Question

find the zeros of quadratic polynomial x square - 3 x minus 10 and verify the relationship between zeros and coefficients

Answer 1

quadratic polynomial=
${x}^{2} - 3x - 10$

$\alpha + \beta = - b \div a$
$\alpha + \beta = - ( - 3) \div 1$
$\alpha + \beta = 3$
NOW,
$\alpha \beta = c \div a$
$\alpha \beta = - 10 \div 1$
$\alpha \beta = - 10$
hence verified

Answer 2

Given:

A quadratic polynomial x² - 3x - 10.

To find:

The zeroes of a given quadratic polynomial. Also, verify the relationship between zeroes and coefficients.

Solution:

We will find the zeroes of a given polynomial by middle splitting term. So,

As given, we have,

${x}^{2} - 3x - 10$

This can be written as

${x}^{2} - 5x + 2x - 10$

After grouping the terms and taking factors common, we get

$x(x - 5) + 2(x - 5)$

$(x + 2)(x - 5) = 0$

$x = 5 \: and \: - 2$

Hence, two zeroes of the given quadratic polynomial are 5 and -2.

Now,

We know that in a quadratic polynomial ax² + bx + c

Sum of zeroes = -b/a

Product of zeroes = c/a

In the given polynomial x² - 3x - 10

a = 1, b = -3 and c = -10

So,

Sum of zeroes

$= 5 + ( - 2)$

$= 3$

and

$\frac{ - b}{a} = \frac{ - ( - 3)}{1} = 3$

Also,

Product of zeroes

$= 5 \times ( - 2)$

$= - 10$

And

$\frac{c}{a} = \frac{ - 10}{1} = - 10$

Hence, the relationship between zeroes and coefficients is verified.