Question

Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) $2x^2 + kx + 3 = 0$ (ii) $kx (x – 2) + 6 = 0$

Answer 1

We know that if an equation $ax^2 + bx + c = 0$ has two equal roots, its discriminant $(b^2 − 4ac)$ will be 0.

(i) $2x^2 + kx + 3 = 0$

Comparing equation with $ax^2 + bx + c = 0,$ we obtain
a = 2, b = k, c = 3

Discriminant = $b^2 − 4ac$ = $(k)^2− 4(2) (3)$ =$k^2 − 24$
For equal roots, Discriminant = 0
$k^2 − 24$ = 0
$k^2 = 24$
$k = ±\sqrt{24} = ±2\sqrt6$

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(ii) $kx (x − 2) + 6 = 0$
or $kx^2 − 2kx + 6 = 0$

Comparing this equation with $ax^2 + bx + c = 0,$ we obtain
a = k, b = −2k, c = 6

Discriminant = $b^2 − 4ac$ = (− 2k)^2 − 4 (k)$$(6) = $4k^2 − 24k$
For equal roots,
$b^2 − 4ac$= 0
$4k^2 − 24k$ = 0
$4k (k − 6)$= 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6

However, if k = 0, then the equation will not have the terms $‘x^2’$ and ‘x’.

Therefore, if this equation has two equal roots, k should be 6 only.