Question

8.
Find the zeroes of the quadratic polynomial y square - 8y - 20, and verify the
relationship between the zeroes and co-efficients.​

Answer 1

Given Polynomial: y² – 8y – 20.

We've to find out the zeroes of the given Quadratic polynomial and verify the relationship b/w the zeroes & Coefficients.

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• S P L I T T I N G⠀T H E⠀M I D D L E⠀T E R M :

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$:\implies\sf y^2 - 8y - 20\\\\\\:\implies\sf y^2 - 10y + 2y - 20= 0\\\\\\:\implies\sf y(y - 10) +2(y - 10) = 0\\\\\\:\implies\sf (y - 10) (y + 2) = 0\\\\\\:\implies\underline{\boxed{\pmb{\frak{\red{y = 10\;\&\;-2}}}}}\;\bigstar$

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∴ Therefore, the zeroes of the polynomial are α = 10 and β = – 2 respectively.

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☆ On Comparing the given polynomial with the (ax² + bx + c = 0) —

• a = 1
• b = – 8
• c = – 20

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★ V E R I F I C A T I O N :

» For any Quadratic polynomial the sum and product of the roots are Given by :

• (α + β) = –b/a
• (αβ) = c/a

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${\qquad\maltese\:\:\bf{Sum \;of\; Zeroes :}} \\\\\twoheadrightarrow\sf \alpha + \beta = \dfrac{-b}{\;a} \\\\\\\twoheadrightarrow\sf 10 + (-2) = \dfrac{8}{1} \\\\\\\twoheadrightarrow{\pmb{\sf{8 = 8}}}$

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${\qquad\maltese\:\:\bf{Product \;of \;Zeroes :}} \\\\\twoheadrightarrow\sf \alpha \;\beta = \dfrac{c}{a}\\\\\\\twoheadrightarrow\sf 10 (-2) = \dfrac{-20}{1} \\\\\\\twoheadrightarrow{\pmb{\sf{-20 = -20}}}$

⠀⠀⠀⠀⠀⠀$\qquad\therefore{\underline{\textsf{\textbf{Hence, Verified!}}}}$⠀⠀

Answer 2

Given :-

y² - 8y - 20

To Find :-

Zeroes and co-efficients.​

Solution :-

y² - 8y - 20

Spillting middle term

y² - (10y - 2y) - 20 = 0

y² - 10y + 2y - 20 = 0

y(y - 10) + 2(y - 10) = 0

(y - 10)(y + 2) = 0

Either

y - 10 = 0

y = 10

Or,

y + 2 = 0

y = -2

Now,

Verification

α + β = -b/a

10 + (-2) = -(-8)/1

10 - 2 = 8/1

8 = 8

αβ = c/a

10 × (-2) = -20/1

-20 = -20