Question

How do you factor and solve ${{x}^{2}}-9x=-20$ ?

Answer 1

We are given ${{x}^{2}}-9x=-20$ to solve this we learn about the type of Equation we are given then learn the number of solutions of 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 problem ${{x}^{2}}-9x=-20$ .
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 equations we will change it into standard form $a{{x}^{2}}+bx+c=0$ we have ${{x}^{2}}-9x=-20$ .
We will add 20 both side, we get –
$\Rightarrow {{x}^{2}}-9x+20=-20+20$
Simplifying we get –
$\Rightarrow {{x}^{2}}-9x+20=0$ (As -20+20=0)
Now we have to solve the equation ${{x}^{2}}-9x+20=0$ .
To solve this equation we first take the greatest common factor possibly available to the terms.
As we can see that in ${{x}^{2}}-9x+20=0$ .
-9 and 20 have nothing in common, So equation remains the same .
${{x}^{2}}-9x+20=0$
Now ,we will solve this using a method which is called by factoring, we factor using middle term split.
For quadratic equations $a{{x}^{2}}+bx+c=0$ we will find such a pair of terms whose products are the same as $a\times c$ and whose sum or difference will be equal to the ‘b’ .
Now in ${{x}^{2}}-9x+20=0$
We have $a=1,b=-9\text{ and }c=20$ .
So $a\times c=1\times 20=20$ .
We can see that $\left( -4 \right)\times \left( -5 \right)=20$ and their sum $-4+\left( -5 \right)=-9$
So, we use this to split
${{x}^{2}}-9x+20=0$
So we get ${{x}^{2}}+\left( -4-5 \right)x+20=0$ .
Opening brackets we get
${{x}^{2}}-4x-5x+20=0$
Taking common in first two and last two terms, we get –
$x\left( x-4 \right)-5\left( x-4 \right)=0$ .
As $x-4$ is common so simplifying further .
$\left( x-4 \right)\left( x-5 \right)=0$
Using zero product rules which says if two terms product is zero that either one of them is zero so either $x-4=0\text{ or }x-5=0$ .
So we get $x=4\text{ and }x=5$
Hence solutions are $x=4\text{ and }x=5$.

Note: Remember that when the sign of ‘a’ and ‘c’ in the $a{{x}^{2}}+bx+c=0$ is same then middle term is splits using addition of two terms and if the sign of a and c is different than b is obtained. By the Subtraction of two numbers we can cross check the solution by putting the value of $x=4$ and 5 one by one Back into the equation.
Now we put $x=4$ in ${{x}^{2}}-9x+20=0$
We get
$\begin{aligned} & \Rightarrow {{4}^{2}}-9\times 4+20=0 \\\ & 16-36+20=0 \\\ \end{aligned}$
Simplifying we get –
$0=0$
So $x=4$ satisfies the equation
Now we put $x=5$ in
We get –
$\begin{aligned} & \Rightarrow {{5}^{2}}-9\times 5+20=0 \\\ & 25-45+20=0 \\\ \end{aligned}$
Simplifying we get
$0=0$
So $x=5$ satisfies the equation
Mean we have got the correct solution.