Question

# Find the zeroes of the following quadratic polynomials and verify the relationship Between the zeroes and the coefficients.$\left( {\text{i}} \right){x^2} - 2x - 8{\text{ }}\left( {{\text{ii}}} \right)4{s^2} - 4s + 1\left( {{\text{iii}}} \right)6{x^2} - 3 - 7x \\ \left( {{\text{iv}}} \right)4{u^2} + 8u\left( {\text{v}} \right){t^2} - 15\left( {{\text{vi}}} \right)3{x^2} - x - 4 \\$

Hint:-To find zeros of quadratic polynomial first you have to make it in factor form and then you can find itâ€™s zeroes and to verify the relationship between the zeroes and the coefficients use sum of zeroes is $\frac{{ - {\text{b}}}}{a}$and product of zeroes $\frac{{\text{c}}}{a}$.

$\left( {\text{i}} \right){x^2} - 2x - 8$
To convert in factor form we will write it as
${x^2} - 4x + 2x - 8 = 0 \\ x\left( {x - 4} \right) + 2\left( {x - 4} \right) = 0 \\ \left( {x - 4} \right)\left( {x + 2} \right) = 0 \\$
Now this equation is in itâ€™s factor form so,
$x = 4,x = - 2$
Now we have to verify the relationship between the zeroes and the coefficients.
Sum of zeroes is equal to $\frac{{ - b}}{a}$ on comparing with $a{x^2} + bx + c = 0$, we get $\left( {b = - 2,a = 1} \right)$
That means $\frac{{ - b}}{a} = 2$and sum of zeroes $\left( {4 + \left( { - 2} \right) = 2} \right)$
$\because$Both are equal, hence the relationship is verified.
Now, product of zeroes is equal to $\frac{c}{a}$$\left( {\because c = - 8,a = 1} \right)$
Product of zeroes is $\left( {4 \times - 2 = - 8} \right)$same as $\frac{{ - c}}{a} = - 8$, both are equal hence verified.
$\left( {{\text{ii}}} \right)4{s^2} - 4s + 1 = 0$
To convert it in factor form we will write it as
$4{s^2} - 2s - 2s + 1 = 0 \\ 2s\left( {2s - 1} \right) - 1\left( {2s - 1} \right) = 0 \\ \left( {2s - 1} \right)\left( {2s - 1} \right) = 0 \\$
Now this is a factor form so we can easily find $s$from here
$s = \frac{1}{2}$ here both zeroes are same that is $\frac{1}{2}$
Now we have to verify the relationship between zeroes and the coefficients.
Here $\frac{{ - b}}{a} = \frac{4}{4} = 1$and sum of zeroes is $\frac{1}{2} + \frac{1}{2} = 1$ both are equal hence verified.
Here $\frac{c}{a} = \frac{1}{4}$and product of zeroes is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$both are equal hence verified.
$\left( {{\text{iii}}} \right)6{x^2} - 3 - 7x = 0$
Now we have to convert it in factor form, so we will write it as
$6{x^2} - 9x + 2x - 3 = 0 \\ 3x\left( {2x - 3} \right) + 1\left( {2x - 3} \right) = 0 \\ \left( {2x - 3} \right)\left( {3x + 1} \right) = 0 \\$
So $x = \frac{3}{2},x = \frac{{ - 1}}{3}$
Now we have to verify the relationship between zeroes and coefficients.
Here $\frac{{ - b}}{a} = \frac{7}{6}$and sum of zeroes $\frac{3}{2} + \frac{{ - 1}}{3} = \frac{7}{6}$both are equal hence verified.
Here $\frac{c}{a} = \frac{{ - 1}}{2}$and product of zeroes $\frac{3}{2} \times \frac{{ - 1}}{3} = \frac{{ - 1}}{2}$both are equal hence verified.
$\left( {{\text{iv}}} \right)4{u^2} + 8u = 0$
Now we have to convert it in factor form
$4u\left( {u + 2} \right) = 0 \\ \therefore u = - 2,u = 0 \\$
Here $\frac{{ - b}}{a} = - 2$and sum of zeroes is $- 2 + 0 = - 2$both are the same hence verified.
Here $\frac{c}{a} = 0$and product of zeroes is $- 2 \times 0 = 0$both are the same hence verified.
$\left( v \right){t^2} - 15 = 0$
We have to convert it in factor form
${t^2} - {\left( {\sqrt {15} } \right)^2} = 0 \\ \left( {t - \sqrt {15} } \right)\left( {t + \sqrt {15} } \right) = 0 \\$
$t = \sqrt {15} ,t = \sqrt { - 15}$
Here $\frac{{ - b}}{a} = 0$ and sum of zeroes $\sqrt {15} - \sqrt {15} = 0$both are the same hence verified.
Here $\frac{c}{a} = - 15$and product of zeroes $\sqrt {15} \times \sqrt { - 15} = - 15$both are the same hence verified.
$\left( {{\text{vi}}} \right)3{x^2} - x - 4$
We will convert it in factor form
$3{x^2} - x - 4 = 0 \\ 3{x^2} - 4x + 3x - 4 = 0 \\ 3x\left( {x + 1} \right) - 4\left( {x + 1} \right) = 0 \\ \left( {x + 1} \right)\left( {3x - 4} \right) = 0 \\ x = - 1,x = \frac{4}{3} \\$
Here $\frac{{ - b}}{a} = \frac{1}{3}$and sum of zeroes $\frac{4}{3} - 1 = \frac{1}{3}$both are same hence verified.
Here $\frac{c}{a} = \frac{{ - 4}}{3}$and product of zeroes $\frac{4}{3} \times - 1 = \frac{{ - 4}}{3}$ both are same hence verified.

Note:-Whenever you get this type of question the key concept of solving is first you have to make a factor form using you basic mathematics and then using properties of quadratic equation you have to check sum of roots or zeros or product of roots with coefficients of polynomial.