Question

How to find the inverse of a quadratic equation?

Answer 1

Hint: In order to find the inverse of a quadratic equation, we first convert it into a perfect square, then we solve it further to bring it in the form of $f\left(x\right)=a{\left(x-h\right)}^2+k$ . We find the domain and range from here. After which, we represent our function $f\left(x\right)$ as $y$ and interchange the positions of $x$ and $y$. We solve it further and thus get our required answer.

Complete step-by-step solution:
Let us demonstrate the process of finding the inverse of a quadratic equation with the following example:
Let us take the quadratic equation as: $x^2-6x+2$ …. equation (A)
This is in the form of $f\left(x\right)=a{x}^2+bx+c$
Now, we need to change this into a perfect square which is denoted by ${\left(a+b\right)}^2=a^2+2ab+b^2$
Here in the quadratic equation, the first term is already a perfect square which is $x^2$ , we take half of the coefficient of the middle term and square it. Here the coefficient of the middle term is: $6$
Therefore, half of $6={\displaystyle \frac{1}{2}}\times 6=3$
In order to make equation (A), a perfect square, we add ${\left(3\right)}^2$ to it. So, the equation becomes:
$x^2-6x+{\left(3\right)}^2+2$
Since we have added ${\left(3\right)}^2$ to the equation from outside, we need to balance it. Thus, we also subtract the same amount from the above equation. Thus the equation becomes:
$x^2-6x+{\left(3\right)}^2+2-{\left(3\right)}^2$
Now we can also write the equation in the form: ${\left(x-3\right)}^2+2-9$ as according to the formula ${\left(a+b\right)}^2=a^2+2ab+b^2$, therefore$x^2-6x+3^2={\left(x-3\right)}^2$
Thus, $f\left(x\right)={\left(x-3\right)}^2-7$
Now, the given quadratic equation is in the form of $f\left(x\right)=a{\left(x-h\right)}^2+k$
Thus, $a=1,h=3,k=-7$
Domain of the original function: $x⩾3$ as domain is the set of x-values that can be used input in the function and $x⩾h$
Range of the original function: $y⩾-7$ as the range is the set of y-values that can be the outcome and $y⩾k$
Now let the inverse function of $f\left(x\right)$ be y
Therefore, the equation becomes: $y={\left(x-3\right)}^2-7$
Let’s interchange the $x$ with $y$ , thus we have:
$x={\left(y-3\right)}^2-7$
On adding $7$ to both sides of the equation we get:
$x+7={\left(y-3\right)}^2$
Now, let us take square root on both the sides of the equation:
$\pm \sqrt{x+7}=y-3$
On adding $3$ to both sides of the equation, we get:
$\pm \sqrt{x+7}+3=y$
Thus, $y=3\pm \sqrt{x+7}$
Now our domain is $x+7⩾0\Rightarrow x⩾-7$ (as we take the positive root $x+7$ )
Our range is $y⩾3$
Thus, we see that the values of our domain and range have gotten inverted.
This is our required inverse form.

Note: All quadratic functions may not have an inverse. We can find the inverse when the domain and the range are well defined. The first thing to notice while finding the inverse of a quadratic equation is the coefficient of $a$ in $a{x}^2+bx+c$ . If we plot the inverse function on a graph, then we get a parabola. This parabola has the following conditions:
If $a>0$ , then the equation defines a parabola whose ends point upwards.
If $a<0$ , then the equation defines a parabola whose ends point downwards.
$a\ne 0$ , since then the function will become linear and not quadratic.