Question

Find the zeroes of quadrtaical polynomial And verify the relationship between the zeroes and coefficients of x2 - 2x - 8​

Answer 1

Given :-

Quadratic polynomial :- x² - 2x - 8

Required to find :-

• Zeroes of the quadratic polynomial

• Verify the relationship between the zeroes and the coefficients

Solution :-

Given information :-

Quadratic polynomial :- x² - 2x - 8

we need to find the zeroes of the polynomial and verify the relationship between the zeroes and the coefficients.

Consider the given polynomial ;

p ( x ) = x² - 2x - 8

we need to Factorise this polynomial and equal the factors to zero in order to find the zeroes .

➦ x² - 2x - 8

➦ x² - 4x + 2x - 8

➦ x ( x - 4 ) + 2 ( x - 4 )

➦ ( x - 4 ) ( x + 2 )

The factors of p ( x ) are ( x - 4 ) , ( x + 2 )

___________________________

This implies ;

=> x - 4 = 0

=> x = 4

Thus ,

α = 4 ( alpha = 4 )

Similarly,

=> x + 2 = 0

=> x = - 2

Thus ,

β = - 2 ( beta = - 2 )

( here the zeroes are represented with alpha & beta )

Now we need to verify the relationship between the zeroes zeroes and the coefficients

So,

The relationship between the sum of the zeroes and the coefficients is ;

$\tt{ \green {\bf{ \alpha + \beta = \dfrac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }}}$

Sum of the zeroes ;

α + β =

=> 4 + ( - 2 )

=> 4 - 2

=> 2

α + β = 2

But,

$\rm \dfrac{ - \: coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$\tt \dfrac{ - ( - 2)}{1} = \dfrac{2}{1} \implies2$

Hence,

$\: \tt{ \boxed{\bf{ \alpha + \beta = \dfrac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }}}$

Similarly,

The relationship between the product of the zeroes and the coefficients is ;

$\tt{ \pink{ \bf{ \alpha \beta = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} } }}}$

product of the zeroes

α β =

=> 4 x - 2

=> - 8

But,

$\tt{ \dfrac{constant \: term}{coefficient \: of \: {x}^{2}}}$

$\tt \rightarrowtail \dfrac{ - 8}{1} \\ \tt \implies - 8$

Hence,

$\tt{ \boxed{ \bf{ \alpha \beta = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} } }}}$

Therefore,

The relationship between the zeroes and the coefficients is verified .

Answer 2

p{x}=x²-2x-8

x²-2x-8=0

x²-2x-8=0x²-4x+2x-8=0

x²-2x-8=0x²-4x+2x-8=0x{x-4}+2{x-4}=0

x²-2x-8=0x²-4x+2x-8=0x{x-4}+2{x-4}=0{x+2}{x-4}=0

So, x=-2or+4

so, α=-2,β=4

α+β=-{coefficient of x}/{coefficient of x²}

α+β=-{coefficient of x}/{coefficient of x²}

-2+4=-{-2}/1

so, 2=2.

αβ=constant/coefficient of x²

αβ=constant/coefficient of x²

-2×4=-8/1

So, -8=-8.