Question

3√3x2 - 19x + 10√3 find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficient

Answer 1

$\large{\underline{\bf{\purple{Given:-}}}}$

• ✦ p(x) = 3√3x² - 19x + 10√3

$\large{\underline{\bf{\purple{To\:Find:-}}}}$

• ✦ we need to find the zeroes of the given polynomial and also find the relationship between the zeroes and coefficients.

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

• p(x) = 3√3x² - 19x + 10√3

$\leadsto \:\:$3√3x² - 9x - 10x + 10√3

$\leadsto$ 3√3x(x-√3) - 10(x-√3)

$\leadsto \:\:$(3√3x - 10)(x - √3)

$\leadsto \:\:$x = 10/3√3 or x = √3

Now relationship between the zeroes coefficients:-

• Let α = 10/3√3
• and β =√3

• a = 3√3
• b = -19
• c = 10√3

sum of zeroes:- - b/a

$\leadsto \:\rm \: \frac{10}{3 \sqrt{3} } + \sqrt{3} \: \: \: = \frac{ - ( - 19)}{3 \sqrt{3} } \\ \\\leadsto \:\rm \: \frac{10 + 9}{3 \sqrt{3} } = \frac{19}{3 \sqrt{3} } \\ \\ \leadsto \:\bf \: \frac{19}{3 \sqrt{3} } = \frac{19}{3 \sqrt{3} }$

product of zeroes = c/a

$\leadsto \rm\: \frac{10}{3 \sqrt{3} } \times \sqrt{ 3} \: \: = \frac{10 \sqrt{3} }{3 \sqrt{3} } \\ \\ \leadsto \rm\: \frac{10}{3{ \cancel{ \sqrt{3} }}} \times { \cancel{ \sqrt{3} }} = \frac{10{ \cancel{ \sqrt{3} }}}{3{ \cancel{ \sqrt{3} }}} \\ \\ \leadsto \bf\: \frac{10}{3} = \frac{10}{3}$

LHS = RHS

hence relationship is verified.

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Answer 2

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{\frac{10}{3\sqrt3} \ and \sqrt3 \ are \ the \ zeroes.}$

$\bold\blue{Explanation}$

$\bold{The \ given \ quadratic \ polynomial \ is}$

$\bold{=>3\sqrt3x^{2}-19x+10\sqrt3}$

$\bold{Here, \ a=3\sqrt3, \ b=-19, \ c=10\sqrt3}$

$\bold{a×c=3\sqrt3×10\sqrt3}$

$\bold{a×c=90}$

$\bold{we, \ should \ split \ middle \ term \ such \ that}$

$\bold{their \ multiplication \ is \ 90 \ and }$

$\bold{addition \ is \ -19.}$

$\bold\orange{Given:}$

$\bold{The \ given \ polynomial \ is}$

$\bold{=>3\sqrt3x^{2}-19x+10\sqrt3}$

$\bold\pink{To \ find:}$

$\bold{Zeroes \ of \ the \ polynomial.}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{The \ given \ quadratic \ polynomial \ is}$

$\bold{=>3\sqrt3x^{2}-19x+10\sqrt3}$

$\bold{=>3\sqrt3x^{2}-9x-10x+10\sqrt3}$

$\bold{=>3\sqrt3x(x-\sqrt3)-10(x-\sqrt3)}$

$\bold{=>(3\sqrt3x-10)(x-\sqrt3)}$

$\bold{\therefore{x=\frac{10}{3\sqrt3} \ or \ \sqrt3}}$

$\bold\purple{\frac{10}{3\sqrt3} \ and \sqrt3 \ are \ the \ zeroes.}$

$\bold\blue{Verification:}$

$\bold{Let \ zeroes \ be \ \alpha \ and \ \beta}$

$\bold{Sum \ of \ zeroes=\frac{-b}{a}}$

$\bold{Product \ of \ zeroes=\frac{c}{a}}$

__________________________________

$\bold{\alpha+\beta=\frac{10}{3\sqrt3}+\sqrt3}$

$\bold{\therefore{\alpha+\beta=\frac{10+9}{3\sqrt3}}}$

$\bold{\alpha+\beta=\frac{19}{3\sqrt3}...(1)}$

$\bold{\frac{-b}{a}=\frac{-(-19)}{3\sqrt3}}$

$\bold{\therefore{\frac{-b}{a}=\frac{19}{3\sqrt3}...(2)}}$

$\bold{from \ (1) \ and \ (2)}$

$\bold{Sum \ of \ zeroes=\frac{-b}{a}}$

__________________________________

$\bold{\alpha×\beta=\frac{10}{3\sqrt3}×\sqrt3}$

$\bold{\alpha×\beta=\frac{10\sqrt3}{3\sqrt3}...(3)}$

$\bold{\frac{c}{a}=\frac{10\sqrt3}{3\sqrt3}...(4)}$

$\bold{from \ (3) \ and \ (4)}$

$\bold{Product \ of \ zeroes=\frac{c}{a}}$

$\bold{Hence \ verified}$