Question

Find the values of k for which the quadratic equation 2x^2 − (k + 2)x + k = 0 has real and equal roots.

Answer 1

k = 2

Explanation:

A quadratic equation has two equal real roots if a discriminant is equal 0.

`ax^2+bx+c=0`

Discriminant
`b^2-4ac`

We have the equation

`2x^2-(k+2)x+k=0\to a=2,\ b=-(k+2),\ c=k`

Substitute:

`b^2-4ac=\bigg(-(k+2)\bigg)^2-4(2)(k)\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\=k^2+2(k)(2)+2^2-8k=k^2+4k+4-8k=k^2-4k+4\\\\b^2-4ac=0\iff k^2-4k+4=0\\\\k^2-2k-2k+4=0\\\\k(k-2)-2(k-2)=0\\\\(k-2)(k-2)=0\\\\(k-2)^2=0\iff k-2=0\qquad\text{add 2 to both sides}\\\\k=2`

Answer 2

k = 2

Explanation:

If the roots are real and equal then the condition for the discriminant is

b² - 4ac = 0

For 2x² - (k + 2)x + k = 0 ← in standard form

with a = 2, b = - (k + 2) and c = k, then

(- (k + 2))² - (4 × 2 × k ) = 0

k² + 4k + 4 - 8k = 0

k² - 4k + 4 = 0

(k - 2)² = 0

Equate factor to zero and solve for k

(k - 2)² = 0 ⇒ k - 2 = 0 ⇒ k = 2