Question

find the values of k for which the roots are real and equal: x²-2x(1+3k)+7(3+2k)=0

Answer 1

hey
here is ur answer
▶▶▶▶▶▶▶▶▶▶
We have to find the value of 'k ' so that the quadratic equation x2 -2x(1+3k) + 7(3+2k) = 0 has equal roots.

x2 − (2 + 6k)x + 7(3+2k) = 0
Now, a = 1; b = −(2+6k); c = 7(3+2k)
Now, D = b^2−4ac
=(6k+2)^2 − 4 (1)× 7(3+2k)
=36k^2 + 4 + 24k − 84 − 56k
=36k^2 − 32k − 80 
Now, for equal roots,      D = 0
⇒36k^2 − 32k − 80 = 0
⇒9k^2 − 8k − 20 = 0
⇒9k^2 − 18k + 10k − 20 = 0
⇒9k(k−2) + 10(k−2) = 0
⇒(9k+10)(k−2) = 0
⇒k = −109  or  k = 2

▶▶▶▶▶▶▶▶▶
I hope this will help
#Prem✴✴✌✴✌✴✌

Answer 2

According to given sum,

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

$= > \: {x}^{2} - (2 + 6k)x + 7(3 + 2k) = 0.$

Given that, Roots are real and equal.

So, the discriminate will be zero.

$= > \: {b}^{2} - 4ac = 0$

$= > \: ( { - 2 - 6k})^{2} - 4(1)(21 + 14k) = 0$

$= > \: 4 + 36 {k}^{2} + 24k - 84 - 56k = 0.$

$= > \: 36 {k}^{2} - 32k - 80 = 0$

$= > \: 4(9 {k}^{2} - 8k - 20) = 0$

$= > \: 9 {k}^{2} - 8k - 20 = 0.$

$= > \: 9 {k}^{2} - 18k + 10k - 20 = 0$

$= > \: 9k(k - 2) + 10(k - 2) = 0$

$= > \: (k - 2)(9k + 10) = 0$

So, the value of k will be 2 or -10/9.

:-)Hope it helps u.