How do you find the value of the discriminant and determine the nature of the roots $x^{2} + 3 x = 40$ $x^2 + 3x = 40$ ?

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How do you find the value of the discriminant and determine the nature of the roots $x^{2} + 3 x = 40$ $x^2 + 3x = 40$ ?

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Answer 1 · 2016-07-27T19:17:35

HI,discriminant is
$b^{2} - 4 a c$ $b^2-4ac$ for quadratic $a x^{2}$ $ax^2$ +bx+c=0

this can be
1.greater then zero
2.equal to zero
3.less than zero

It depends on the sign of $x^{2}$ $x^2$ term
if $x^{2}$ $x^2$ is positive $b^{2} - 4 a c$ $b^2-4ac$
thrn $b^{2} - 4 a c \geq 0$ $b^2-4ac>=0$
hence
b=3 ,a=1
c=-40
then ....
9-4(1)(-40)= is the value of determinant

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How do you find the value of the discriminant and determine the nature of the roots $x^{2} + 3 x = 40$ $x^2 + 3x = 40$ ?

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How do you find the value of the discriminant and determine the nature of the roots x^2 + 3x = 40 x2+3x=40x^2 + 3x = 40 ?

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How do you find the value of the discriminant and determine the nature of the roots $x^{2} + 3 x = 40$ $x^2 + 3x = 40$ ?