Question

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

Answer 1

Hey!

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x^2 + 7x + 10

=> x^2 + 5x + 2x + 10

=> (x^2 + 5x) + (2x + 10)

=> x (x + 5) + 2 (x + 5)

=> (x+2) (x+5)

Zeroes -

x + 2 = 0
x = -2

x+5 = 0
x = -5

Verifying the relationship -:

Sum of zeroes = - 2 + (-5) = -2 - 5 = -7 = Coefficient of x / coefficient of x^2

Product of zeroes = - 2 × - 5 = 10 = Constant / Coefficient of x^2


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Hope it helps...!!!

Answer 2

Hey there

We have zeroes of the Quadratic Polynomial

x² + 7x + 10 = 0

⇒ x² + 5x + 2x + 10 = 0

[ Here a = x² , b = 7x and c = 10 ]


⇒ x ( x + 5 ) + 2 ( x + 5 ) = 0

⇒ ( x + 2) ( x + 5 ) =0

when ,( x + 2 ) = 0

Then , x = -2

when , ( x + 5 ) =0

Then , x = -5


$so \: here \: \\ \alpha \: = \: \: - 2 \: and \: \: \: \: \beta \: \: = - 5 \\ \\ \\ verification \: > \\ \\ \\ sum \: of \: the \: zeroes \: = \: \frac{ - b}{a} \\ \\ = > \alpha + \beta = \frac{ - 7}{1} \\ \\ = > - 2 +( - 5 \: ) \: = - 7 \\ \\ = > - 7 = - 7 \\ \\ and \\ \\ \: product \: of \: the \: zeroes \: = \frac{c}{a} \\ \\ = > \alpha \times \beta \: \: \: \: \: \: \: \: \: = 10 \\ \\ = > - 2 \times - 5 = 10 \\ \\ = > 10 = 10$
Since LHS = RHS

Therefore it is verified


Hope this helps ya ☺