Question

find the quadric polynomial whose zeroes are 2 and -6.and verify the relationship between the zeroes of the polynomial​

Answer 1

Answer:-

Your answer is $\bf x^2+4x-12.$

Explanation:-

Given:-

• Zeroes of a polynomial are 2 and -6.

To Find:-

• The Polynomial.

• Also to verify the relationship between the zeroes of the polynomial.

Concept Used:-

A quadratic polynomial is always in the form,

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

Where,

• $\rm\alpha\:\:And\:\:\beta$ are zeroe of the polynomial.

Solution.

So Here,

$\\ \tt\mapsto \alpha = 2.$

$\\ \tt\mapsto \beta = - 6.$

Therefore,

$\\ \tt\mapsto \alpha + \beta = 2 + ( - 6).$

$\\ \bf\mapsto \boxed{ \bf \alpha + \beta = - 4.}$

And,

$\\ \tt\mapsto \alpha \beta = 2 \times ( - 6).$

$\\ \bf\mapsto \boxed{ \bf \alpha \beta = - 12}.$

So The Polynomial will be,

$\\ \tt : \implies p(x) = {x}^{2} - ( \alpha + \beta )x + \alpha \beta .$

$\\ \tt : \implies p(x) = {x}^{2} - ( - 4)x + ( - 12).$

$\\ \large\bf : \implies \boxed{ \bf p(x) = {x}^{2} + 4x - 12.}$

$\bf\therefore$ The required polynomial is $\bf x^2+4x-12.$

_________________________

VerificaTion:-

We know that,

$\\ \tt\mapsto \alpha + \beta = - \frac{b}{a} \\ \\ \rm and \\ \\ \tt\mapsto \alpha \beta = \frac{c}{a} \: \: \: \: \: \: \: \: \: \: \: \:$

Where,

• a = Coefficient of x².
• b = Coefficient of x.
• c = Constant term.

So Here,

• a = 1.
• b = 4.
• c = -12.

Sum Of Zeroes,

$\\ \tt\mapsto \alpha + \beta = - 4.$

Also

$\\ \tt\mapsto - \frac{b}{a} = - 4.$

Hence verified....

Product Of Zeroes,

$\\ \tt\mapsto \alpha \beta = - 12.$

Also

$\\ \tt\mapsto \frac{c}{a} = - 12.$

Hence Verified....

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Answer 2

• $\large\boxed{\sf{\purple{Polynomial = x^2 + 4x - 12}}}$

Explanation :

$\underline{\underline{\bf{Given \::-}}}$

• Zeroes of quadratic polynomial = 2 and -6

$\underline{\underline{\bf{To\:Find \::-}}}$

• Quadratic polynomial = ?
• Also verify the relationship between the zeroes and coefficients.

$\underline{\underline{\bf{Solution \::-}}}$

• Let zeroes of polynomial = α and β
• So, α = 2 and β = -6
• Sum of zeroes = α + β = 2 + (-6) = 2 - 6 = -4
• Product of zeroes = α × β = 2 × (-6) = -12

★ We know that :-

$\qquad\leadsto\quad\sf p(x) = x^2 - (\alpha + \beta)x + \alpha\beta$

$\qquad\leadsto\quad\sf p(x) = x^2 - (-4)x + (-12)$

$\qquad\leadsto\quad{\boxed{\frak{\green{p(x) = x^2 + 4x - 12}}}}\:\red{\bigstar}$

$\quad\red{\therefore}\:{\underline{\sf{Polynomial = \bf{x^2 + 4x - 12}\:\sf{respectively.}}}}$

★ Verifying relation b/w zeroes and coefficients :-

• Sum of zeroes = -b/a
• Product of zeroes = c/a

$\qquad\leadsto\quad\small\sf \alpha + \beta = \dfrac{-b}{a}\:,\:\alpha\beta = \dfrac{c}{a}$

★ Values that we have :-

• α + β = -4
• αβ = -12
• a = coefficient of x² = 1
• b = coefficient of x = 4
• c = constant term = -12

★ Putting all known values :-

$\qquad\leadsto\quad\small\sf -4 = \dfrac{-4}{1}\:,\:-12 = \dfrac{-12}{1}$

$\qquad\leadsto\quad\small\sf -4 = -4\:,\:-12 = -12$

$\qquad\leadsto\quad\small\sf \cancel{-}4 = \cancel{-}4\:,\:\cancel{-}12 = \cancel{-}12$

$\qquad\leadsto\quad\small\bf{\red{4} = \red{4}\:,\:\red{12} = \red{12}}$

$\qquad\qquad\underline{\underline{\bf{Hence,\:Verified!}}}$

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

$\underline{\underline{\bf{More\:to\:know \::-}}}$

ㅤㅤ✰ Sridhara Acharya's formula ✰

$\qquad\blue\bigstar\:{\underline{\boxed{\bf{\orange{x = \dfrac{-b \:\pm\:\sqrt{b^2 - 4ac}}{2a}}}}}}$

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