Question

Find the zeros of \[\text{6}{{\text{x}}^{\text{2}}}\text{-3-7x=0}\]
(a) \[\dfrac{3}{2},\,\dfrac{1}{3}\]
(b) \[-\dfrac{3}{2},-\,\dfrac{1}{3}\]
(c) \[-\dfrac{3}{2},\,\dfrac{1}{3}\]
(d) \[\dfrac{3}{2},\,-\dfrac{1}{3}\]

Answer 1

Hint: We will be using the concept of quadratic equation to tackle this question. Formula used for solving quadratic equation of the form \[\text{f(x)=a}{{\text{x}}^{\text{2}}}\text{+bx+c}\] is \[\text{( }\!\!\alpha\!\!\text{ , }\!\!\beta\!\!\text{ )}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step-by-step answer:
Before proceeding with the question, we should know about quadratic equations. Quadratic equations are the polynomial equations of degree 2 in one variable of type \[\text{f(x)=a}{{\text{x}}^{\text{2}}}\text{+bx+c}\] where a, b, c, \[\in R\] and \[\text{a}\ne \text{0}\]. It is the general form of a quadratic equation where a is called the leading coefficient and c is called the absolute term. The values of x satisfying the quadratic equation are the roots of the quadratic equation \[\text{( }\!\!\alpha\!\!\text{ , }\!\!\beta\!\!\text{ )}\].
The quadratic equation will always have two roots(zeros). The nature of roots may be either real or imaginary.
Formula for solving quadratic equation of the form \[\text{f(x)=a}{{\text{x}}^{\text{2}}}\text{+bx+c}\] is \[\text{( }\!\!\alpha\!\!\text{ , }\!\!\beta\!\!\text{ )}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.....(1)\]
And the given equation is \[\text{6}{{\text{x}}^{\text{2}}}\text{-3-7x=0}........(2)\].
Now we can write equation (2) as
\[\text{6}{{\text{x}}^{\text{2}}}\text{-7x-3=0}.......\text{(3)}\]
Now comparing equation (3) with \[\text{f(x)=a}{{\text{x}}^{\text{2}}}\text{+bx+c}\], we get, \[\text{a=6}\], \[\text{b= -7}\] and \[\text{c =}\,-3\].
Now substituting the values of a, b and c in equation (1) we get,
 \[\text{y}=\dfrac{7\pm \sqrt{{{(-7)}^{2}}-4\times 6\times (-3)}}{2\times 6}.....(4)\]
Now simplifying and rearranging equation (4) we get,
 \[\text{y}=\dfrac{7\pm \sqrt{121}}{12}.....(5)\]
Solving for y in equation (5) we get,
 \[\text{y}=\dfrac{7\pm 11}{12}\]
We get two values of y which are \[{{\text{y}}_{1}}=\dfrac{7+11}{12}=\dfrac{18}{12}=\dfrac{3}{2}\] and \[{{\text{y}}_{2}}=\dfrac{7-11}{12}=\dfrac{-4}{12}=-\dfrac{1}{3}\].
Hence we get \[\dfrac{3}{2},\,-\dfrac{1}{3}\] as the answer. So the correct answer is option (d)

Note: Remembering the formula for solving the quadratic equation is the key here. An alternate method for solving quadratic equations is by factoring the equation.
\[\text{6}{{\text{x}}^{\text{2}}}\text{-7x-3=0}........\text{(1)}\]
Simplifying equation (1) and then solving for x we get,
 \[\begin{align}
  & \Rightarrow \,6{{x}^{\text{2}}}\text{-7x-3=0} \\
 & \Rightarrow \text{6}{{\text{x}}^{\text{2}}}\text{-9x+2x-3=0} \\
 & \Rightarrow 3x\text{(2x-3)+1(2x-3)=0} \\
 & \Rightarrow \text{(2x-3) (3x+1)=0} \\
 & \Rightarrow x\text{=}\,\dfrac{3}{2},\,-\dfrac{1}{3} \\
\end{align}\]