Question

For what value of $k$, $\left( {4 - k} \right){x^2} + \left( {2k + 4} \right)x + \left( {8k + 1} \right) = 0$ is a perfect quadratic equation.

Answer 1

Hint:- Here, we will use the discriminant method in order to ensure that the given equation is a perfect quadratic equation.

Complete step-by-step solution -
Given quadratic equation is $\left( {4 - k} \right){x^2} + \left( {2k + 4} \right)x + \left( {8k + 1} \right) = 0$
As we know that any quadratic equation $a{x^2} + bx + c = 0$ is said to be perfect quadratic equation, if the given quadratic equation can be represented as a perfect square or we can say that if both the roots of the quadratic equation are equal.
According to discriminant method,
Discriminant $d = \sqrt {{b^2} - 4ac} $
For roots of the quadratic equations to be equal, the value of the discriminant will be equal to zero.
i.e., $
   \Rightarrow \sqrt {{{\left( {2k + 4} \right)}^2} - 4\left( {4 - k} \right)\left( {8k + 1} \right)} = 0 \Rightarrow \sqrt {4{k^2} + 16 + 16k - 4\left( {31k - 8{k^2} + 4} \right)} = 0 \\
   \Rightarrow \sqrt {4\left[ {{k^2} + 4 + 4k - \left( {31k - 8{k^2} + 4} \right)} \right]} = 0 \Rightarrow 2\sqrt {9{k^2} - 27k} = 0 \Rightarrow 2\sqrt {9k\left( {k - 3} \right)} = 0 \\
   \Rightarrow 6\sqrt {k\left( {k - 3} \right)} = 0 \Rightarrow \sqrt {k\left( {k - 3} \right)} = 0 \\
 $
Now squaring both sides, we get
$ \Rightarrow k\left( {k - 3} \right) = 0$
$\therefore $ Either $k = 0$ or $k = 3$
Therefore, the values of $k$ for which the given quadratic equation is a perfect quadratic equation are 0 and 3.

Note- In these types of problems in order to make the given quadratic equation as a perfect quadratic equation, the discriminant of the given quadratic equations is equated to zero. From this, the value of any existing unknown is determined.