Question

The sum and the product of zeroes of a quadratic polynomial are -8 and 12 respectively. Find the zeroes of polynomial. *​

Answer 1

$\huge\tt{\red{\underline{Given:}}}$

★The sum of zeroes of a quadratic polynomial is -8.

★The product of zeroes of a quadratic polynomial is 12.

$\huge\tt{\red{\underline{To\:\:Find:}}}$

★The zeroes of the quadratic polynomial.

$\huge\tt{\red{\underline{Concept\:\:Used:}}}$

★We will use the formula for finding the quadratic polynomial and then find it's zeroes.

$\huge\tt{\red{\underline{Answer:}}}$

Let

• $\alpha\:as \:the\: first \:zero.$
• $\beta\:as\:the\:second\:zero.$

Now, by question,

$\longrightarrow \alpha+\beta=-8$

$\longrightarrow \alpha\beta=12$

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So, we can find the polynomial as

$\large\green{\boxed{p(x)= k[x^{2}-(\alpha+\beta)x+\alpha\beta]}}$

where

• k is constant
• alpha & beta are the zeroes
• p(x) is the polynomial.

$\implies p(x) k[x^{2}-(\alpha+\beta)x+\alpha\beta$

$\implies p(x) = k[x^{2}-(-8)x+(12)]$

$\implies p(x) =k[x^{2}+8x+12]$

${\underline{\boxed{\red{.°.p(x) =k[x^{2}+8x+12]}}}}$

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Now for finding zeroes equate it with 0.

$\implies k[x^{2}+8x+12]=0$

$\implies \cancel{k}[x^{2}+8x+12]=0$

$\implies x^{2}+8x+12=0$

$\implies x^{2}+6x+2x+12=0$

$\implies x(x+6) +2(x+6) =0$

$\implies (x+2) (x+6) =0$

${\underline{\boxed{\orange{.°. x= -2, -6}}}}$

Therefore the zeroes are -2 & -6.

Answer 2

Answer:

alpha+beta= -8 = -b/a

alpha beta = 12 = c/a

by the formula

ax^+bx+c =0

a=1;b=8;c=12

so the quadratic equation is 1x^+8+12=0

soving quadratic equations

1x^+8x+12=0(by middle term splitting method)

firstly multiply last term and first term that is 12×1x^=12x^

now when we factorise 12x^ by 6x ×2x we get 12x^ and when we add the multiples 6 and 2 we should get 8.So now the equation is 1x^+6x+2x+12=0

now take out common term

x( x+6)+ 2( x+6)=0

(x+2) (x+6)

x+2=0. x+6=0

x= -2 x= -6

So the zeores of polynomial is

-2. and -6