We simply have to solve a quadratic equation [tex]ax^2+bx+c=0[/tex] using the quadratic formula
[tex]x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In your case, [tex]a=b=1,\ c=-42[/tex]. So, the formula becomes
[tex]x_{1,2} = \dfrac{-1\pm\sqrt{1+168}}{2} = \dfrac{-1\pm 13}{2}[/tex]
So, if we choose the two signs, we have
[tex]x_1 = \dfrac{-1-13}{2}=-7,\quad x_2 = \dfrac{-1+13}{2}=6[/tex]
Answer:
x = -7, x = 6
Step-by-step explanation:
f(x) = x² + x − 42
You can use the quadratic formula to find the roots (x-intercepts), or, if it's "factorable", you can use the AC method.
Here, a = 1, b = 1, and c = -42.
The product a times c is -42.
Factors of ac that add up to b are +7 and -6.
So the quadratic factors to:
f(x) = (x + 7) (x − 6)
To find the x-intercepts, we set this equal to 0:
0 = (x + 7) (x − 6)
x + 7 = 0, x − 6 = 0
x = -7, x = 6
