The set of solutions for the equation (x^2-25)^2=x^2-10 x+25 contains how many real numbers?

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GMAT Panda · Accepted Answer

3 We notice that the right-hand side is the square of x minus 5. (x^2 - 25)^2 = (x - 5)^2 We move all terms to one side to create a difference of squares. (x^2 - 25)^2 - (x - 5)^2 = 0 ((x^2 - 25) - (x - 5))((x^2 - 25) + (x - 5)) = 0 We set the first factor equal to zero and solve for x. ((x^2 - 25) - (x - 5)) = 0 x^2 - x - 20 = 0 (x-5)(x+4) = 0 x = 5, -4 We set the second factor equal to zero and solve for x. ((x^2 - 25) + (x - 5)) = 0 x^2 + x - 30 = 0 (x+6)(x-5) = 0 x = -6, 5 We collect the distinct real solutions. x = -6, -4, 5

Explanation	Calculations	Help
$We notice that the right-hand side is the square of x minus 5.$	$(x^{2} - 25)^{2} = (x - 5)^{2}$
$We move all terms to one side to create a difference of squares.$	$(x^{2} - 25)^{2} - (x - 5)^{2} = 0 ((x^{2} - 25) - (x - 5)) \times ((x^{2} - 25) + (x - 5)) = 0$
$We set the first factor equal to zero and solve for x.$	$((x^{2} - 25) - (x - 5)) = 0 x^{2} - x - 20 = 0 (x - 5) (x + 4) = 0 x = 5, - 4$
$We set the second factor equal to zero and solve for x.$	$((x^{2} - 25) + (x - 5)) = 0 x^{2} + x - 30 = 0 (x + 6) (x - 5) = 0 x = - 6, 5$
$We collect the distinct real solutions.$	$x = - 6, - 4, 5$

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