Question

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

Answer 1

The roots of the equation are the $x$-intercepts. By definition, the y-coordinate on the $x$-axis is $0$. Therefore, to find the roots of a quadratic function, we set $f\left( x \right) = 0$, and solve the equation, $a{x^2} + bx + c = 0$.

Complete step-by-step answer:
It is given that we have to find the value of $k$ so that they have equal roots.
The equation to find the roots of a quadratic equation, $a{x^2} + bx + c = 0$ in general is given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
If the roots are equal then the value of the discriminant, ${b^2} - 4ac$ will be equal to $0$,
Hence ${b^2} = 4ac$
Here, the equal roots will be $x = \dfrac{{ - b}}{{2a}}$
(i) $2{x^2} + kx + 3 = 0$
We will equate this with the general equation.
That is we can write that $a{x^2} + bx + c = 0$
Now, we get $a = 2{\text{ }},{\text{ }}b = k{\text{ }}and{\text{ }}c = 3$
The roots are equal therefore the value of discriminant will be equal to$0$.
${b^2} - 4ac = 0$
Substituting the values of $a,{\text{ }}b{\text{ and }}c$ we get,
${k^2} - 4ac = 0$
${k^2} - 4 \times 2 \times 3 = 0$
${k^2} - 24 = 0$
${k^2} = 24$
$k = \pm 2\sqrt 6$
Therefore, the values of k are $\pm 2\sqrt 6$ for the roots to be equal.
(ii) $kx\left( {x - 2} \right) + 6 = 0$
The above equation can be written as,
$k{x^2} - 2kx + 6 = 0$
We will equate this with the general equation. i.e. $a{x^2} + bx + c = 0$
Therefore we get $a = k{\text{ }},{\text{ }}b = - 2k{\text{ and }}c = 6$
The roots are equal therefore the value of discriminant will be equal to 0.
${b^2} - 4ac = 0$
Equating the values of a, b and c
${b^2} - 4ac = 0$
${( - 2k)^2} - 4 \times k \times 6 = 0$
$4k2 - 24k = 0$
$4k(k - 6) = 0$
$4k = 0,{\text{or, }}k - 6 = 0$
$k = 0, 6$
Therefore, the values of $k$ are $0, 6$ for the roots to be equal. However, the value of$\;k = 0$ is not valid because on substituting this in the question, $k{x^2} - 2kx + 6 = 0$, the equation will become$0 + 6 = 0$, which is not be valid.

Note: In other words, the quadratic formula is simply just $a{x^2} + bx + c = 0$ in terms of $\;x$. So the roots of $a{x^2} + bx + c = 0$ would just be the quadratic equation, which is: $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
If two roots are equal real numbers, then we can write that ${b^2} - 4ac = 0$
If two roots are different real numbers, then we can write that ${b^2} - 4ac > 0$
If two roots are imaginary real numbers, then we can write that ${b^2} - 4ac < 0$