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How do you solve ${{x}^{2}}-10x+24=0$

Last updated date: 23rd Feb 2024
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IVSAT 2024
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Hint: We are given ${{x}^{2}}-10x+24=0$ 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 given Equation and check whether they are same or not

Complete step by step answer:
Given is, ${{x}^{2}}-10+24=0$
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 an ‘n’ solution.
Now as it is a quadratic equation we will change it into standard form $a{{x}^{2}}+bx+c=0$
As we look closely our problem is already in standard form
Now we have to solve the equation ${{x}^{2}}-10x+24=0$
To solve this equation we first take the greatest common factor possibly available to the terms. As we can see that in ${{x}^{2}}-10x+24=0$
1,-10 and 24 has nothing in common so
So Equation remains same
Now to solve our problem we will use Middle term split method
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 of difference will be equal to the b
Now in ${{x}^{2}}-10x+24=0$
We have $a=1$ , $b=-10$ and $c=24$
So $a\times c=1\times 24$
We can see that $-6\times -4=24$ and
Their sum is $-6-4=-10$
So we use this to split
${{x}^{2}}+\left( -6-4 \right)x+24=0$
Opening brackets we get
Taking common in first two and last two terms
We get
$x\left( x-6 \right)-4\left( x-6 \right)=0$
Simplifying further
$\left( x-4 \right)\left( x-6 \right)=0$
Using zero product rule which says if two terms product is zero that either one of them is zero So either $x-4=0$ or $x-6=0$
So we get $x=4$ and $x=6$
Hence solution are $x=4$ and 6

We can cross check that our solution $x=4$ and $x=6$ is correct or not.
To do so we will find $x=6$ and $x=4$ one by one in ${{x}^{2}}-10x+24=0$ if they satisfy the equation then the solutions are correct.
Firstly putting $x=6$ in ${{x}^{2}}-10x+24=0$ we get-
  & {{6}^{2}}-10\times 6+24=0 \\
 & 36+24-60=0 \\
 & 60-60=0 \\
 & 0=0 \\
So $x=6$ satisfy the equation.
Similarly putting $x=4$ , we get –
  & {{4}^{2}}-4\times 10+24=0 \\
 & 16+24-40=0 \\
 & 40-40=0 \\
 & 0=0 \\
So $x=4$ also satisfies the equation.
Hence, we got the correct solution.