**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 ax^{2} +bx +c = 0 , a ≠ 0 are found using the formula x = [-b ± √(b^{2} - 4ac)]/2a

Here, b^{2} - 4ac called as the discriminant (which is denoted by D ) of the quadratic equation, decides the nature of roots as follows

Value of discriminant Δ = b Δ > 0 Δ = 0 Δ < 0 |
Nature of roots Real and unequal roots Real and equal roots No real roots |

**Question 1 :**

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

(i) (5k −6)x^{2} + 2kx + 1 = 0

**Solution :**

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

**ax ^{2} + bx + c = 0**

**a = 5k-6, b = 2k and c = 1**

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

**Δ =** **b ^{2}** -

**(2k)^{2} - 4(5k-6)1 = 0**

**4k ^{2} - 20k + 24 = 0**

**k ^{2} - 5k + 6 = 0**

**k ^{2} - 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) kx^{2} +(6k + 2)x + 16 = 0

**a = k, b = 6k + 2 and c = 16**

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

**Δ =** **b ^{2}** -

**(6k + 2)^{2} - 4(k)(16) = 0**

**(6k) ^{2} + 2(6k)(2) + 2^{2} - 64k = 0**

**36k ^{2} + 24k + 4 - 64k = 0**

**36k ^{2} - 40k + 4 = 0**

**Divide by 4, we get **

**9****k ^{2} - 10k + 1 = 0**

**9****k ^{2} - 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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