Question

find a quadratic polynomial the sum and product of whose zeroes are √2 and -3/2 respectively. find the zeroes also​

Answer 1

Given

$\sf \to \: sum \: of \: zeroes \: = \sqrt{2}$

$\sf \to \: product \: of \: zeroes \: = \dfrac{ - 3}{2}$

To Find

$\sf \to \: quadratic \: \: polynomial \: \: and \: zeroes$

We have

$\sf \to( \alpha + \beta ) = \sqrt{2}$

$\sf \to( \alpha \beta ) = \dfrac{ - 3}{2}$

General Equation of polynomial

$\to \sf {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0$

Put the value

$\sf \to \: {x}^{2} - \sqrt{2} x - \dfrac{3}{2} = 0$

Now Simplify the equation

$\sf \to 2 {x}^{2} - 2\sqrt{2} x - 3 = 0$

Now Find the zeroes

$\sf \to \: 2 {x}^{2} - 3 \sqrt{2} x + \sqrt{2} x - 3 = 0$

$\sf \to \: \sqrt{2} x( \sqrt{2} x - 3) + 1( \sqrt{2} x - 3) = 0$

$\sf \to \: ( \sqrt{2} x + 1)( \sqrt{2} x - 3) = 0$

$\sf \to \: \sqrt{2} x + 1 = 0 \: and \: \sqrt{2} x - 3 = 0$

$\sf\to \: x = \dfrac{ - 1}{ \sqrt{2} } \: \: and \: x = \dfrac{3}{ \sqrt{2} }$

answer

$\sf\to \: x = \dfrac{ - 1}{ \sqrt{2} } \: \: and \: x = \dfrac{3}{ \sqrt{2} }$

Answer 2

Answer:

Given :-

Sum of zeroes = √2

Product of zeroes = -3find a quadratic polynomial the sum and product of whose zeroes are √2 and -3/2 respectively. find the zeroes also/2

To Find :-

• Quardtic polynomial
• Zeroes

Solution :-

We know that

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

Here,

$\sf \: \alpha + \beta = \sqrt{2}$

$\sf \: \alpha \beta = \dfrac{c}{a}$

$\sf \: \alpha \beta = \dfrac{ - 3}{2}$

Now,

By putting the value

$\tt \: {x}^{2} - ( \sqrt{2} )x + \dfrac{ - 3}{2}$

$\tt \: {x}^{2} - \sqrt{2x } + \dfrac{ - 3}{2}$

$\sf \: 2 {x}^{2} -\sqrt{2} x- 3$

Now

Finding zeroes

2x² - 3√2x + √2x - 3 = 0

√2x(√2x - 3) + 1(√2x - 3) = 0

Now,

By taking √2x as common

(√2x + 1) (√2x - 3)

So,

Either

x = -1/√2

or,

x = 3/√2