Question

How do you solve ${{x}^{2}}-5x=-9$ using the quadratic formula?

Answer 1

We are given ${{x}^{2}}-5x=-9$ to solve this, we learn about the type of equation then we learn the number of solutions of the equation. we will learn how to factor the quadratic equation, we will use the middle term split to factor the term and we will simplify by taking common terms out. We also use zero product rules to get our answer. To be sure about your answer we can also check by putting the acquired value of the solution in the given equation and check whether they are the same or not.

Complete step by step solution:
We are asked to solve the given equation ${{x}^{2}}-5x=-9$ .
First we observe that it has a maximum power of ‘2’ so it is a quadratic equation.
Now we should know that a quadratic equation has 2 solutions or we say an equation of power ‘n’ will have ‘n’ solutions.
Now as it is a quadratic equation we will change it into standard form $a{{y}^{2}}+by+c=0$ .
We have ${{x}^{2}}-5x=-9$ .
We will add ‘9’ on both sides, we get –
$\Rightarrow {{x}^{2}}-5x+9=-9+9$ .
By simplifying, we get –
${{x}^{2}}-5x+9=0$ ($-9+9=0$ )
Now we have to solve the equation ${{x}^{2}}-5x+9=0$ .
To solve this equation we first take the greatest common factor possible available to the terms.
As we can see that in ${{x}^{2}}-5x+9=0$ .
$1,-5\text{ and }9$ has nothing in common . So, the equation remains the same.
$\Rightarrow {{x}^{2}}-5x+9=0$
Now, we will use a quadratic formula to solve our problem.
For $a{{x}^{2}}+bx+c=0$ .
Solution using quadratic formula is given as –
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
For ${{x}^{2}}-5x+9=0$ , we have $a=1,b=-5\text{ and }c=9$ .
So, using these value, we get –
$\Rightarrow x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 1\times 9}}{2\times 1}$ .
By simplifying, we get –
$\Rightarrow x=\dfrac{5\pm \sqrt{25-36}}{2}$
As $25-36=-11$
So, we get –
$\Rightarrow x=\dfrac{5\pm \sqrt{-11}}{2}$ .
Now as we know that $\sqrt{-1}=i$ in a complex plane, so $\sqrt{-11}=\sqrt{11}\times i$ .
Hence we get solution as –
$\Rightarrow x=\dfrac{5\pm \sqrt{11}i}{2}$
So, solutions are –
$x=\dfrac{5+\sqrt{11}i}{2}$ and $x=\dfrac{5-\sqrt{11}i}{2}$.

Note: Remember, $\sqrt{-1}$ is not valid in real axis, so $\sqrt{{{b}^{2}}-4ac}<0$ then it has no solution on real axis but on complex plane $\sqrt{-1}=i$ , so we can just find solution for the case when $\sqrt{{{b}^{2}}-4ac}<0$ in the complex plane only. Also remember if $\sqrt{{{b}^{2}}-4ac}>0$ then the both root are real and distinct and if the $\sqrt{{{b}^{2}}-4ac}=0$ then the root are real but same.
Then ${{b}^{2}}-4ac$ is known as discriminate and it helps us in knowing the nature of the root.