How do you find the discriminant $2 x^{2} + 3 x - 9 = 0$ $2x^2+3x-9=0$ ?

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How do you find the discriminant $2 x^{2} + 3 x - 9 = 0$ $2x^2+3x-9=0$ ?

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Answer 1 · 2017-02-14T10:20:09

Discriminant is $81$ $81$

Explanation:

$2 x^{2} + 3 x - 9 = 0$ $2x^2+3x-9=0$
Comparing with general quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ we get $a = 2, b = 3, c = - 9$ $a=2 ,b= 3 , c= -9$
Disciminant is $D = b^{2} - 4 a c or D = 3^{2} - 4 \cdot 2 \cdot (- 9) = 9 + 72 = 81$ $D=b^2-4ac or D= 3^2 -4 * 2* (-9) = 9 +72 = 81$

If $D > 0$ $D >0$ , there is 2 real solution.
If $D = 0$ $D =0$ , there is 1 real solution.
If $D < 0$ $D<0$ , there is no real solution , roots are imaginary.
Discriminant is $81$ $81$ [Ans]

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How do you find the discriminant $2 x^{2} + 3 x - 9 = 0$ $2x^2+3x-9=0$ ?

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How do you find the discriminant $2 x^{2} + 3 x - 9 = 0$ $2x^2+3x-9=0$ ?