Question

find the value of'k' for which the roots are real and equal
4x2-2(k+1)X+(k+1)=0​

Answer 1

Answer:

4

Step-by-step explanation:

use discriminant formula for finding the value of k

Answer 2

Answer:

$\sf{The \ value \ of \ k \ is \ 1+\sqrt3 \ or}$

$\sf{1-\sqrt3.}$

Given:

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

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

$\sf{The \ equation \ has \ real \ and \ equal \ roots.}$

To find:

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

Solution:

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

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

$\sf{Here, \ a=4, \ b=-2(k+1) \ and \ c=k+1}$

$\sf{If \ the \ equation \ has \ real \ and \ equal \ roots}$

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

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

$\sf{\leadsto{4(k+1)^{2}-16(k+1)=0}}$

$\sf{\leadsto{4(k^{2}+2k+1)-16k-16=0}}$

$\sf{\leadsto{4k^{2}+8k+8-16k-16=0}}$

$\sf{\leadsto{4k^{2}-8k-8=0}}$

$\sf{\leadsto{4(k^{2}-2k-2)=0}}$

$\sf{\leadsto{k^{2}-2k-2=0}}$

$\sf{Here, \ a=1, \ b=-2 \ and \ c=-2}$

$\sf{Now, \ b^{2}-4ac=(-2)^{2}-4(1)(-2)}$

$\sf{\therefore{b^{2}-4ac=4+8}}$

$\sf{\therefore{b^{2}-4ac=12}}$

$\sf{By \ quadratic \ formula}$

$\sf{\leadsto{k=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}}$

$\sf{\leadsto{k=\dfrac{2\pm\sqrt{12}}{2}}}$

$\sf{\leadsto{k=\dfrac{2\pm2\sqrt3}{2}}}$

$\sf{\leadsto{k=1\pm\sqrt3}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ k \ is \ 1+\sqrt3 \ or}}}$

$\sf\purple{\tt{1-\sqrt3.}}$