Question

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

Answer 1

Solution:-
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.

Answer 2

Answer:

Step-by-step explanation:

Solution :-

For Equation x² + kx + 64 = 0

b² - 4ac = 0

⇒ k² - 4 × 1 × 64 = 0

⇒ k² - 256 = 0

⇒ k = ± 16 ... (i)

And for equation, x² - 8x + k = 0

b² - 4ac = 0

⇒ (- 8)² - 4 × 1 × k = 0

⇒ 64 = 4k

⇒ k = 64/4 ... (ii)

From Eq (i) and (ii), we get

For K = 16, is the given equation have equal roots.

Nature of roots of a quadratic equation:-

(i). If b² - 4ac > 0, the quadratic equation has two distinct real roots.

(ii). If b² - 4ac = 0,  the quadratic equation has two equal real roots.

(iii). If b² - 4ac < 0, the quadratic equation has no real roots.