How do you find the roots, real and imaginary, of $y= 12x^2 + 35x + 72$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y= 12x^2 + 35x + 72$ using the quadratic formula?

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Answer 1 · 2018-02-13T13:50:04

$x = -35/24+-sqrt(2231)/24i$

Explanation:

The equation:

$y = 12x^2+35x+72$

is a quadratic in standard form:

$y = ax^2+bx+c$

with $a=12$ , $b=35$ and $c=72$

It has zeros given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$color(white)(x) = (-(color(blue)(35))+-sqrt((color(blue)(35))^2-4(color(blue)(12))(color(blue)(72))))/(2(color(blue)(12)))$

$color(white)(x) = (-35+-sqrt(1225-3456))/24$

$color(white)(x) = (-35+-sqrt(-2231))/24$

$color(white)(x) = -35/24+-sqrt(2231)/24i$

Note that $2231 = 23 * 97$ has no square factors, so $sqrt(2231)$ cannot be simplified.

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How do you find the roots, real and imaginary, of $y= 12x^2 + 35x + 72$ using the quadratic formula?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of y= 12x^2 + 35x + 72 y= 12x^2 + 35x + 72 using the quadratic formula?

1 Answer

Explanation:

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Impact of this question

How do you find the roots, real and imaginary, of $y= 12x^2 + 35x + 72$ using the quadratic formula?