How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given $2x-5=-x^2$ ?

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Answer 1 · 2017-06-03T15:29:15

Roots are $-1+sqrt6$ and $-1-sqrt6$

For an equation $ax^2+bx+c=0$ , discriminant is $b^2-4ac$

and quadratic formula gives roots as $x=(-b+-sqrt(b^2-4ac))/(2a)$

Note that if $a,b$ and $c$ are rational number, as we have in the given example

if $b^2-4ac$ is a perfect square (i.e. positive as well), roots are rational

if $b^2-4ac$ is positive but not a perfect square, roots are irrational and conjugate i.e. of the type $p+-sqrtq$

if $b^2-4ac$ is negative, roots are complex conjugate i.e. of the type $p+-iq$

Here we have $2x-5=-x^2$ i.e. $x^2+2x-5=0$

and as $a=1$ , $b=2$ and $c=-5$ , the discriminant is

$b^2-4ac=2^2-4xx1xx(-5)=4+20=24$

and hence roots are irrational and conjugate.

These are $(-2+-sqrt24)/2=-2/2+-(2sqrt6)/2=-1+-sqrt6$

How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given 2x-5=-x^22x-5=-x^2?