Question

find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficient
xsquare-5x​

Answer 1

Answer:

Correct Question :-

• Find the zeroes of the following quadratic equation polynomial x² - 5 and verify the relationship between the zeroes and co-efficient.

Given :-

• x² - 5

To Find :-

• What is the zeroes of the quadratic equation.
• Verify the relationship between the zeroes and co-efficient.

Formula Used :-

$\clubsuit$ Sum of roots :

$\longmapsto \: \sf\boxed{\bold{\pink{Sum\: of\: roots\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}$

$\clubsuit$ Product of roots :

$\longmapsto \sf \boxed{\bold{\pink{Product\: of\: roots\: (\alpha\beta) =\: \dfrac{c}{a}}}}$

Solution :-

Given :

$\dashrightarrow \sf\bold{\green{x^2 - 5}}$

$\implies \sf p(x) =\: x^2 - 5$

$\implies \sf x^2 - 5 =\: 0$

$\implies \sf x^2 =\: 5$

$\implies \sf x =\: \sqrt{5}$

$\implies \sf\bold{\red{x =\: \sqrt{5}}}$

And,

$\implies \sf\bold{\red{x =\: - \sqrt{5}}}$

$\therefore$ The zeroes of quadratic polynomial is √5 and - √5.

Hence,

• α = √5
• β = - √5

$\rule{150}{2}$

$\bigstar$ Verify the relationship between the zeroes and co-efficient :

Given equation :

$\dashrightarrow \sf x^2 - 5$

where,

• a = 1
• b = 0
• c = - 5

$\leadsto \sf\bold{Sum\: of\: roots\: :-}$

$\implies \sf \sqrt{5} + (- \sqrt{5}) =\: \dfrac{- 0}{1}$

$\implies \sf {\cancel{\sqrt{5}}} - {\cancel{\sqrt{5}}} =\: 0$

$\implies \sf\bold{\red{0 =\: 0}}$

$\leadsto \sf\bold{Product\: of\: roots}$

$\implies \sf \sqrt{5} \times (- \sqrt{5}) =\: \dfrac{- 5}{1}$

$\implies \sf - \sqrt{25} =\: - 5$

$\implies \sf\bold{\red{- 5 =\: - 5}}$

Hence, Verified.

Answer 2

✯✯ QUESTION ✯✯

Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and coefficients of x²-9..

━━━━━━━━━━━━━━━━━━━━

✰✰ ANSWER ✰✰

$\longrightarrow{{x}^{2}-9}$

Using Identity : -

$\longrightarrow{{a}^{2}-{b}^{2}=(a+b)(a-b)}$

$\longrightarrow{({x}^{2}+(-9)({x}^{2}-(-9)}$

$\longrightarrow{({x}^{2}-9)({x}^{2}+9)}$

Now ,

$\longrightarrow{{x}^{2}-9=0}$

$\longrightarrow{{x}^{2}=\sqrt{9}}$

$\red\longmapsto\:\large\underline{\boxed{\bf\green{x}\orange{=}\purple{3}}}$

$\longrightarrow{{x}^{2}+9=0}$

$\longrightarrow{{x}^{2}=\sqrt[-]{9}}$

$\red\longmapsto\:\large\underline{\boxed{\bf\orange{x}\green{=}\blue{-3}}}$

So , 3 and -3 are the zeroes of polynomials x²-9..

Now ,

➥Sum of Zeroes : -

$\longrightarrow{\alpha+\beta=\dfrac{-b}{a}}$

$\longrightarrow{3+(-3)=\dfrac{-(-0)}{1}}⟶3+(−3)$

$\longrightarrow{0=0}$

➥Product of Zeroes : -

$\longrightarrow{\alpha\beta=\dfrac{c}{a}}$

$\longrightarrow{3\times{-3}=\dfrac{-9}{1}}$

$\longrightarrow{-9=-9}$

$\pink\longmapsto\:\large\underline{\boxed{\bf\purple{L.H.S}\green{=}\red{R.H.S}}}$