Question

How do you use a quadratic formula to solve $3{{x}^{2}}-2x+7=0$ ?

Answer 1

Hint: As per question, it is given a quadratic equation in the form of $a{{x}^{2}}+bx+c=0$. We have to solve the given equation in order to find the roots or zeros. Roots or zeroes are the values which satisfy the equation and result in zero after placing the roots. To solve this type of equation, we need to apply the formula in order to find the roots which are called quadratic formulas. Quadratic Formula is:
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Complete step by step answer:
Now, let’s solve the question.
We already know that the equation which has the highest degree as 2, and which is in the form of $a{{x}^{2}}+bx+c=0$ is called quadratic equation where ‘a’ and ‘b’ are coefficients and c is the constant term. Roots or zeros are those values which satisfy the equation and give result as zero on solving the equation. Common method of finding roots is middle term splitting but in question it is mentioned to solve by using quadratic formula. Quadratic formula is the formula which is specially used to find the roots.
So the quadratic formula is:
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Write the equation given in question.
$\Rightarrow 3{{x}^{2}}-2x+7=0$
Here, the value of a = 3, b = -2 and c = 7. Put all these values in the quadratic formula:
$\Rightarrow x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 3 \right)\left( 7 \right)}}{2\left( 3 \right)}$
Solve further:
$\Rightarrow x=\dfrac{2\pm \sqrt{4-84}}{6}$
Solve the under root now:
$\Rightarrow x=\dfrac{2\pm \sqrt{-80}}{6}$
Now just focus on $\sqrt{-80}$. -80 is made up of -1 and 80. When we break it into these two numbers, then see what we will get:
$\Rightarrow x=\dfrac{2\pm \sqrt{-1}.\sqrt{80}}{6}$
As we have already studied that $\sqrt{-1}=i$. So will replace it with iota and find the LCM of 80:
$\Rightarrow x=\dfrac{2\pm i.\sqrt{80}}{6}$
$\begin{align}
  & 2\left| \!{\underline {\,
  80 \,}} \right. \\
 & 2\left| \!{\underline {\,
  40 \,}} \right. \\
 & 2\left| \!{\underline {\,
  20 \,}} \right. \\
 & 2\left| \!{\underline {\,
  10 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
LCM of 80 is: $2\times 2\times 2\times 2\times 5$now, we have to break $\sqrt{80}$ into $\sqrt{16}$ and $\sqrt{5}$ because 16 has the root value as 4 but 5 does not have any root value in whole number. So place these values in the equation.
$\Rightarrow x=\dfrac{2\pm i.\sqrt{16}.\sqrt{5}}{6}$
Now place the root of 16:
$\Rightarrow x=\dfrac{2\pm i.4\sqrt{5}}{6}$
Let’s write the roots of the equation by reducing all the terms:
$\therefore x=\dfrac{1}{3}+\dfrac{2}{3}i\sqrt{5},\dfrac{1}{3}-\dfrac{2}{3}i\sqrt{5}$
So this is the final answer.

Note:
This question is not as complex as the solution looks like. The main point to be noted is, if you find any such number whose direct square root is not there, you don’t have to panic or leave the answer there itself, you will solve that root by taking LCM and at the end, reduce the terms as much as possible.