Question

2.
Find the sum of the zeroes of quadratic polynomial
x2 + 7x + 10.​

Answer 1

$\star\;{\underline{\underline{\sf{\pink{Solution:-}}}}}$

$\rule{200}{3}$

$\implies$ x² + 7x + 10 = 0

$\implies$ x² + 2x + 5x + 10 = 0

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

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

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

Either (x + 2) or (x + 5) is equals to 0,

$\implies$ (x + 2) = 0 ; (x + 5) = 0

Therefore, x = -2 , -5

We know that,

★ Sum of Zeroes ( $\sf \alpha + \beta$ ) = $\sf \dfrac{-b}{a}$

$\implies$ ( $\sf \alpha + \beta$ ) = -2 + (-5) = -7

$\implies$ $\sf \dfrac{-b}{a}$ = $\sf \dfrac{-7}{1}$

$\;\;\;\;{\underline{\underline{\sf{\purple{\dag\;Hence\; Solved!!}}}}}$

$\rule{200}{3}$

Answer 2

$\sf{\huge{\underline{\boxed{\pink{\mathfrak{ Question}}}}}}$

Find the sum of the zeroes of quadratic polynomial

$x^2 + 7x + 10.$

$\sf{\huge{\underline{\boxed{\pink{\mathfrak{ solutionn}}}}}}$

$\implies p(x) = x^2 + 7x + 10$

• zeros of the polynomial is tge value of x where p(x) = 0
• putting p(x) = 0

$\implies x^2 + 7x + 10 = 0$

• we can find the roots using middle term split method .

$\implies \sf x^2 + 5x + 2x + 10 = 0$

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

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

so x = -2 , -5

Therefore ;

α = -2 , β = -5 are the zeros of the Polynomial .

• comparing eq with $ax^2 + bx + c$

⠀$\begin{tabular}{|c|c|}\cline{1-2}\sf a &\sf 1 \\\cline{1-2}\sf b &\sf 7 \\\cline{1-2}\sf c &\sf 10\\\cline{1-2}\end{tabular}$

sum of zeros = $\sf\dfrac{-b}{a}$

α + β = $\sf\dfrac{-7}{1}$

-2 + (- 5) = - 7

- 7 = - 7

$\sf{\bold{\underline{\boxed{\red{sum\: of \: zeros = \: -7 }}}}}$

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