How do you find the zeros, real and imaginary, of $y = 5 x^{2} - x - 2$ $y=5x^2-x-2$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y = 5 x^{2} - x - 2$ $y=5x^2-x-2$ using the quadratic formula?

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Answer 1 · 2016-01-11T21:47:52

The zeroes are given by y= $\frac{1 (\frac{+}{-}) \sqrt{41}}{10}$ $(1 (+/-) sqrt(41))/10$

Explanation:

The quadratic equation, which gives the roots of a second degree polynomial of the form $a x^{2} + b x + c$ $ax^2+bx+c$ , is:

$\frac{- b (\frac{+}{-}) \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b (+/-) sqrt(b^2 - 4ac))/(2a)$ .

Substituting for a the value 5, for b the value -1, and for c the value -2, gives $\frac{- (- 1) (\frac{+}{-}) \sqrt{{(- 1)}^{2} - 4 (5) (- 2)}}{2 (5)}$ $(-(-1) (+/-) sqrt((-1)^2 - 4(5)(-2)))/(2(5))$ . This gives the stated values.

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How do you find the zeros, real and imaginary, of $y = 5 x^{2} - x - 2$ $y=5x^2-x-2$ using the quadratic formula?

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Explanation:

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Science

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How do you find the zeros, real and imaginary, of y=5x^2-x-2y=5x2−x−2y=5x^2-x-2 using the quadratic formula?

1 Answer

Explanation:

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How do you find the zeros, real and imaginary, of $y = 5 x^{2} - x - 2$ $y=5x^2-x-2$ using the quadratic formula?