The x intercepts of the parabola y=2x^2+13x-7-x^2-6x
are at (p,0) and (q,0) Find p and q
Enter your answer in the form "p, q ".
+9519 To find the x-intercepts, we let y = 0 and solve for x. If x = a is a possible value of x, that corresponds to the x-intercept (a, 0).
But before that, we need to simplify it first.
\(y = 2x^2 +13x - 7 - x^2 - 6x \\y= x^2 + 7x - 7\)
Then let y = 0, and solve the equation using quadratic formula:
\(x^2 + 7x - 7 = 0\\ x= \dfrac{-7 \pm \sqrt{7^2 - 4(1)(-7)}}{2(1)}\\ x = \boxed{\phantom{\dfrac{-7 \pm \sqrt{7^2 - 4(1)(-7)}}{2(1)}}}\)
Each solution corresponds to one x-intercept, as I explained above.
