Question

Find the value of k for which given quadratic
equations are real and equal roots.
k^2 x^2– 2(k–1)x + 4 = 0

Answer 1

Answer:

Correct option is true

k

k 2

k 2 x

k 2 x 2

k 2 x 2 −2(2k−1)x+4=0

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)]

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k 2

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k 2 −16k+4−16k

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k 2 −16k+4−16k 2

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k 2 −16k+4−16k 2 =0

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k 2 −16k+4−16k 2 =0⇒ −16k=−4

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k 2 −16k+4−16k 2 =0⇒ −16k=−4∴ k=

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k 2 −16k+4−16k 2 =0⇒ −16k=−4∴ k= 4

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k 2 −16k+4−16k 2 =0⇒ −16k=−4∴ k= 41

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k 2 −16k+4−16k 2 =0⇒ −16k=−4∴ k= 41

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k 2 −16k+4−16k 2 =0⇒ −16k=−4∴ k= 41

k 2 x 2 −2(2k−1)x+4=0⇒ Here, a=k 2 ,b=−2(2k−1),c=4⇒ It is given that roots are real and equal.∴ b 2 −4ac=0⇒ [2(2k−1)] 2 −4(k 2 )(4)=0⇒ 4(4k 2 −4k+1)−16k 2 =0⇒ 16k 2 −16k+4−16k 2 =0⇒ −16k=−4∴ k= 41 ∴ We can see value of k given in question is correct.