How do you solve ${(8 x)}^{\frac{1}{2}} + 6 = 0$ $(8x)^(1/2)+6=0$ ?

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How do you solve ${(8 x)}^{\frac{1}{2}} + 6 = 0$ $(8x)^(1/2)+6=0$ ?

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Answer 1 · 2016-07-27T11:13:46

$x = \frac{9}{2}$ $x=9/2$

$x = 4.5$ $x=4.5$

Explanation:

${(8 x)}^{\frac{1}{2}} + 6 = 0$ $(8x)^(1/2)+6=0$

Get rid of 6 from left side
For that subtract 6 on both sides

${(8 x)}^{\frac{1}{2}} = - 6$ $(8x)^(1/2)=-6$

Squaring on both sides

$8 x = 36$ $8x=36$

$x = \frac{36}{8}$ $x=36/8$

$x = \frac{9}{2}$ $x=9/2$

$x = 4.5$ $x=4.5$

Answer 2 · 2016-07-28T06:47:47

There are no values of $x$ $x$ which satisfy this equation.

Explanation:

${(8 x)}^{\frac{1}{2}} + 6 = 0$ $(8x)^(1/2)+6=0$

Subtract $6$ $6$ from both sides to get:

${(8 x)}^{\frac{1}{2}} = - 6$ $(8x)^(1/2) = -6$

Square both sides, noting that squaring may introduce spurious solutions:

$8 x = 36$ $8x = 36$

Divide both sides by $8$ $8$ to get:

$x = \frac{36}{8} = \frac{9}{2}$ $x = 36/8 = 9/2$

Check:

${(8 x)}^{\frac{1}{2}} + 6 = {(8 \cdot \frac{9}{2})}^{\frac{1}{2}} + 6 = 36^{\frac{1}{2}} + 6 = 6 + 6 = 12$ $(8x)^(1/2)+6 = (8*9/2)^(1/2)+6 = 36^(1/2)+6 = 6+6 = 12$

So this $x$ $x$ is not a solution of the original equation.

The problem is that while $36$ $36$ has two square roots (viz $\pm 6$ $+-6$ ), $36^{\frac{1}{2}} = \sqrt{36} = 6$ $36^(1/2) = sqrt(36) = 6$ denotes the principal, positive square root.

So the original equation has no solutions (Real or Complex).

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How do you solve ${(8 x)}^{\frac{1}{2}} + 6 = 0$ $(8x)^(1/2)+6=0$ ?

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