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`x^2-6 = sqrt(x+6)` How to solve for x?

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ogidi | Student | (Level 1) Honors

Posted November 2, 2012 at 11:25 AM via web

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`x^2-6 = sqrt(x+6)`

How to solve for x?

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mjripalda | High School Teacher | (Level 1) Senior Educator

Posted November 2, 2012 at 2:11 PM (Answer #1)

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`x^2-6=sqrt (x+6)`

To solve for x, eliminate the square root. To do so, square both sides.

`(x^2-6)^2 = (sqrt(x+6))^2`

`(x^2-6)^2=x+6`

Then, expand left side. Use FOIL method.

`(x^2-6)(x^2-6)=x+6`

`x^4-6x^2-6x^2+36=x+6`

`x^4-12x^2+36=x+6`

Then, set right side equal to zero. So, subtract both sides by x + 6.

`x^4-12x^2+36-(x+6)=x+6-(x+6)`

`x^4-12x^2-x+30=0`

Then, factor left side. Using synthetic division, the resulting factor is:

`(x+2)(x-3)(x^2+x-5) =0`

Then, set each factor to zero and solve for x.

`x+2=0 `              `x-3=0`            and            `x^2+x - 5 = 0`  

    `x=-2`                 `x=3`                

To solve the values of x in the third factor, apply quadratic formula.

`x=[-b+-sqrt(b^2-4ac))/(2a) = (-1+-sqrt(1^2-4(1)(-5)))/(2*1) = (-1+-sqrt21)/2`

So the other values of x are:

`x=(-1+sqrt21)/2 = 1.7913`          and           `x=(-1-sqrt21)/2=-2.7913`

Since the given equation has a radical, it is necessary to substitute the values of x to the original equation `x^2-6 = sqrt(x+6)` .

`x= 3` ,                     `3^2-6 = sqrt(3+6)`

                                  `9 - 6 = sqrt9`

                                       `3 = 3`               (True)

`x= -2` ,              `(-2)^2 - 6 = sqrt(-2+6)`

                                   `4 - 6 = sqrt4`

                                      `-2 = 2`                (False)

`x=1.7913`         `1.7913^2 - 6 = sqrt(1.7913+6)`

                                   `-2.8 = 2.8`            (False)

`x=-2.7913`   `(-2.7913)^2-6=sqrt(-2.7913+6)`

                                      `1.8 = 1.8`             (True)

Since only tow values of x result to a true condition when plug-in to the given equation, hence the solutions are `x = 3`  and `x=-2.7913` .

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