Question

How do you solve $3{{x}^{2}}+5x=2$ by using the quadratic formula?

Answer 1

Hint: In this question, we are given a quadratic equation and we need to find the values of x which satisfy this equation using the quadratic formula. For this, we will first compare the given equation with $a{{x}^{2}}+bx+c=0$ to get the values of a, b and c. After that, we will use the quadratic formula, according to which $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This will give us two values of x which is our required solution.

Complete step by step answer:
Here we are given an equation of degree 2 (quadratic equation) as $3{{x}^{2}}+5x=2$. We need to apply a quadratic formula to it to find the value of x which satisfies this equation. For this, let us understand the quadratic formula.
According to the quadratic formula, if an equation is of the form $a{{x}^{2}}+bx+c=0$ then its roots (values of x) can be calculated using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
So here let us first rearrange the given equation in the form of $a{{x}^{2}}+bx+c=0$ we have $3{{x}^{2}}+5x=2$.
Taking 2 to the other side we have $3{{x}^{2}}+5x-2=0$.
Comparing the equation with $a{{x}^{2}}+bx+c=0$ we get a = 3, b = 5 and -c = 2.
We have found the values of a, b and c. Let us put them in the quadratic formula we get $x=\dfrac{-5\pm \sqrt{{{\left( 5 \right)}^{2}}-4\left( 3 \right)\left( -2 \right)}}{2\left( 3 \right)}$.
We know ${{\left( 5 \right)}^{2}}=5\times 5=25$ so we have $x=\dfrac{-5\pm \sqrt{25+24}}{6}\Rightarrow x=\dfrac{-5\pm \sqrt{49}}{6}$.
We know that ${{\left( 7 \right)}^{2}}=7\times 7=49$ so we can write $\sqrt{49}$ as 7. We get $x=\dfrac{-5\pm 7}{6}$.
The two values of x become $x=\dfrac{-5+7}{6}\text{ and }x=\dfrac{-5-7}{6}$.
Simplifying the numerator we get $x=\dfrac{2}{6}\text{ and }x=\dfrac{-12}{6}$.
Cancelling common factors from the numerator and denominator in both values we get $x=\dfrac{1}{3}\text{ and }x=-2$.
Therefore, the required values of x are $\dfrac{1}{3}$ and -2.

Note:
Students can check their answers by putting both the values back in the original equation to verify equality. Students should take care of signs while comparing, they often make mistakes of c as 2 only rather than -2. Keep in mind the formula for the general equation for solving this sum.