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Question: Find the roots of the quadratic equation \(3{{x}^{2}}-5x-12=0\) by using the quadratic formula. A....

Find the roots of the quadratic equation 3x2−5x−12=03{{x}^{2}}-5x-12=0 by using the quadratic formula.
A. −23,3-\dfrac{2}{3},3
B. −43,−3\dfrac{-4}{3},-3
C. −−43,−3-\dfrac{-4}{3},-3
D. −43,3\dfrac{-4}{3},3

Explanation

Solution

For solving this question you should know about the quadratic equations and to find the roots of them. This b2−4ac{{b}^{2}}-4ac is a part of the quadratic formula which is denoted at place of discriminant. Find this discriminant here denoted as ‘D’. If we have to find the roots of any quadratic equation, then we use this formula. And if b2−4ac≥0{{b}^{2}}-4ac\ge 0, then the quadratic equation has real roots.

Complete step-by-step solution:
According to the question we have to find the roots of the quadratic equation 3x2−5x−12=03{{x}^{2}}-5x-12=0, by quadratic formula. So, as we know that the quadratic equations of the format ax2+bx+ca{{x}^{2}}+bx+c always contains roots, which can be imaginary or real roots. But in many equations, we cannot determine the roots easily. So, then we use the formula for finding the roots. If the equation is in ax2+bx+ca{{x}^{2}}+bx+c form, then,
Quadratic formula : −b±b2−4ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
But discriminant D=b2−4acD={{b}^{2}}-4ac
So, quadratic formula : −b±D2a\dfrac{-b\pm \sqrt{D}}{2a}
And here we get the two roots as:
−b+b2−4ac2a\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} and −b−b2−4ac2a\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}
If we take an example to understand it clearly then:
Example 1: Find the roots of x2−4x+6=0{{x}^{2}}-4x+6=0.
The discriminant is used to determine how many different solutions and what type of solutions will a quadratic equation have. So, here according to the question,
a=1,b=−4,c=6 ⇒D=(−4)2−4(1)(6) =16−24 =−8 \begin{aligned} & a=1,b=-4,c=6 \\\ & \Rightarrow D={{\left( -4 \right)}^{2}}-4\left( 1 \right)\left( 6 \right) \\\ & =16-24 \\\ & =-8 \\\ \end{aligned}
This indicates that this equation contains 2 imaginary solutions (with ii, the square root of -1). If the answer had been positive (assume 9), then the equation would have 2 real solutions (with real numbers; they might not be rational solutions, but they are real). And if the answer was 0, then the equation would have 1 solution (it would be square root of something).
As our question is,
3x2−5x−12=0 x=5±(−5)2−(4)(3)(−12)2(3) x=5±25+1446 x=5±136 \begin{aligned} & 3{{x}^{2}}-5x-12=0 \\\ & x=\dfrac{5\pm \sqrt{{{\left( -5 \right)}^{2}}-\left( 4 \right)\left( 3 \right)\left( -12 \right)}}{2\left( 3 \right)} \\\ & x=\dfrac{5\pm \sqrt{25+144}}{6} \\\ & x=\dfrac{5\pm 13}{6} \\\ \end{aligned}
So, if we take them as:
x=5+136 ⇒x=186 ⇒x=3 \begin{aligned} & x=\dfrac{5+13}{6} \\\ &\Rightarrow x=\dfrac{18}{6} \\\ &\Rightarrow x=3 \\\ \end{aligned}
And,
x=5−136 ⇒x=−86 ⇒x=−43 \begin{aligned} & x=\dfrac{5-13}{6} \\\ &\Rightarrow x=\dfrac{-8}{6} \\\ &\Rightarrow x=\dfrac{-4}{3} \\\ \end{aligned}
So, the roots are 3,−433,\dfrac{-4}{3}.
Hence the correct option is D.

Note: If we want to calculate the roots of any equation, then always try to reduce that to a form of ax2+bx+ca{{x}^{2}}+bx+c because it will be very easy to determine the roots by this. And it gives us exact roots, so follow this method always. Here for solving the Quadratic equations we can also use the factorisation method and completing the square method as well.