Question

if the quadratic equation (k+1)x²- 2(k-1) x + 1=0 has real and equal roots than find the value of k
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Answer 1

Given :

Quadratic equation

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

has equal and real roots

To find :

• value of k

Knowledge required :

A quadratic equation of the form

$\sf{a {x}^{2} + bx + c = 0}$

will have equal and real roots

when,

$\red{ \star} \: \boxed{ \sf{discriminant = {b}^{2} - 4ac = 0}}$

Solution :

Comparing

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

with

$\sf{a {x}^{2} + bx + c = 0}$

we will get,

• a = k + 1
• b = -2 ( k - 1 )
• c = 1

Using Formula for discriminant

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

putting values

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

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

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

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

$\longrightarrow\sf{4{k}^2 - 12 k = 0 }$

$\longrightarrow\sf{4k-12=0}$

$\longrightarrow \sf{ 4 k = 12}$

$\longrightarrow \overbrace{\underbrace{\boxed{ \red{\sf{ k = 3}}}}}$

therefore,

• Value of k is 3 .

Answer 2

$\red{\square}~~ \sf {\underline{(k+1)x²- 2(k-1) x + 1 = 0}}$

$\red{\circ}~~ \boxed{{\sf Discriminent = 0}}~~\red{\circ}$

$\sf b²-4ac = 0$

$\sf b² = 4ac$

$\sf {-2(k-1)}² = 4×(k+1)(1)$

$\sf (-2k+2)² = 4k + 4$

$\sf 4k²+4-8k = 4k+4$

$\sf 4k² - 12k = 0$

$\sf 4k -12=0$

$\sf 4k=12$

$\boxed{\orange{\sf {k=3}}}$