Question

Hey brainliers!!

Find k in following equation to have equal and real roots :-

${x}^{2} + k(4x + k - 1) + 2 = 0$
Pls explain

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Answer 1

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Answer 2

Answer:

⇒ $k= \frac{2}{3}$ (or) ⇒ $k= -1$$[/tex]</strong></p><p><strong>Step-by-step explanation:</strong></p><p><u><strong>Given</strong></u><strong> :</strong></p><p><strong>Find k, if the fgiven equation have real & equal roots :</strong></p><p><strong>[tex]x^2 + k(4x + k - 1)+2=0$

Solution :

The given equation can also be written as,

$x^2 + k(4x + k - 1)+2=0$

⇒ $x^2 + 4xk+ k^2 - k+2=0$

⇒ $x^2 + 4xk+ (k^2 - k+2)=0$

For a quadratic equation of the form :

ax² + bx + c = 0

for real & equal roots from a quadratic equation,.

b² - 4ac = 0

a = 1

b = 4k,

c = k² - k + 2 = 0

⇒ b² - 4ac = 0

⇒ $(4k)^2 - 4(1)(k^2-k+2) = 0$

⇒ $16k^2 - 4k^2+4k-8 = 0$

⇒ $12k^2+4k-8 = 0$

⇒ $12k^2+12k - 8k-8 = 0$

⇒ $12k(k+1) - 8(k+1) = 0$

⇒ $(12k-8)(k+1) = 0$

⇒ $(12k-8)= 0$ (or) ⇒ $(k+1) = 0$

⇒ $12k= 8$ (or) ⇒ $k= -1$

⇒ $k= \frac{8}{12}$ (or) ⇒ $k= -1$

⇒ $k= \frac{2}{3}$ (or) ⇒ $k= -1$