819$

How do you solve (8x−9)2=6(8x-9)^2=6?