How do you solve using the quadratic formula $x^{2} + x - 42 = 0$ $x^2 + x - 42 = 0$ ?

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How do you solve using the quadratic formula $x^{2} + x - 42 = 0$ $x^2 + x - 42 = 0$ ?

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Answer 1 · 2015-04-30T16:54:42

x=6, -7

Compare the given equation with $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ and determine a=1, b=1 and c= -42. Plug these values in the quadratic formula x= $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$

x= $\frac{- 1 \pm \sqrt{1 + 168}}{2}$ $(-1+-sqrt(1+168))/2$

= $\frac{- 1 \pm 13}{2}$ $(-1+-13)/2$

x=6, -7

Answer 2 · 2015-05-02T04:12:14

There is another way. Use the new AC Method.
y = x^2 + x - 42.
Compose factor pairs of c = -42. Proceed: (-1, 42)(-2, 21)(-3, 14)(-6, 7).
This last sum is -6 + 7 = 1 = b. Then, the 2 real roots are: 6 and -7.

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How do you solve using the quadratic formula $x^{2} + x - 42 = 0$ $x^2 + x - 42 = 0$ ?

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How do you solve using the quadratic formula $x^{2} + x - 42 = 0$ $x^2 + x - 42 = 0$ ?