How do you find the value of the discriminant and state the type of solutions given $- 2 x^{2} - x - 1 = 0$ $-2x^2-x-1=0$ ?

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Answer 1 · 2017-10-18T19:05:58

no real roots; $Delta<0$

$-2x^2-x-1=0$ is already in $ax^2+bx+c$ form, so the $a$ , $b$ and $c$ -values can be used.

$a=-2$
$b=-1$
$c=-1$

$Delta=b^2-4ac$
$=(-1)^2 - (4*(-2*-1))$
$=1-8$
$-7$

$-7<0 therefore Delta<0$

hence, $-2x^2-x-1=0$ has no real roots.

this can also be seen by graphing $-2x^2-x-1=0$ :

![https://www.desmos.com/calculator]( useruploads.socratic.org )

this parabola does not meet the $x$ -axis on a graph for real numbers, so there are no (real) roots.

How do you find the value of the discriminant and state the type of solutions given -2x^2-x-1=0−2x2−x−1=0-2x^2-x-1=0?