Question

1. Find the zerves of the following quadratic polynomials and verify the relationship between
the zeroes and the coefficients.
1)X^2-2X-8

Answer 1

Step-by-step explanation:

x² - 2x - 8

x² - 4x + 2x - 8 = 0

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

( x + 2) ( x - 4)

x = -2,4

Therefore, Alpha = -2 and Beta = 4

Sum of zeroes

LHS

Alpha + Beta = -b/a

-2 + 4

LHS = 2

RHS

-b/a

-(-2)/1

RHS = 2

LHS = RHS

Now, product of zeroes

Alpha*Beta = c/a

LHS

-2(4) = -8

LHS = -8

RHS = c/a

= -8/1

RHS = -8

LHS = RHS

Hence Verifed

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Answer 2

Solution :

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

The quadratic polynomial is x² - 2x - 8

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

The zeroes of the polynomial & verify the relationship between zeroes and coefficient.

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

We have p(x) = x² - 2x - 8

Zero of the polynomial is p(x) = 0

So;

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

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

Now;

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

• a = 1
• b = -2
• c = -8

$\star\:\underline{\large{\orange{\mathcal{SUM\:OF\:THE\:ZEROES\::}}}}$

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

$\star\:\underline{\large{\orange{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}}}$

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

Thus;

Relationship between zeroes and coefficient is verified .