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

Hint: 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 \[{b^2} - 4ac < 0\]