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# Find the zeros of the quadratic polynomial $\left( 6{{x}^{2}}-7x-3 \right)$.

Last updated date: 18th Mar 2023
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Hint: First of all, compare the given quadratic equation by general quadratic equation $a{{x}^{2}}+bx+c=0$ and get the values of a, b and c. Then find the zeros of the equation by using
$X=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\text{ or }X=\dfrac{-b\pm \sqrt{D}}{2a}$

Here we have to find the zeros of a quadratic polynomial $\left( 6{{x}^{2}}-7x-3 \right)$. Before proceeding with this question, we must know what the root of an equation means. First of all, roots or zeros of any equation are the values of the variable which satisfy the equation. In other words, if by substituting some value of the variable, the equation becomes zero, then that value of the variable is the root of the equation. For example, 5x + 10 = 0 has a root – 2 because by substituting x = – 2, this equation becomes 0.
For any general quadratic equation $a{{x}^{2}}+bx+c=0$ where $a,b,c\in R$ and $a\ne 0$, then, the solution or zeroes of the quadratic equation is given by:
$X=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
That means, $X=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\text{ }$
And $X=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\text{ }$
The expression ${{b}^{2}}-4ac=D$ is called the discriminant of the quadratic equation. So, we can also write zeroes as,
$X=\dfrac{-b+\sqrt{D}}{2a}\text{ }$
And, $X=\dfrac{-b-\sqrt{D}}{2a}\text{ }$
Now, let us consider the quadratic polynomial given in the question,
$Q\left( x \right)=6{{x}^{2}}-7x-3=0$

By comparing the above equation by general quadratic equation $a{{x}^{2}}+bx+c=0$, we get,
a = 6, b = – 7 and c = – 3
Let us find the discriminant of the above equation, we get,
$D={{b}^{2}}-4ac$
By substituting the values of a, b and c, we get,
\begin{align} & D={{\left( -7 \right)}^{2}}-4\left( 6 \right)\left( -3 \right) \\ & =49+72 \\ & D=121 \\ \end{align}
We know that the zeroes of the quadratic equation are:
$X=\dfrac{-b+\sqrt{D}}{2a}\text{ }$
And, $X=\dfrac{-b-\sqrt{D}}{2a}\text{ }$
By substituting the values of a, b and D, we get,
$X=\dfrac{-\left( -7 \right)+\sqrt{121}}{2\times \left( 6 \right)}$
And, $X=\dfrac{-\left( -7 \right)-\sqrt{121}}{2\times \left( 6 \right)}$
We know that $\sqrt{121}=11$, so we get,
$X=\dfrac{7+11}{12}=\dfrac{18}{12}=\dfrac{3}{2}$
And $X=\dfrac{7-11}{12}=\dfrac{-4}{12}=\dfrac{-1}{3}$
So, we get the roots of the equation $6{{x}^{2}}-7x-3=0$ as $\dfrac{3}{2}$ and $\dfrac{-1}{3}$.

Note:
Students can also find the zeroes of the quadratic equation $6{{x}^{2}}-7x-3$ as follows:
$6{{x}^{2}}-7x-3=0$
Here, we can write 7x = 9x – 2x, we get,
$6{{x}^{2}}-\left( 9x-2x \right)-3=0$
Or, $6{{x}^{2}}-9x+2x-3=0$
We can also write the above equation as,
$3x\left( 2x-3 \right)+\left( 2x-3 \right)=0$
By taking (2x – 3) common, we get,
$\left( 2x-3 \right)\left( 3x+1 \right)=0$
So, we get $\left( 2x-3 \right)=0$ and $\left( 3x+1 \right)=0$
$x=\dfrac{3}{2}\text{ and }x=\dfrac{-1}{3}$
So, we get the roots of the quadratic equation as $\dfrac{3}{2}$ and $\dfrac{-1}{3}$.