Question

Find the quadratic polynomial, for the zeroes alpha, beta given in each case 1) 2,-1 2.√3, -√3 3.1/4, -1 4.1/2, 3/2​

Answer 1

Answer:

$1. \: x {}^{2} - x - 2 = 0$

$2. \: x {}^{2} - 3 = 0$

$3. \: 4x {}^{2} + 3x - 1 = 0$

$4. \: 4x {}^{2} - 8x + 3 = 0$

hope this helps u

Answer 2

Concept

If α , β are zeroes of a quadratic polynomial, the polynomial is given by

$(x - \alpha )(x - \beta )$

$or \: {x}^{2} - ( \alpha + \beta)x + \alpha \beta$

Solutions

1) (α , β) = (2, -1)

The required polynomial is

${x}^{2} - (2 - 1)x + 2 \times ( - 1)$

$= {x}^{2} - x - 2$

2) (α , β) = (√3, -√3)

The required polynomial is

${x}^{2} - ( \sqrt{3} - \sqrt{3} )x + \sqrt{3} \times ( - \sqrt{3} )$

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

3) (α , β) = (1/4, -1)

The required polynomial is

${x}^{2} - ( \frac{1}{4} - 1)x + \frac{1}{4} .( - 1)$

$= {x}^{2} + \frac{3}{4} x - \frac{1}{4}$

4) (α , β) = (1/2, 3/2)

The required polynomial is

${x}^{2} - ( \frac{1}{2} + \frac{3}{2} )x + \frac{1}{2} . \frac{3}{2}$

$= {x}^{2} - 2x + \frac{3}{4}$