Question

Q.4 Find the zeros of the given polynomial and verify the relationship between the zeros and their coefficients. qquad P(x)=x^(2)+4sqrt(3)x-15.(2marks)​

Answer 1

P(x)=x²+4✓3x-15.

P(x)=(x-✓3)(x+5✓3)

P(x)=0

•°• x=✓3,x= -5✓3

The Zeroes are ✓3, -5✓3

Sum of Zeroes=✓3+(-5✓3) =-6.9

by the definition

Sum of Zeroes= -coefficient of x/coefficient of x²

= -4✓3 =-6.9

Product of Zeroes=✓3 x -5✓3 = -15

by the definition

Product of Zeroes=constant term/coefficient of x²

=-15

Hence Verified

Answer 2

$\bf \LARGE\color{pink}Hola!$

GiveN :

$\sf \mapsto \: p(x) = {x}^{2} + 4 \sqrt{3} \: x \: - 15$

To FinD :

$\mapsto \sf \: zeros \: \: of \: \: the \: \: given \: polynomial$

$\sf \mapsto \: Relation \: \: between \: \: coefficients \: and\: Zeros$

SolutioN :

$\sf \mapsto \: p(x) = {x}^{2} + 4 \sqrt{3} x - 15$

$\sf \: \: \: \: For \: \: it \: \: to \: \: be \: \: zero$

$\tt \star \: \: \: \underline {Zeros} :$

$\sf \implies \: {x}^{2} + 4 \sqrt{3} x - 15 = 0$

$\: \: \: \: \: \: \: \: \: \: \: \tt {\tt { \underline{ \boxed{ \tt{ \: x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a} }}}}} \: \: \: \: \bf[ Formula ]$

$\implies \sf \: x = \frac{ - 4 \sqrt{3} ± \sqrt{48 + 60} }{2}$

$\implies \sf \: x = - 2\sqrt{3} ± \sqrt{27}$

$\: \: \: \: \: \: \: \therefore { \underline{ \boxed{ \sf \: x = (- 2\sqrt{3} + \sqrt{27}) \: \: \: \: \: \: \: \: \: \:; \: \: \: \: \: \: \: \: \: \: \: \: x = ( - 2 \sqrt{3} + \sqrt{27}) }}}$

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$\star \: \: \: \: \tt \underline{Relation \: \: between \: \: coefficients \: and \: Zeros} :$

let, ax²-bx+c=0 is a quadratic equations

$\sf \: We \: \: know \: \: for \: \: any \: \: quadratic \: \: equation,$

$\mapsto \: \: \: \: \sf \underline{\underline{ sum \: \: of \: \: roots = - \frac{b}{a} }}$

$\mapsto \: \: \: \: \sf \underline{\underline{ product \: \: of \: \: roots = - \frac{c}{a} }}$

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HOPE THIS IS HELPFUL...