Question

Find the zeroes of the following quadratic polynomials and verify the relationship between

the zeroes and the coefficients.

x2
– 2x – 8​

Answer 1

Answer:

hey mate here is your answer....

Step-by-step explanation:

x²-2x-8=x²-4x+2x-8

= x(x-4) +2(x-4)

= (x-4) (x+2)

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Answer 2

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

The quadratic polynomial are x² - 2x - 8

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The zeroes and the verify relationship between zeroes and coefficient.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We have p(x) = x² - 2x - 8

Zero of the polynomial is p(x) = 0

So;

$\leadsto\sf{x^{2} -2x-8=0}\\\\\leadsto\sf{x^{2} +2x-4x-8=0}\\\\\leadsto\sf{x(x+2)-4(x+2)=0}\\\\\leadsto\sf{(x+2)(x-4)=0}\\\\\leadsto\sf{x+2=0\:\:Or\:\:x-4=0}\\\\\leadsto\sf{\green{x=-2\:\:Or\:\:x=4}}$

∴ The α = -2 and β = 4 are the zeroes of the polynomial.

As the given quadratic polynomial as we compared with ax² + bx + c = 0

• a = 1

• b = -2

• c = -8

So;

$\underline{\underline{\bf{Sum\:of\:the\:zeroes\::}}}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:(x)^{2} }{Coefficient\:of\:(x)} }\\\\\\\mapsto\sf{-2+4=\dfrac{-(-2)}{1} }\\\\\\\mapsto\sf{\orange{2=2}}$

$\underline{\underline{\bf{Product\:of\:the\:zeroes\::}}}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:(x)} }\\\\\\\mapsto\sf{-2\times 4=\dfrac{-8}{1} }\\\\\\\mapsto\sf{\orange{-8=-8}}$

Thus;

Relationship between zeroes and coefficient is verified .