Question

Find the zeroes of the polynomial x²-2x-8
& verify the relationship between the zeroes
& coefficients​

Answer 1

${\huge{\bf{\red{\underline{Solution:}}}}}$

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

$\star \: \: \: {\sf{ {x}^{2} - 2x + 8}}$

${\bf{\blue{\underline{To\:Find:}}}}$

• Relationship between the zeros and Coeffiicient

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{ {x}^{2} - 2x - 8 }}$

$: \implies{\sf{ {x}^{2} - 4x + 2x - 8 }}$

$: \implies{\sf{ x(x - 4) + 2(x - 4) }}$

$: \implies{\sf{ (x - 4) ( x + 2) }}$

Take ,

◕ x- 4 = 0. and. x+2 =0

$: \implies{\sf \boxed{ x = 4}} \: \: \: \: {\sf \boxed{ x = - 2}}$

• So the values of x²-2x-8 is zero when x=4 and x=-2.
• Therefore,The zeros of x²-2x-8 are 4 and -2.

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Now,

$\dagger \: \: \: \boxed{\sf{ { \purple{Sum \: of \: zeros = \frac{ - Coeffiicient \: of \: x}{Coeffiicient \: of \: {x}^{2} } }}}}$

$: \implies{\sf{( 4) + ( - 2) = - \frac { - 2}{1} }}$

$: \implies{\sf{4-2= \frac { 2}{1} }}$

$: \implies{\sf{2= 2 }}$

Hence verified.

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$\dagger \: \: \: \boxed{\sf{ { \purple{Product \: of \: zeros = \frac{constant\: term }{coeffiicient \: of \: {x}^{2} } }}}}$

$: \implies{\sf{ (4) \times ( - 2) = \frac{ - 8}{1} }}$

$: \implies{\sf{ -8 = \frac{ - 8}{1} }}$

$: \implies{\sf{ -8=-8 }}$

Hence Verified.

Answer 2

S O L U T I O N :

We have p(x) = x² - 2x - 8

Zero of the polynomial p(x) = 0

So;

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

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

As we know that given polynomial compared with ax² + bx + c;

• a = 1
• b = -2
• c = -8

Now;

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

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

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

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

Thus;

Relationship between zeroes and coefficient is verified .