# 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 \\

$

Answer

Verified

365.1k+ views

Hint:-To find zeros of quadratic polynomial first you have to make it in factor form and then you can find it’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 it’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.

$\left( {\text{i}} \right){x^2} - 2x - 8$

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 it’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.

Last updated date: 29th Sep 2023

•

Total views: 365.1k

•

Views today: 7.65k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE