Question

Find the zeros of the quadratic polynomial $6{x^2} - 3 - 7x$ and verify the relationship between the zeros and the coefficient of the polynomial.

Answer 1

Hint: In this question a quadratic polynomial is given and we have to find the zeros (roots) of the polynomial or the point where the value of $x$ and the coefficients of the polynomial and then verify the relationship between the roots and the coefficients of the polynomial by calculating the value of the sum and product of the roots of the polynomial.

Complete step-by-step answer:
Given:
Let us assume the $f\left( x \right)$ is the given polynomial then –
$f\left( x \right) = 6{x^2} - 7x - 3$
Now the roots or the zeros of the polynomial are obtained when the value of the polynomial is equal to the zero.
So,
$
f\left( x \right) = 0\\
6{x^2} - 7x - 3 = 0
$
The first part of the polynomial is $6{x^2}$ and its coefficient is $6$ .
The middle part of the polynomial is $ - 7x$ and its coefficient is $ - 7$ .
The last part of the polynomial is the constant, and its value is $ - 3$ .
We can find the roots of the polynomial by using the Grouping Method. In this method first we split the middle part of the polynomial in two parts. To split the middle part first find the product of the first part and the constant of the polynomial. Then we have to split the middle part in such a way that the sum of the two parts should be equal to the middle part and the product should be equal to the product of the first part and the constant part of the polynomial.
Then,
$6 \times - 3 = - 18$
So, the sum of the two numbers should be equal to $ - 7$ and the product should be equal to $ - 18$ . the numbers are –
$
 - 9 \times 2 = - 18\\
 - 9 + 2 = - 7
$
So, we can split the polynomial in the following way –
$
6{x^2} - 7x - 3 = 0\\
6{x^2} - 9x + 2x - 3 = 0\\
3x\left( {2x - 3} \right) + 1\left( {2x - 3} \right) = 0\\
\left( {3x + 1} \right)\left( {2x - 3} \right) = 0
$
So, the two roots of the polynomial are –
$
\left( {3x + 1} \right) = 0\\
3x = - 1\\
x = \dfrac{{ - 1}}{3}
$
And,
$
\left( {2x - 3} \right) = 0\\
2x = 3\\
x = \dfrac{3}{2}
$
So, the two roots of the polynomial are $\alpha = \dfrac{{ - 1}}{3}$ and $\beta = \dfrac{3}{2}$ .
Also, comparing the given polynomial with the standard quadratic equation $a{x^2} + bx + c = 0$ we have,
$
a = 6\\
b = - 7\\
c = - 3
$
Now to verify the relationship we have,
$
\Rightarrow {\text{Sum of the roots}} = - \dfrac{{{\text{coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}\\
\Rightarrow \alpha + \beta = - \dfrac{b}{a}
$
Substituting the values, we get,
$
\dfrac{{ - 1}}{3} + \dfrac{3}{2} = - \dfrac{{\left( { - 7} \right)}}{6}\\
\dfrac{{ - 2 + 9}}{6} = \dfrac{7}{6}\\
\dfrac{7}{6} = \dfrac{7}{6}\\
L.H.S. = R.H.S
$
Hence, the relationship is verified for the sum of the roots.
Similarly,
$
\Rightarrow {\text{Product of the roots}} = \dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}\\
\alpha \times \beta = \dfrac{c}{a}
$
Substituting the values, we get,
$
\Rightarrow \dfrac{{ - 1}}{3} \times \dfrac{3}{2} = \dfrac{{ - 3}}{6}\\
\dfrac{{ - 1}}{2} = \dfrac{{ - 1}}{2}\\
L.H.S. = R.H.S
$
Hence, the relationship is also verified for the product of the roots.
Therefore, the relationship between the roots (zeros) and the coefficients of the polynomial is verified.

Note: The two formulas used to verify the relationship between the roots and the coefficients of the polynomial is only valid for the quadratic equations because only a quadratic equation has two roots possible. Also, we can use these formulas to calculate the two roots of the polynomial.