Question

✴️CONTENT QUALITY ANSWER WITH PROPER STEPS ✴️

➡️STANDARD MARKS 3⬅️

Find the positive value of K for which the equations x²+Kx+64=0 and x²-8x+K=0 will have real roots.

❌NO LAZY ANSWERS❌

Answer 1

Let D₁ and D₂ be the discriminant of the given equations and these will have equal roots only if D₁, D₂ ≥ 0
Or, if D₁ = (k² -4 × 64) ≥ 0  and D₂ = (64 - 4k) ≥ 0
Or,  if k² ≥ 256  and  4k ≤ 64
Or, if k ≥ 16  and  k ≤ 16
Or, k = 16
Hence the positive value of k is 16

Answer 2

if the equaion x2+kx+64=0 has real roots, then D>=0

k2-256>=0  => k2>=256  => k2>=(16)2  => k>=16.- (1)

if the equation x2-8x+k=0 has real roots.

then D>=0  => 64-4k>=0  => 4k<=64  => k<=16 - (2)

from (1) and (2) we get k=16