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`x^2-6 = sqrt(x+6)` How to solve for x?
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High School Teacher
To solve for x, eliminate the square root. To do so, square both sides.
`(x^2-6)^2 = (sqrt(x+6))^2`
Then, expand left side. Use FOIL method.
Then, set right side equal to zero. So, subtract both sides by x + 6.
Then, factor left side. Using synthetic division, the resulting factor is:
Then, set each factor to zero and solve for x.
`x+2=0 ` `x-3=0` and `x^2+x - 5 = 0`
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)
`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` .
Posted by mjripalda on November 2, 2012 at 2:11 PM (Answer #1)
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