Question

Find value of K for which the quadratic equation (2K+1)x²-(7K+2)x+(7K-3)=0 has equal roots

Answer 1

So the given polynomial is;

$(2k + 1) {x}^{2} - (7k + 2) x+ (7k - 3) = 0$


As for equal roots,
we have,
$4ac = {b}^{2}$
Using this relation in the given polynomial where,

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


$4(2k + 1)(7k - 3) = {( - (7k + 2))}^{2} \\ 4(14 {k}^{2} + k - 3) = (49 {k}^{2} + 28k + 4) \\56 {k}^{2} + 4k - 12 = 49 {k}^{2} + 28k + 4 \\ 56 {k}^{2} - 49 {k}^{2} - 28k + 4k - 12 - 4 = 0 \\ 7 {k}^{2} - 24k - 16 = 0 \\ 7 {k}^{2} - 28k + 4k - 16 = 0 \\ 7k(k - 4) + 4(k - 4) = 0 \\ (7k + 4)(k - 4) = 0$
Therefore,
$7k + 4 = 0 \\ 7k = - 4 \\ k = \frac{ - 4}{7}$
or
k-4=0
k= 4

Answer 2

Answer:

put the value of a, b, c, then see the image

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