Question

Find a quadratic polynomial,the sum and product of whose zeroes are -7 and 10 respectively.

Answer 1

It is given that :

• Sum of zeroes = - 7
• Product of zeroes = 10

We know that,

Quadratic polynomial :

$\sf = K [ x^2 - ( sum\:of\:zeroes )x + ( Product\:of\:zeroes)]$

$\sf = k [ x^2 - ( - 7 )x + ( 10 ) ]$

$\sf = k [ x^2 + 7x + 10]$

Here, K = 1

Hence,

Quadratic polynomial : $\sf k [x^2 + 7x + 10]$

$\rule{200}2$

Substitute the given values.

We get,

Answer 2

The required quadratic equation is x²+7x+(10) = 0.

Given:

The sum and product of zeroes of a quadratic equation are -7 and 10 respectively.

To Find:

The quadratic equation whose sum and product of zeroes of a quadratic equation are -7 and 10 respectively.

Solution:

A quadratic equation is a polynomial of degree 2 and has the standard form ax2+bx+c = 0, where a and b are the coefficients of x²

And x respectively, and c is the constant term (a≠0).

The solution of a quadratic equation is called its roots or zeros.

We are given that the sum and product of zeroes of a quadratic equation are -7 and 10 respectively.

Now, in a quadratic equation x²+bx+c = 0:

The sum of its roots = –b/a = -7

The product of its roots = c/a = 10.

Now,

x²+bx+c = 0

x²+(b/a)x+(c/a) = 0   …………………on dividing ‘a’ to both sides.

x²-(-b/a)x+(c/a) = 0

x²-(the sum of zeros)x+(the product of zeros) = 0

Substituting the given values, we have:

x²-(-7)x+(10) = 0

x²+7x+(10) = 0

Hence, our required quadratic equation is x²+7x+(10) = 0.

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