Question

For what value of k the quadratic equation x2 -2(1+3k)x +7(3+2k) = 0 has equal roots.

Answer 1

Step By Step Solution:-

=> To find Value Of K

=> x²-2(1+3k)x+7(3+2k) = 0

=> we know that D = b²-4ac and D = 0 If Roots are equal.......

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

=> substituting the values for D......

=> [-2(1+3k)]²-4(1){7(3+2k)}

=> 4(1+6k+9k²) - 4(21+14k) = 0

=>4[ 1+6k+9k²-21-14k] = 0

=> 4[ 9k² - 8k -20 ] = 0

=> Shifting 4 to L.H.S

=> 9k² -8k -20 = 0

=> Using Middle Term Split,

=> 9k²-18k+10k -20 = 0

=> 9k(k-2) + 10(k-2) = 0

=> (9k +10)(k-2) = 0

So, K = -10/9 , 2 Ans...

Hope It Helps You......... :)

Answer 2

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{The \ value \ of \ k \ is \ 2 \ or \ \frac{-10}{9}}$

$\sf\orange{Given:}$

$\sf{The \ given \ quadratic \ equation \ is}$

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

$\sf{\implies{Equation \ has \ equal \ roots}}$

$\sf\pink{To \ find:}$

$\sf{The \ value \ of \ k.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{The \ given \ quadratic \ equation \ is}$

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

$\sf{Here, \ a=1, \ b=-2(1+3k) \ and \ c=7(3+2k)}$

$\sf{If \ equation \ has \ equal \ roots,}$

$\sf{then \ the \ value \ of \ the \ Determinant}$

$\sf{will \ be \ zero.}$

$\sf{\therefore{b^{2}-4ac=0}}$

$\sf{[-2(1+3k)]^{2}-4(1)[7(3+2k)]=0}$

$\sf{4(1+3k)^{2}-28(3+2k)=0}$

$\sf{4(1+6k+9k^{2})-84-56k=0}$

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

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

$\sf{Dividing \ equation \ by \ 4 \ throughout}$

$\sf{9k^{2}-8k-20=0}$

$\sf{Here, \ a=9, \ b=-8\ and \ c=-20}$

$\sf{b^{2}-4ac=-8^{2}-4(9)(-20)}$

$\sf{=64+720=784}$

$\sf{By \ formula \ method}$

$\sf{k=\frac{-b+\sqrt{b^{2}-4ac}}{2a} \ or \ \frac{-b-\sqrt{b^{2}-4ac}}{2a}}$

$\sf{k=\frac{8+\sqrt784}{18} \ or \ \frac{8-\sqrt784}{18}}$

$\sf{k=\frac{8+28}{18} \ or \ \frac{8-28}{18}}$

$\sf{k=\frac{36}{18} \ or \ \frac{-20}{18}}$

$\sf{k=2 \ or \ \frac{-10}{9}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ k \ is \ 2 \ or \ \frac{-10}{9}}}}$