Question

How do you solve $3{x^2} + 6x = 8$ ?

Answer 1

Hint: In this problem we have a quadratic equation with some unknown $x$. And we are asked to solve the given quadratic equation. We can solve the given quadratic equation by factoring and then we can find the value of $x$. Suppose we cannot factorize the given equation then we have to use the quadratic formula to solve the given equation.

Formula used: Quadratic formula: $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , where $a,b,c$ are constants and $a \ne 0$ and $x$ is the unknown.

Complete step by step answer:
Given quadratic equation is $3{x^2} + 6x = 8$
To solve this given quadratic equation we are going to use a quadratic formula. A quadratic formula can be used only when the given quadratic equation is equal to zero.
So, we can rewrite the given equation as $3{x^2} + 6x - 8 = 0$
General form of a quadratic equation is $a{x^2} + bx + c = 0$ , now compare this with the given equation.
Then $a = 3,b = 6,c = - 8$ , apply these values in the quadratic formula.
We get, $x = \dfrac{{ - 6 \pm \sqrt {36 + 96} }}{6}$ , simplifying this we get
$x = - 1 \pm \dfrac{{\sqrt {132} }}{6}$ , taking square root in the numerator we get
$x = - 1 \pm \dfrac{{2\sqrt {33} }}{6}$, cancelling numbers we get
$x = - 1 \pm \dfrac{{\sqrt {33} }}{3}$

There are two values of $x$ . They are $x = - 1 + \dfrac{{\sqrt {33} }}{3}$ or $x = - 1 - \dfrac{{\sqrt {33} }}{3}$ . This is a required solution.

Additional information:
Quadratic equations are actually used in everyday life, as when calculating areas, determining a product’s profit or formulating the speed of an object. Quadratic equation refers to equations with at least one squared variable.

Note: We must know why we are using a particular method to solve and not the other one. To solve this problem, first we have to make the equation equal to zero. By using the factorization method we can easily solve the quadratic equation. But there is not a pair of numbers that multiply to give us $ - 8$ that will also add up to $6$ . So, we cannot find the factors for the given equation. So, we use quadratic formulas to solve.