Question

5x² - 20x + k = 0 has exactly one real solution, show that k=20​

Answer 1

x = 2

Explanation:

5x^2 - 20x + 20 = 0

5 (x^2 - 4x + 4) = 0

x^2 - 4x + 4 = 0

x -2

x -2

(x - 2)(x - 2) = 0

-------------------------

x - 2 = 0

x = 2

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5x² - 20x + k = 0

k = - 5x² + 20x

k = - 5(2)² + 20(2)

k = -20 + 40

k = 20

Answer 2

Answer + Step-by-step explanation:

Consider the equation :

5x² - 20x + k = 0

∆ = b² − 4bc is the discriminant of the quadratic equation.

In our equation :

a = 5 ; b= -20 ; c = k

Then

∆ = (-20)² − 4×5×k

= 20² − 20k

= 20×(20 − k) (Factoring)

The equation 5x² - 20x + k = 0 has exactly one real solution

if and only if ∆ = 0.

Solving the equation ∆ = 0 :

∆ = 0

⇔ 20×(20 − k) = 0

⇔ (20 − k) = 0 (Zero product property)

⇔ k = 20