Question

3.
For what values of k, the given equations have real and equal root k2x2 - 2(2k-1)x + 4 = 0​

Answer 1

if this equation has real and equal roots hence Discriminant(D)=0

Given:-

here a = k²,b = -2(2k-1),c = 4

To Find:-

value of k = ?

Solution:-

As we know,D=b²-4ac

so, b²-4ac = 0

$\implies$[-2(2k-1)]² - 4×k²×4

$\implies$(-4k+2)²-16k²=0

$\implies$(-4k+2)² = 16k²

$\implies$-4k+2 = √16k²

$\implies$-4k+2 = 4k

$\implies$2 = 8k

$\implies$k = $\frac{2}{8}$

$\implies$k = $\frac{1}{4}$ ......ans.......

Answer 2

Answer:

Given:-

here a = k²,b = -2(2k-1),c = 4

To Find:-

value of k = ?

Solution:-

As we know,D=b²-4ac

so, b²-4ac = 0

\implies⟹ [-2(2k-1)]² - 4×k²×4

\implies⟹ (-4k+2)²-16k²=0

\implies⟹ (-4k+2)² = 16k²

\implies⟹ -4k+2 = √16k²

\implies⟹ -4k+2 = 4k

\implies⟹ 2 = 8k

\implies⟹ k = \frac{2}{8}82

\implies⟹ k