Question

For what values of k, the roots of the quadratic equation $(k + 4){x^2} + (k + 1)x + 1 = 0$ are equal?

Answer 1

Hint: We know that if we have an equation $a{x^2} + bx + c = 0$, then $D = {b^2} - 4ac$ is the discriminant of the equation. Using this and putting D =0, we will have the required result.

Complete step-by-step answer:
Let us consider the general equation $a{x^2} + bx + c = 0$.
The discriminant of this equation is given by $D = {b^2} - 4ac$ and if D < 0, then we have imaginary roots; if D > 0, then we have both real and distinct roots; and if D = 0, then both the roots are real and distinct.
This happens because the roots of a quadratic equation is given by $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$.
Now if D < 0, then $\sqrt D $ is always imaginary.
If D > 0, then $\sqrt D $ is always positive.
If D = 0, then $\sqrt D $ will be 0.
Now, we need to find some conditions on the given equation $(k + 4){x^2} + (k + 1)x + 1 = 0$ such that it has equal roots. It will have equal roots when D = 0.
So, we need $D = {b^2} - 4ac = 0$.
Comparing the general equation to the equation in question, we will have:-
a = k + 4, b = k + 1 and c = 1.
So, $D = {b^2} - 4ac = {(k + 1)^2} - 4 \times (k + 4) \times 1 = 0$
Simplifying by using the formula: ${(a + b)^2} = {a^2} + {b^2} + 2ab$, we will have:-
$ \Rightarrow D = {(k + 1)^2} - 4 \times (k + 4) \times 1 = {k^2} + 1 + 2k - 4k - 16 = 0$
Simplifying further by combining the like terms with each other, we will get:-
$ \Rightarrow D = {k^2} + 1 + 2k - 4k - 16 = {k^2} - 2k - 15 = 0$
Now, equating the simplified value of D with 0, we will have:-
$ \Rightarrow {k^2} - 2k - 15 = 0$
Now, we will use the factorization method to make factors of this quadratic equation as follows:-
$ \Rightarrow {k^2} + 3k - 5k - 15 = 0$
$ \Rightarrow k(k + 3) - 5(k + 3) = 0$
$ \Rightarrow (k + 3)(k - 5) = 0$.
Hence, if k = -3 or k = 5, then we will have the roots equal.

Note: The students must note that the conditions in the question can be different. Use the formula and condition accordingly.
The students can also solve this by finding both the roots first and then equating them. We will still get the same result.
Fun Fact:- Do you see $D = {b^2} - 4ac$ in the formula above? It is called the Discriminant, because it can "discriminate" between the possible types of answer:
When $D = {b^2} - 4ac$ is positive, we get two Real solutions.
When it is zero we get just ONE real solution (both answers are the same).
When it is negative we get a pair of Complex solutions.