Question

2xsquaer -(2+k)x+k=0 solution

Answer 1

A. By quadratic formula :

The given equation is

$2x^{2}-(2+k)x+k=0$

Comparing it with the standard quadratic equation $ax^{2}+bx+c=0,\:a\neq 0$, we get

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

By quadratic formula, we get

x = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$=\frac{2+k\pm \sqrt{(2+k)^{2}-4.2.k}}{2.2}$

$=\frac{2+k\pm \sqrt{4+4k+k^{2}-8k}}{4}$

$=\frac{2+k\pm \sqrt{2^{2}-4k+k^{2}}}{4}$

$=\frac{2+k\pm \sqrt{(2-k)^{2}}}{4}$

$=\frac{2+k\pm (2-k)}{4}$

$=\frac{2+k+2-k}{4},\:\frac{2+k-2+k}{4}$

$=\frac{4}{4},\:\frac{2k}{4}$

$=1,\:\frac{k}{2}$,

which is the required solution.

B. By middle term formula :

Now, 2x² - (2+k)x + k = 0

⇒ 2x² - 2x - kx + k = 0

⇒ 2x (x - 1) - k (x - 1) = 0

⇒ (x - 1) (2x - k) = 0

Either x - 1 = 0 or, 2x - k = 0

      ⇒ x = 1 , x = k/2

Therefore, the required solution be

                   x = 1 , k/2