Question

find the Quadratic polynomial,whose zeroes are -3 and 4​

Answer 1

Quadratic polynomial = x² - 1x - 12 = 0

Given :-

• The quadratic polynomial, whose zeroes are -3 & 4 .

To Find :-

• The quadratic polynomial.

Solution :-

Let,

• The zeroes of quadratic polynomial be α and β .

.°. α = -3 and β = 4 .

=> Sum of zeroes = α + β

=> Sum of zeroes = -3 + 4

=> Sum of zeroes = 1 .

______________

=> Product of zeroes = αβ

=> Product of zeroes = -3×4

=> Product of zeroes = -12

______________

As we know that,

Quadratic polynomial = x²-(sum of zeroes)x + (Product of zeroes) = 0

=> Quadratic polynomial = x²-(1) x+(-12)= 0

=> Quadratic polynomial = x² - 1x -12 = 0 .

_______________

Hence,

• The quadratic polynomial is x²-1x - 12 = 0.

Answer 2

Given:

Zeroes are (-3) and (4)

To be found:

The quadratic formula whose zeroes are (-3) and (4)

So,

The formula for forming a quadratic polynomial if zeroes are given.

$\boxed{\bf \implies k(x^{2}- (sum\ of\ zeroes)x + (Product\ of\ zeroes))}$

Where k is a real number .

So,

Method 1

• Find the zeroes separately and put in the formula to get the required polynomial.

Sum of zeroes

= (-3)+(4) = 1

And

Product of zeroes

= (-3)(4) = (-12)

Now,

The required polynomial is

$\bf \implies k(x^{2}- (1)x + (-12))$

$\implies \bf k(x^{2} - 1x - 12)$

(Here k = 1)

∴ Required polynomial = x² - x - 12

- - -

Method 2

• Put the zeroes in the formula to get the answer directly.

$\boxed{\bf \implies k(x^{2}- (sum\ of\ zeroes)x + (Product\ of\ zeroes))}$

$\bf \implies k(x^{2}- (-3 + 4)x + (-3 \times 4))$

$\bf \implies k(x^{2}- (1)x + (-12))$

$\bf \implies k(x^{2}- x -12)$

(Here k = 1)

∴ x² - x - 12 is the required quadratic polynomial.