Question

find k,if has real nd equal roots x^2+k(2x+k-1)+2=0​

Answer 1

Question :-

Find value of k if it has real and equal roots ,

•x² +k(2x + k - 1) + 2 = 0

Answer :-

→ k = 2

To find :-

→ Value of k

Formula used :-

Discriminant = (b² - 4ac)

Explanation :-

If given any Quadratic p(x) = ax²+ b x +x

c= 0 ,has equal and real roots then it's discriminant (D) is equal to zero(0) .

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Given equation is ,

→x² + k(2x + k -1) +2 = 0

→ x² +2kx +( k² - k +2) = 0

Comparing this equation with given p(x)

we get,

→ a = 1 , b = 2k and c = k² - k +2

Now ,apply the given formula for discriminant ,

→ D = 0

→ b² - 4ac = 0

→(2k)² -4 ×1(k² - k +2 ) = 0

→ 4k² - 4k² + 4k - 4×2 = 0

→ 4k - 8 = 0

→ 4k = 8

divided by 4 on both sides ,

→ k = 2

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Some similar questions for practice :-

Q.1 If "a" and "b" are the roots of the equation x² + 4x -12 = 0 then find -

→ Sum of roots

→Product of roots

→Discriminant

→Nature of roots .

Q.2 If roots of the polynomial 2x² + 2a+2a(x + 2) + 4 = 0 are real and equal then find value of "a" .

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