If $a{x^2} + bx + c = 0$ has equal roots, then c is equal to ______.
(A) $\dfrac{{ - b}}{{2a}}$
(B) $\dfrac{b}{{2a}}$
(C) $\dfrac{{ - {b^2}}}{{4a}}$
(D) $\dfrac{{{b^2}}}{{4a}}$
Answer
645.9k+ views
Hint:
Use the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ in the given quadratic equation. Put both roots equal to each other and then simplify the equation. This will result in establishing a relation between $'a'$ , $'b'$ and $'c'$ . Now express $'c'$ in terms of $'a'$ and $'b'$.
Complete step by step solution:
Here in this problem, a quadratic equation is given as $a{x^2} + bx + c = 0$ , which is in its general form. For this equation, we need to find the value for c if the roots are equal for this equation.
This can be solved by writing the two roots for this equation using the quadratic formula and then equation both of them to establish a relationship.
The Quadratic Formula uses the $'a'$ , $'b'$ and $'c'$ from $a{x^2} + bx + c = 0$ , where $'a'$ , $'b'$ , and $'c'$ are just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to solve.
As we know that the Quadratic formula for the quadratic equation in general forms $a{x^2} + bx + c = 0$ where $a \ne 0$ is given by:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
This represents that the two roots of the equations are: ${x_1} = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}{\text{ and }}{x_2} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
But it is given that these two roots are equal. Therefore, we can write that:
$ \Rightarrow {x_1} = {x_2} \Rightarrow \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
Now we can further simplify this by multiplying both sides by $2a$ . This will give us:
$ \Rightarrow - b + \sqrt {{b^2} - 4ac} = - b - \sqrt {{b^2} - 4ac} \Rightarrow 0 + 2\sqrt {{b^2} - 4ac} = 0$
Let’s now square both sides of the above equation to remove the radical sign
$ \Rightarrow 2\sqrt {{b^2} - 4ac} = 0 \Rightarrow 4\left( {{b^2} - 4ac} \right) = 0$
We can now transpose $c$ to the other side and get the required expression:
$ \Rightarrow 4\left( {{b^2} - 4ac} \right) = 0 \Rightarrow {b^2} = 4ac \Rightarrow c = \dfrac{{{b^2}}}{{4a}}$
Therefore, we get: $c = \dfrac{{{b^2}}}{{4a}}$
Hence, the option (D) is the correct answer.
Note:
In a question like this, the use of the quadratic formula is the most crucial part of the solution. And the expression $\sqrt {{b^2} - 4ac} $ is called discriminant and is used to determine the nature and behaviour of the roots of a quadratic equation. An alternative approach for this problem can be to equate discriminant $\sqrt {{b^2} - 4ac} $ with zero.
Use the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ in the given quadratic equation. Put both roots equal to each other and then simplify the equation. This will result in establishing a relation between $'a'$ , $'b'$ and $'c'$ . Now express $'c'$ in terms of $'a'$ and $'b'$.
Complete step by step solution:
Here in this problem, a quadratic equation is given as $a{x^2} + bx + c = 0$ , which is in its general form. For this equation, we need to find the value for c if the roots are equal for this equation.
This can be solved by writing the two roots for this equation using the quadratic formula and then equation both of them to establish a relationship.
The Quadratic Formula uses the $'a'$ , $'b'$ and $'c'$ from $a{x^2} + bx + c = 0$ , where $'a'$ , $'b'$ , and $'c'$ are just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to solve.
As we know that the Quadratic formula for the quadratic equation in general forms $a{x^2} + bx + c = 0$ where $a \ne 0$ is given by:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
This represents that the two roots of the equations are: ${x_1} = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}{\text{ and }}{x_2} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
But it is given that these two roots are equal. Therefore, we can write that:
$ \Rightarrow {x_1} = {x_2} \Rightarrow \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
Now we can further simplify this by multiplying both sides by $2a$ . This will give us:
$ \Rightarrow - b + \sqrt {{b^2} - 4ac} = - b - \sqrt {{b^2} - 4ac} \Rightarrow 0 + 2\sqrt {{b^2} - 4ac} = 0$
Let’s now square both sides of the above equation to remove the radical sign
$ \Rightarrow 2\sqrt {{b^2} - 4ac} = 0 \Rightarrow 4\left( {{b^2} - 4ac} \right) = 0$
We can now transpose $c$ to the other side and get the required expression:
$ \Rightarrow 4\left( {{b^2} - 4ac} \right) = 0 \Rightarrow {b^2} = 4ac \Rightarrow c = \dfrac{{{b^2}}}{{4a}}$
Therefore, we get: $c = \dfrac{{{b^2}}}{{4a}}$
Hence, the option (D) is the correct answer.
Note:
In a question like this, the use of the quadratic formula is the most crucial part of the solution. And the expression $\sqrt {{b^2} - 4ac} $ is called discriminant and is used to determine the nature and behaviour of the roots of a quadratic equation. An alternative approach for this problem can be to equate discriminant $\sqrt {{b^2} - 4ac} $ with zero.
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