Question

How do you factor and solve $$x_{}^2+5x-24=0$$?

Answer 1

Hint: Factorisation is the process in which an expression is broken into factors which can be multiplied to get back the original expression. To factorise the given expression compare the given expression with the algebraic equation $$x_{}^2+(a+b)x-(ab)=0$$. Then find $$a$$ and $$b$$ which are the required factors.

Complete step by step solution:
We have the quadratic equation $$x_{}^2+5x-24=0$$
We will use the algebraic identity $$x_{}^2+(a+b)x-(ab)=0$$ to factorize
Now after comparing above two equations –
$(a+b)=5$
$ab=-24$
Now we need to find two numbers whose sum is 5 and product is -24
So $a=8$and $b=-3$
Now putting these values in the expression
$$x_{}^2+(8-3)x-24=0$$
$$x_{}^2+8x-3x-24=0$$
Now paring as 2 and factorize each pair
$$x(x+8)-3(x+8)=0$$
Now $(x+8)$is common factor so we can write as
$$(x+8)(x-3)=0$$
So factors are $$(x+8)(x-3)$$
Now to get the value of$x$,
$$(x+8)=0$$ and $$(x-3)=0$$
$x=-8$ and 3
So we get two solutions -8 and 3

Note: To factorize any quadratic equation we can also use the formula $$\left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)$$, if the equation can be expressed as a difference of squares of two terms. Also we can use the Sridharacharya formula - $$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$$ to solve any quadratic equation of the form $$a{x}^2+bx+c=0$$.