Question

Find a
quadretic polynomial the
sum and praduct of whose
are e 2 and 3/2
vrespectively. Also find their Zeroes
​

Answer 1

Answer:

Step-by-step explanation:

Let the zeroes of the polynomial be α and β

Now, we know that,

a quadratic equation has,

x² + (sum of zeroes)x + (product of zeroes)

Thus the quadratic equation is

x² + 2x + 3/2

Now,

α + β = 2

αβ = 3/2

Answer 2

Answer:

1) x² - 2x + 3/2

2)  (2+√-2)/2 , (2-√-2)/2

Step-by-step explanation:

Given ,

Sum of the zeroes of the polynomial = 2

Product of the zeroes of the polynomial = 3/2

To Find :-

1) The Quadratic Polynomial

2) Zeroes of the polynomial

How To Do :-

As they gave the sum and product of the zeroes we need to find the quadratic polynomial by using the formula in terms of sum and product of the zeroes. After obtaining the quadratic polynomial we need to find the zeroes by applying  the quadratic formula.

Formula Required :-

Quadratic polynomial :-

x² - (sum of zeroes of the polynomial)x + (Product of zeroes of the polynomial)

Quadratic formula :-

If 'ax² + bx + c'  is the general form of quadratic polynomial then :-

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Solution :-

Finding the quadratic polynomial :-

x² - (2)x + (3/2)

x² - 2x + 3/2

∴ 'x² - 2x + 3/2 ' is the required quadratic polynomial .

Finding the zeroes of the quadratic polynomial :-

Comparing 'x² - 2x + 3/2' with the general form of the quadratic polynomial 'ax² + bx + c' :-

→ a = 1 , b = - 2 , c = 3/2

Substituting in the formula :-

$x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)\left(\dfrac{3}{2}\right)}}{2(1)}$

$=\dfrac{2\pm\sqrt{4-2(3)}}{2}$

$=\dfrac{2\pm\sqrt{4-6}}{2}$

$=\dfrac{2\pm\sqrt{-2}}{2}$

$=\dfrac{2+\sqrt{-2}}{2},\dfrac{2-\sqrt{-2}}{2}$

∴ Zeroes of the polynomial =  (2+√-2)/2 , (2-√-2)/2