Question

Find the zeroes of quadratic polynomial ${x}^{2} + 7x + 10$and find relationship between zeroes and co efficients.​

Answer 1

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$\longrightarrow$$\sf{x}^{2} + 7x + 10$

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$\sf{by\:splitting\:the\:middle\:term}$

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

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$\longrightarrow$$\sf{x(x+5)+2(x+5)}$

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$\longrightarrow$$\sf{(x+2)(x+5)}$

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$\sf{x+2=0}$

$\sf{x=-2}$

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$\sf{x+5=0}$

$\sf{x=-5}$

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$\sf{Sum\:of\:zeroes=-2+(-5)}$ ⠀⠀⠀$\sf\frac{ - (coeff.of \: x)}{coeff.of {x}^{2} }$

$\sf{-2-5}$ =⠀$\sf\frac{-7}{1}$

$\sf{-7}$

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━━━━━━━━━━━━━━━

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$\sf{Product\:of\:zeroes=(-2) ×(-5)=}$

$\sf\frac{constant \: term}{coeff. \: of \: {x}^{2} }$

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$\longrightarrow$$\sf\Large{10=10}$

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Answer 2

Step-by-step explanation:

x

2

−7x=−10

⇒x

2

−7x+10=0

⇒x

2

−2x−5x+10=0

⇒x(x−2)−5(x−2)=0

⇒(x−2)(x−5)=0

∴x=2,5

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