Question

Find the zeroes of the quadratic polynomial x2 + 7x +10 and verify the relation between the zeroes

and coefficients​

Answer 1

Given Polynomial: x² + 7x + 10.

We've to find out the zeroes of the given Quadratic polynomial x² + 7x + 10. & also, verify the relationship b/w it's zeroes and coefficients.

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$\underline{\bigstar\:{\pmb{\sf{By\; Splitting\;the\;Middle\;term\; Method :}}}}$

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$\dashrightarrow\sf x^2 + 7x + 10 = 0 \\\\\\\dashrightarrow\sf x^2 + 2x + 5x + 10 = 0 \\\\\\\dashrightarrow\sf x(x + 2)+ 5(x + 2) = 0\\\\\\\dashrightarrow\sf (x + 2) (x + 5) = 0\\\\\\\dashrightarrow\underline{\boxed{\pmb{\frak{\red{x = -2\;or\;-5}}}}}\;\bigstar$

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Hence, the zeroes of the given polynomial are, α = – 2 & β = – 5 respectively.

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⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀

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

• a = 1
• b = 7
• c = 10

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¤ Let's verify, the relationship b/w the zeroes and the coefficients —

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⠀⠀${\qquad\maltese\:\:\textsf{Sum of Zeroes :}} \\\\\twoheadrightarrow\sf \alpha + \beta = \dfrac{-b}{\;a} \\\\\\\twoheadrightarrow\sf -2 + (-5) = \dfrac{- 7}{1} \\\\\\\twoheadrightarrow\sf - 2 - 5 = - 7\\\\\\\twoheadrightarrow{\pmb{\sf{-7 = -7}}}$

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⠀⠀${\qquad\maltese\:\:\textsf{Product of Zeroes :}} \\\\\twoheadrightarrow\sf \alpha \;\beta = \dfrac{c}{a}\\\\\\\twoheadrightarrow\sf -2 \times (-5) = \dfrac{10}{1}\\\\\\\twoheadrightarrow{\pmb{\sf{10 = 10}}}$

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

Answer 2

Given :-

x² + 7x + 10

To Find :-

Zeroes

Solution :-

At first spiltting the middle term

$\sf x^2+7x+10$

$\sf x^2+(5x+2x)+10$

$\sf x^{2} +5x+2x+10$

$\sf x(x + 5) + 2 (x + 5) = 0$

$\sf (x + 5) (x + 2) = 0$

Either

x + 5 = 0

x = 0 - 5

x = -5

or

x + 2 = 0

x = 0 - 2

x = -2

Now

Sum of zeroes

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

$\sf -2+(-5) = \dfrac{-7}1$

$\sf -2-5=\dfrac{-7}{1}$

$\sf -7=-7$

Product of zeroes

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

$(-2)(-5)=\dfrac{10}{1}$

$\sf 10 = 10$