Question

Find the value of k for which the following quadratic equation has equal roots x² -2x(1+3k) + 7(3+2k)

Answer 1

Step-by-step explanation:

Given: x² -2x(1+3k) + 7(3+2k) quadratic equation has equal roots

To find: value of k

Solution:

if a quadratic equation has equal roots then and its discriminant is equal to zero.

i.e.

$\bold{\blue{D= {b}^{2} - 4ac = 0}}$

On comparison with standard quadratic equation

$\bold{ax^2+bx+c=0}$

Here

$a = 1 \\ b = - 2(1 + 3k) \\ c = 7(3 + 2k)$

Put these values of a,b and c in D

$( { - 2(1 + 3k))}^{2} - 4(1)(7(3 + 2k) = 0 \\ \\ 4( {1 + 3k) }^{2} - 28(3 + 2k) = 0 \\ \\ 4(1 + 9 {k}^{2} + 6k) - 84 - 56k = 0 \\ \\ 4 + 36 {k}^{2} + 24k - 84 - 56k = 0 \\ \\ 36 {k}^{2} - 32k - 80 = 0 \\ \\ or \\ \\ 9 {k}^{2} - 8k - 20 = 0$

After solving this quadratic equation in k,we will find the value of k

Try to factorise

$9 {k}^{2} - 8k - 20 = 0 \\ \\ 9 {k}^{2} - 18k + 10k - 20 = 0 \\ \\ 9k(k - 2) + 10(k - 2) = 0 \\ \\ (k - 2)(9k + 10) = 0 \\ \\\bold{\red{ k = 2}} \\ \\ or \\ \\ \bold{\green{k = \frac{ - 10}{9}}}$

Thus,

For equation to be both roots equal,value of k is to be

2 and -10/9

Hope it helps you.

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Answer 2

Answer:

2 and -10/9 are the values of k.