# FIND MISSING VALUE IN A QUADRATIC EQUATION USING NATURE OF ROOTS

Find Missing Value in a Quadratic Function Using Nature of Roots :

Here we are going to see some example problems  of finding nature of roots of a quadratic equation.

The roots of the quadratic equation ax2 +bx +c = 0 , a  0 are found using the formula x = [-b ± √(b2 - 4ac)]/2a

Here, b2 - 4ac called as the discriminant (which is denoted by D ) of the quadratic equation, decides the nature of roots as follows

 Value of discriminantΔ = b2 - 4acΔ > 0Δ = 0Δ < 0 Nature of rootsReal and unequal rootsReal and equal rootsNo real roots

## Find Missing Value in a Quadratic Function Using Nature of Roots - Questions

Question 1 :

Find the value(s) of ‘k’ for which the roots of the following equations are real and equal.

(i)  (5k −6)x2 + 2kx + 1 = 0

Solution :

By comparing the given quadratic equation with the general form of quadratic equation.

ax2 + bx + c = 0

a = 5k-6, b = 2k  and c = 1

If the roots are real and equal, then Δ = 0

Δ = b2 - 4ac

(2k)2 - 4(5k-6)1  =  0

4k2 - 20k + 24  =  0

k2 - 5k + 6  =  0

k2 - 2k - 3k + 6  =  0

k(k - 2) - 3(k - 2)  =  0

(k - 3)(k - 2)  =  0

k - 3  =  0, k - 2  =  0

k = 3, k = 2

Hence the values of k are 2 or 3.

(ii)  kx2 +(6k + 2)x + 16 = 0

a = k, b = 6k + 2  and c = 16

If the roots are real and equal, then Δ = 0

Δ = b2 - 4ac

(6k + 2)2 - 4(k)(16)  =  0

(6k)2 + 2(6k)(2) + 22 - 64k  =  0

36k2 + 24k + 4 - 64k  =  0

36k2 - 40k  + 4  =  0

Divide by 4, we get

9k2 - 10k  + 1  =  0

9k2 - 9k - k  + 1  =  0

9k(k - 1) - 1(k - 1)  =  0

(9k - 1) (k - 1)  =  0

9k - 1  =  0, k - 1  =  0

k = 1/9 or k = 1

Hence the values of k are 1/9 or 1. After having gone through the stuff given above, we hope that the students would have understood, "Find Missing Value in a Quadratic Function Using Nature of Roots".

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