How do you solve $3 \sqrt{x} + x = 10$ $3sqrtx + x = 10$ ?

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How do you solve $3 \sqrt{x} + x = 10$ $3sqrtx + x = 10$ ?

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Answer 1 · 2016-06-24T15:21:06

$x = 4$ $x=4$

Explanation:

'Transpose' $x$ $x$ to the other side of the equation and square both sides.
Expand the binomial on the right side.
'Transpose $9 x$ $9x$ to the right side of the equation.
Use the quadratic formula.

It should look like this:

$3 \sqrt{x} = 10 - x$ $3sqrtx=10-x$

$9 x = {(10 - x)}^{2}$ $9x=(10-x)^2$

$9 x = 100 - 20 x + x^{2}$ $9x=100-20x+x^2$

$0 = 100 - 29 x + x^{2}$ $0=100-29x+x^2$

$x = - 25$ $x=-25$

and

$x = 4$ $x=4$

However, we will not accept $x = - 25$ $x=-25$ as an answer because it will yield an imaginary number in the original equation. The squaring step (step1b) created this extra root.

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How do you solve $3 \sqrt{x} + x = 10$ $3sqrtx + x = 10$ ?

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