How do you find the roots, real and imaginary, of $y = x^{2} - x - 2$ $y= x^2-x-2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = x^{2} - x - 2$ $y= x^2-x-2$ using the quadratic formula?

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Answer 1 · 2016-09-02T03:41:35

The roots are rational (i.e. real) and are $2$ $2$ and $- 1$ $-1$ .

Explanation:

The quadratic formula gives the roots of the general form of quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ as $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

The roots of a quadratic equation thus critically depend on the discriminant $b^{2} - 4 a c$ $b^2-4ac$ .

In $y = x^{2} - x - 2$ $y=x^2-x-2$ , the discriminant is

${(- 1)}^{2} - 4 \times 1 \times (- 2)$ $(-1)^2-4×1×(-2)$

= $1 + 8 = 9$ $1+8=9$

As it is a complete square one can easily take square root and roots of equation will be rational (i.e. real).

The roots are

$\frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 \times 1 \times (- 2)}}{2}$ $(-(-1)+-sqrt((-1)^2-4×1×(-2)))/2$ or

$\frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2}$ $(1+-sqrt9)/2=(1+-3)/2$

i.e. $2$ $2$ and $- 1$ $-1$

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How do you find the roots, real and imaginary, of $y = x^{2} - x - 2$ $y= x^2-x-2$ using the quadratic formula?

1 Answer

Explanation:

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How do you find the roots, real and imaginary, of y=x2−x−2y= x^2-x-2 using the quadratic formula?

1 Answer

Explanation:

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How do you find the roots, real and imaginary, of $y = x^{2} - x - 2$ $y= x^2-x-2$ using the quadratic formula?