Question

How do you solve $6{x^2} - 12x + 1 = 0$ using the quadratic formula?

Answer 1

Hint: Here we will solve the given quadratic equation by using quadratic formula
Solving quadratic equations involves the use of the following formula
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $a,b$ and $c$ are taken from the quadratic equation written in its general form of $a{x^2} + bx + c = 0$.

Complete step-by-step solution:
Given quadratic equation is $6{x^2} - 12x + 1 = 0$
Comparing this equation with the general form of the quadratic equation that is $a{x^2} + bx + c = 0$ where a is the numeral that goes in front of ${x^2}$ ,$b$ is the numeral that goes in front of $x$ , and $c$ is the numeral with no variable next to it.
 We get,
$a = 6$ ,$b = - 12$ and $c = 1$
We have the quadratic formula that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substitute the values of $a,b$ and $c$ in this formula we get,
$\Rightarrow$\[x = \dfrac{{ - \left( { - 12} \right) \pm \sqrt {{{\left( { - 12} \right)}^2} - 4\left( 6 \right)\left( 1 \right)} }}{{2\left( 6 \right)}}\]
Simplifying numerator we get.
$\Rightarrow$$x = \dfrac{{12 \pm \sqrt {144 - 24} }}{{12}}$
Subtracting the term which is in the square root we get,
$\Rightarrow$$x = \dfrac{{12 \pm \sqrt {120} }}{{12}}$
Now here we need perfect square term so we can split the term $120 = 4 \times 30$,
$\Rightarrow$$x = \dfrac{{12 \pm \sqrt {4 \times 30} }}{{12}}$
Take the square root value of $4$ we get,
$\Rightarrow$$x = \dfrac{{12 \pm 2\sqrt {30} }}{{12}}$
Take the common term and Divide we get,
$\Rightarrow$$x = 1 \pm \dfrac{{\sqrt {30} }}{6}$
Therefore the quadratic equation has two solution that is \[x = \left\{ {1 + \dfrac{{\sqrt {30} }}{6},1 - \dfrac{{\sqrt {30} }}{6}} \right\}\]

Because the discriminant ${b^2} - 4ac$ is positive, we get two different real roots.

Note: Many quadratic equations cannot be solved by factoring. This is generally true when the roots, or answers, are not rational numbers. A second method of solving quadratic equations involves the use of the quadratic formula. When using the quadratic formula, you should be aware of three possibilities. These three possibilities are distinguished by a part of the formula called the discriminant. The discriminant is the value of under the radical sign, ${b^2} - 4ac$.
 A quadratic equation with real numbers as coefficients can have the following:
$1)$ Two different real roots if the discriminant ${b^2} - 4ac$ is a positive number
$2)$ One real root if the discriminant ${b^2} - 4ac$ is equal to $0$
$3)$ No real root if the discriminant ${b^2} - 4ac$ is a negative number.