How do you find the zeros of y = -x^2 - 9x + 3 using the quadratic formula?

1 Answer
Shwetank Mauria
Jul 21, 2016

-9/2-sqrt93/2 and -9/2+sqrt93/2,

are real zeros of y=-x^2-9x+3.

Explanation:

Zeros of y=-x^2-9x+3 are those values of x for which y=0. Hence, zeros are given by the solution of quadratic equation -x^2-9x+3=0 or x^2+9x-3=0.

Quadratic formula gives solution of equation ax^2+bx+c=0 as x=(-b+-sqrt(b^2-4ac))/(2a). Hence for x^2+9x-3=0, we have

x=(-9+-sqrt(9^2-4*1*(-3)))/(2*1)

= (-9+-sqrt(81+12))/2

= (-9+-sqrt93)/2

= -9/2-sqrt93/2 and -9/2+sqrt93/2,

are real zeros of y=-x^2-9x+3.

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