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The Positive Value of K for Which the Equation X2 + Kx + 64 = 0 and X2 − 8x + K = 0 Will Both Have Real Roots, is - CBSE Class 10 - Mathematics

ConceptSolutions of Quadratic Equations by Factorization

Question

The positive value of k for which the equation x2 + kx + 64 = 0 and x2 − 8x + k = 0 will both have real roots, is

• 4

• 8

• 12

• 16

Solution

The given quadric equation are  x2 + kx + 64 = 0 and x2 − 8x + k = 0 roots are real.

Then find the value of a.

Here, x2 + kx + 64 = 0 ….. (1)

x2 − 8x + k = 0 ….. (2)

a_1 = 1,b_1 = k and ,c_1 = 64

a_2 = 1,b_2 = -8 and ,c_2 =  k

As we know that D_1 = b^2 - 4ac

Putting the value of a_1 = 1,b_1 = k and ,c_1 = 64

=(k)^2 - 4 xx 1 xx 64

= k^2 - 256

The given equation will have real and distinct roots, if D >0

k^2 - 256 = 0

k^2 = 256

k = sqrt256

 k  = ± 16

Therefore, putting the value of k = 16 in equation (2) we get

 x^2 - 8x + 16 = 0

(x - 4)^2 = 0

x - 4 = 0

x = 4

The value of  k = 16 satisfying to both equations

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Solution The Positive Value of K for Which the Equation X2 + Kx + 64 = 0 and X2 − 8x + K = 0 Will Both Have Real Roots, is Concept: Solutions of Quadratic Equations by Factorization.
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