# Find k For Quadratic Equation?

Find k such that the equation x^2 - kx + 4 = 0 has a repeated real solution.

### 8 Answers

- Engr. RonaldLv 76 months agoBest Answer
x^2 - kx + 4 = 0

apply the discriminant formula

a = 1 , b = - k, c = 4

b^2 - 4ac = 0

(-k)^2 - 4(1)(4) = 0

k^2 - 16 = 0

k^2 = 16

k = √(16)

k = 4 answer//

x^2 - 4x + 4 = 0

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

x = 2

- Iggy RockoLv 76 months ago
We want the determinant to be 0.

b^2 - 4ac = 0

(-k)^2 - 4(1)(4) = 0

k^2 = 16

k = 4 or k = -4

If k = 4, x = 2 with multiplicity 2.

If k = -4, x = -2 with multiplicity 2.

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- Φ² = Φ+1Lv 76 months ago
To form a square term which gives the repeat real solution you use

k = 2√([x²][x⁰]) = 2√(1×4) = 2×2 = 4

Note that [x²] is strictly positive and [x⁰] is strictly non-negative.

x² - 4x + 4 = 0 is (x - 2)² = 0 with the repeated real solution x = 2.

- Steve ALv 76 months ago
(x - a)(x - a) = x^2 - kx + 4

x^2 - 2ax + a^2 = x^2 - kx + 4

-2ax + a^2 = -kx + 4

a = 2

k = 4

- rotchmLv 76 months ago
A repeated real solution implies that it can be written as a(x-r)² . Expanding gives ax² -2arx + r². This must equal your given equation.

From the cooef of x² we see that a=1. Can you see what r can be? So what is k?

Or another way: a repeated real solution means that the discriminant "b² - 4ac" = 0. What can you conclude from that?

Show your work/answers here & we can proceed if need be.