The roots of the equation $\displaystyle \left ( x+3 \right )\left ( x-3 \right )=160$ are
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The roots of the equation $\displaystyle \left ( x+3 \right )\left ( x-3 \right )=160$ are
Expanding the left side of the equation gives x^2 - 9 = 160. Adding 9 to both sides yields x^2 = 169. Taking the square root of both sides gives x = 13 or x = -13.
First, expand the left side of the equation using the difference of squares identity, (x + 3)(x - 3) = x^2 - 9. Setting this equal to 160 gives x^2 - 9 = 160. Adding 9 to both sides results in x^2 = 169. Taking the square root of both sides yields x = 13 or x = -13.