Question

find quaderatic polynomial when sum and product of its zeroes are 6 and 8?​

Answer 1

$\orange{\bold{\underline{\underline{Given:-}}}}$

• Sum of zeroes of quadratic polynomial = 6
• Product of zeroes of quadratic polynomial = 8

$\orange{\bold{\underline{\underline{To \: find:-}}}}$

• Quadratic Polynomial

$\large \underline{ \red{ \boxed{ \bf \red{Solution:-}}}}$

$\sf{Suppose \: \: \: \alpha \: \: \: and \: \: \: \beta \: is \: the \: zeroes \: of \: the \: Polynomial}$

• Sum of zeroes = $\sf{\alpha \: + \: \beta}$
• Product of zeroes = $\sf{\alpha \: \beta}$

Formula Used :

$\tt \underline{ \blue{ \boxed{ \bf \green{{x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0}}}}$

Putting the values, We Get the Polynomial

$\sf\purple{{x}^{2} - 6x + 8 = 0}$

$\rm\pink{Verification:–}$

$\sf{{x}^{2} - 4x - 2x + 8 = 0}$

$\sf{x(x - 4) - 2(x - 4) = 0}$

$\sf{(x - 4)(x - 2) = 0}$

$\sf\purple{x = 4 \: and \: 2}$

• Sum of zeroes = 4 + 2 = 6
• Product of zeroes = 4 × 2 = 8

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Answer 2

ᎯᏁᏕᏯᎬᏒ

Let the zeros of the polynomial be α & β.

α+β = 6

αβ = 8

We know that a polynomial can be written as:-

x² - (α+β)x + αβ = 0

Therefore, the polynomial is:-

x² - 6x + 8 = 0