Question

if the equation x^2- kx+49=0 has only one root real then k=__
plzzzzz..answer my question fast

Answer 1

Answer:

±14

Step-by-step explanation:

Given Quadratic Equation is x² - kx + 49 = 0

On comparing with ax² + bx + c = 0,

We get a = 1, b = -k, c = 49

Given that the equation has only one real root.

∴ Δ = b² - 4ac

⇒ 0 = (-k)² - 4(1)(49)

⇒ 0 = k² - 196

⇒ k² = 196

⇒ k = ±14.

Therefore, the value of k = ±14.

Hope it helps!

Answer 2

this is the answer of math