How do you find the zeros, real and imaginary, of $y=-x^2- x -35$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y=-x^2- x -35$ using the quadratic formula?

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Answer 1 · 2016-05-30T18:18:07

The zeros are $x=-0.5+i5.895$ and $x=-0.5-i5.895$ .

Explanation:

The zeros are the solution of this equations

$-x^2-x-35=0$

first of all we can remove all the minus multiplying left and right for $-1$ , but on the right there is zero that does not change sign.

$x^2+x+35=0$

The quadratic formula tells us that an equation in the form
$ax^2+bX+c=0$ has solutions $x=(-b\pmsqrt(b^2-4ac))/(2a)$ .

For us $a=1$ , $b=1$ and $c=35$ , I substitute

$x=(-1\pmsqrt(1^2-4*1*35))/(2)$
$=(-1\pmsqrt(-139))/2$
We notice that the root is negative and then imaginary. So I rewrite it using the fact that $i=sqrt(-1)$

$x=(-1\pmisqrt(139))/2\approx-1/2\pmi11.79/2=-0.5\pmi5.895$

The two solutions are then $x=-0.5+i5.895$ and $x=-0.5-i5.895$ .

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How do you find the zeros, real and imaginary, of $y=-x^2- x -35$ using the quadratic formula?

1 Answer

Explanation:

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How do you find the zeros, real and imaginary, of $y=-x^2- x -35$ using the quadratic formula?