Question

How do you factor and solve ${{x}^{2}}-14x+42=0$?

Answer 1

Hint: We are given an expression which we have to factorize and solve. We will first find the discriminant which is $D={{b}^{2}}-4ac$ and we get the value greater than 0. So, the equation has real roots. Then we will substitute the known values of a, b and c in the quadratic formula and we will get two values of $x$ for the given equation.

Complete step by step solution:
According to the given question, we have been given a quadratic equation. We will first check if we can factor it, and we see that we can’t do the direct factorization. So, we will use the discriminant.
From the equation we have, that is,
${{x}^{2}}-14x+42=0$------(1)
We get the value of $a=1,b=-14,c=42$,
Discriminant of an equation is denoted by D, and we have,
$D={{b}^{2}}-4ac$
Substituting the value in the discriminant, we get,
$\Rightarrow D={{(-14)}^{2}}-4(1)(42)$
$\Rightarrow D=196-168$
$\Rightarrow D=28>0$
Therefore, the given equation has two real roots. We will find the roots by using quadratic formula and substituting in it the value of discriminant.
We have,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting in the quadratic formula the values known to us, we get,
$\Rightarrow x=\dfrac{-(-14)\pm \sqrt{28}}{2(1)}$
$\Rightarrow x=\dfrac{14\pm \sqrt{28}}{2}$
$\Rightarrow x=\dfrac{14\pm 2\sqrt{7}}{2}$
$\Rightarrow x=\dfrac{2(7)\pm 2\sqrt{7}}{2}$
Dividing the numerator and denominator by 2, we get,
$x=7+\sqrt{7},7-\sqrt{7}$
Therefore, we have the value of $x=7+\sqrt{7},7-\sqrt{7}$.

Note:
The discriminant of an equation can tell us about the type of zeroes as well as the number of zeroes that the polynomials will have, that is,
We know that $D={{b}^{2}}-4ac$
If the $D > 0$, then the equation has two real roots.
If the $D=0$, then the equation has one real root,
And if the $D < 0$, then the equation has two imaginary roots (roots involving iota ($i$)).