Question

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (i) 1/4 , -1 (ii) √2 , 1/3 (iii) 0, √5 (iv) 1,1 (v) -1/4 ,1/4 (vi) 4,1

Answer 1

Answer:

Step-by-step explanation:

1) 1/4 , -1

Given

Alpha + beta =1/4

Alpha ×beta =1

x^2 -(aplha +beta)x + aplha×beta

Putting constant term as K

K(x^2 -(aplha +beta)x + aplha×beta)

Now putting the values

K( x^2-1/4x + (-1)=0

So we get ,

K ( x^2-1/4x -1)=0

Or

By dividing 4 we get

K (4x^2-x-4)=0

Answer 2

We need to recall the following concept for a quadratic equation.

• If $\alpha$ , $\beta$ are the two zeros of a quadratic equation, then the general form of an equation is $x^{2} -(\alpha +\beta )x+\alpha \beta =0$ .

Given:

i) Sum of zeros $=\frac{1}{4}$ , Product of zeros $=-1$

Using the general formula for a quadratic polynomial, we get

The quadratic polynomial is,

$x^{2} - \frac{1}{4} x-1 =0$

$4x^{2} -x-4=0$

ii) Sum of zeros $=\sqrt{2}$ , Product of zeros $=\frac{1}{3}$

Using the general formula for a quadratic polynomial, we get

The quadratic polynomial is,

$x^{2} -\sqrt{2}x +\frac{1}{3} =0$

$3x^{2} -3\sqrt{2}x +1 =0$

iii) Sum of zeros $=0$ , Product of zeros $=\sqrt{5}$

Using the general formula for a quadratic polynomial, we get

The quadratic polynomial is,

$x^{2} -(0)x +\sqrt{5} =0$

$x^{2} +\sqrt{5} =0$

iv) Sum of zeros $=1$ , Product of zeros $=1$

Using the general formula for a quadratic polynomial, we get

The quadratic polynomial is,

$x^{2} -x +1 =0$

v) Sum of zeros $=\frac{-1}{4}$ , Product of zeros $=\frac{1}{4}$

Using the general formula for a quadratic polynomial, we get

The quadratic polynomial is,

$x^{2} -(\frac{-1}{4} )x +\frac{1}{4} =0$

$4x^{2} +x +1 =0$

vi) Sum of zeros $=4$ , Product of zeros $=1$

Using the general formula for a quadratic polynomial, we get

The quadratic polynomial is,

$x^{2} -4x +1 =0$