Math, asked by yashsinari5, 3 months ago


if root of the equation x2 + k (4x + k - 1) + 2 = 0 are real and equal, find the value of k.​

Answers

Answered by Anonymous
12

Answer :-

\implies\sf x^2 + k (4x + k - 1) + 2 = 0

\implies\sf x^2 + 4kx + k^2 - k + 2 = 0

If roots of a quadratic equation are real and equal, then the value of discriminant is equal to zero.

\implies\sf D = b^2 - 4ac = 0

For the given equation :-

  • \sf a = 1
  • \sf b = 4k
  • \sf c = k^2 - k + 2

\implies\sf b^2 - 4ac = 0

\implies\sf (4k)^2 - 4 ( 1 ) ( k^2 - k + 2 ) = 0

\implies\sf 16k^2 - 4k^2 + 4k - 8 = 0

\implies\sf 12k^2 + 4k - 8 = 0

\implies\sf 12k^2 + 12k - 8k - 8 = 0

\implies\sf 12k ( k + 1 ) - 8 ( k + 1 ) = 0

\implies\sf ( 12k - 8 )( k + 1 ) = 0

Either ( 12k - 8 ) is zero or ( k + 1 ) is zero :-

  • \sf 12k - 8 = 0

\implies\sf 12k = 8

\implies\sf k = \dfrac{8}{12}

  • \sf k + 1 = 0

\implies\sf k = -1

\sf Value \:of\: k = -1 ,  \dfrac{2}{3}

Answered by Anonymous
6

Given :

  • x² + k(4x + k - 1) + 2
  • Roots are real and equal

To Find :

The value of k.

Solution :

Using the formula for discriminant and the condition for real and equal roots we can find the answer.

Required Formula :

D = b² - 4ac

Explanation :

x² + k(4x + k - 1) + 2

⇒ x² + 4kx + (k² - k + 2)

We know that if the roots are real and equal then the discriminant is 0.

ATQ,

D = b² - 4ac

where,

  • a = 1
  • b = 4k
  • c = k² - k + 2

Substituting the values,

⇒ D = (4k)² - [4.1.(k² - k + 2)]

⇒ D = 16k² - [4(k² - k + 2)]

⇒ D = 16k² - [4k² - 4k + 8]

⇒ D = 16k² - 4k² + 4k - 8

⇒ D = 12k² + 4k - 8

For equal roots, D = 0;

⇒ 12k² + 4k - 8 = 0

Splitting the middle term,

⇒ 12k² + 12k - 8k - 8 = 0

⇒ 12k(k + 1) - 8(k + 1) = 0

⇒ (k + 1)(12k - 8) = 0

⇒ k + 1 = 0

⇒ k = -1

k = -1.

⇒ 12k - 8 = 0

⇒ 12k = 8

⇒ k = 8/12

⇒ k = 2/3

k = 2/3.

The value of k is -1 or 2/3.

Explore More :

  • If roots are real and distinct, D > 0
  • If roots are not real, D < 0
  • If roots are real, D ≥ 0
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