Question

Find the zeroes of the following quadratic polynomials,
(i)${x^2} - 2x - 8$
(ii)$6{x^2} - 3 - 7x$
Also, verify the relationship between the zeroes and the coefficients.

Answer 1

Hint: In this problem, we have to find the value of zeroes of the quadratic polynomial, here we will find the zeros by factoring and the standard form of the quadratic equation is $a{x^2} + bx + c$ and after finding the zeros, we need to verify the relationship between the zeroes and the coefficients.

Formula Used:
sum of the zeroes$ = \dfrac{ negative \;coefficient\; of \;x }{coefficient \; of x^2}$
Product of zeroes$ = \dfrac{ constant\; term}{coefficient\; of \;x^2}$

Complete step by step answer:
The first quadratic equation given is ${x^2} - 2x - 8$. To factor this, we need to multiply the value of a and c which results into $1 \times 8 = 8$ now, we have to find the numbers which when multiplied gives $8$ and on subtraction/addition it gives the value of b i.e, $2$ , now $4$ and $2$ are those numbers which on multiplying gives $8$ and on subtracting it gives $2$ which is b. Now we can write the equation ${x^2} - 2x - 8$as,
$
   \Rightarrow {x^2} - \left( {4 - 2} \right)x - 8 \\
   \Rightarrow {x^2} - 4x + 2x - 8 \\
 $
Now, we will take $x$ common from the first two terms and $2$ from the last two terms.
$ \Rightarrow x(x - 4) + 2\left( {x - 4} \right)$
On further solving, we get,
$ \Rightarrow \left( {x + 2} \right)\left( {x - 4} \right)$
The value o f${x^2} - 2x - 8$ becomes zero when$x + 2 = 0$ and $x - 4 = 0$ from this the values of zeroes are $ - 2,4$.
Now, the sum of the zeroes$ = 4 + \left( { - 2} \right) = 2$
And from the above formula of sum of zeroes we will find the value,
sum of the zeroes$ = \dfrac{ negative \;coefficient\; of \;x }{coefficient \; of x^2}$
sum of the zeroes$ = $$\dfrac{{ - \left( { - 2} \right)}}{1} = 2$
Now, the product of zeroes$ = 4 \times \left( { - 2} \right) = - 8$
And from the above formula of product of zeroes we will find the value,
Product of zeroes$ = \dfrac{ constant\; term}{coefficient\; of \;x^2}$
Product of zeroes$ = \dfrac{{ - 8}}{1} = - 8$
Hence, the relationship between the zeroes and the coefficients is verified.
Now, the second quadratic equation given is$6{x^2} - 3 - 7x$, firstly, we will rearrange the equation and after rearranging it becomes$6{x^2} - 7x - 3$. To factor this, we need to multiply the value of a and c which results into $6 \times 3 = 18$ now, we have to find the numbers which when multiplied gives $18$ and on subtraction/addition it gives the value of b i.e, $7$ , now $9$ and $2$ are those numbers which on multiplying gives $18$ and on subtracting it gives $7$ which is b. Now we can write the equation $6{x^2} - 7x - 3$ as,
$
   \Rightarrow 6{x^2} - \left( {9 - 2} \right)x - 3 \\
   \Rightarrow 6{x^2} - 9x + 2x - 3 \\
 $
Now, we will take $3x$ common from the first two terms and $1$ from the last two terms.
$ \Rightarrow 3x(2x - 3) + 1\left( {2x - 3} \right)$
On further solving, we get,
$ \Rightarrow \left( {3x + 1} \right)\left( {2x - 3} \right)$
The value of $6{x^2} - 7x - 3$becomes zero when,$3x + 1 = 0$ and $2x - 3 = 0$ from this the values of zeroes are $ - \dfrac{1}{3},\dfrac{3}{2}$.
Now, the sum of the zeroes$ = \dfrac{{ - 1}}{3} + \dfrac{3}{2} = \dfrac{7}{6}$
And from the above formula of sum of zeroes we will find the value,
sum of the zeroes$ = \dfrac{ negative \;coefficient\; of \;x }{coefficient \; of x^2}$
sum of the zeroes$ = $$\dfrac{{ - \left( { - 7} \right)}}{6} = \dfrac{7}{6}$
Now, the product of zeroes$ = \dfrac{{ - 1}}{3} \times \dfrac{3}{2} = \dfrac{{ - 1}}{2}$
And from the above formula of product of zeroes we will find the value,
Product of zeroes$ = \dfrac{ constant\; term}{coefficient\; of \;x^2}$
Product of zeroes$ = \dfrac{{ - 3}}{6} = \dfrac{{ - 1}}{2}$
Hence, the relationship between the zeroes and the coefficients is verified.

Note: A polynomial of degree $2$ is called a quadratic polynomial and an equation which has a quadratic polynomial is called a quadratic equation. Here, we have used the method of factoring to solve the quadratic equation and we have also verified that the sum of the zeroes and product of zeroes found by solving the quadratic equation and by the formula we have used are the same.