Question

please answer this ASAP!​

Answer 1

Given that :

$Polynomial = 4 \sqrt{3} {x}^{2} + 5x - 2 \sqrt{3}$

On comparing with the equation :

$a {x}^{2} + bx + c$

We get : a = 4√3, b = 5 and c = -2√3

Now, from Shridhar-Acharya formula :

$x = \frac{ - b \binom{ + }{ - } \sqrt{ {b}^{2} - 4ac} }{2a}$

On putting the value of a, b and c in the formula :

$x = \frac{ - 5 \binom{ + }{ - } \sqrt{ {5}^{2} - 4(4 \sqrt{3} )( - 2 \sqrt{3} )} }{2 \times 4 \sqrt{3} } \\ \\ = x = \frac{ - 5 \binom{ + }{ - } \sqrt{25 + 96} }{8 \sqrt{3} } \\ \\ = > x = \frac{ - 5 \binom{ + }{ - } \sqrt{121} }{8 \sqrt{3} } = \frac{ - 5 \binom{ + }{ - } 11 }{8 \sqrt{3} }$

On taking (+) sign :

$x = \frac{ - 5 + 11}{8 \sqrt{3} } = \frac{6}{8 \sqrt{3} } = \frac{ \sqrt{3} }{4}$

On taking (-) sign :

$x = \frac{ - 5 - 11}{8 \sqrt{3} } = \frac{ - 16}{8 \sqrt{3} } = \frac{ - 2}{ \sqrt{3} }$

So, the zeroes of the given polynomial are

$\frac{ \sqrt{3} }{4} \: and \: \frac{ - 2}{ \sqrt{3} }$

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Hope it helps ☺

Answer 2

Question :

• Find the zeroes of the quadratic polynomial $\tt\:f(x)~=~4 \sqrt{3} x^{2} +5x-2 \sqrt{3}$ and verify the relation between the zeroes and the coefficient.

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Given :

• Quadratic polynomial $\tt\:f(x)~=~4 \sqrt{3} x^{2} +5x-2 \sqrt{3}$

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To find :

• Zeroes of the polynomial $\tt\:f(x)~=~4 \sqrt{3} x^{2} +5x-2 \sqrt{3}$

• Verify the relation between the zeroes and the coefficient.

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Concept :

Here at first we would find the zeroes of the given polynomial by splitting middle term. Then we would verify the relation between the zeroes and the coefficient by using two formulas. The formulas are as follows :

❒ $\small{\underline{\boxed{\mathrm{Sum~of~the~roots~=~\frac{-b}{a}}}}}$

❒ $\small{\underline{\boxed{\mathrm{Product~of~the~roots~=~\frac{c}{a}}}}}$

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Solution :

The given polynomial is of the form $\tt\:ax^{2}+bx+c$ which is equal to zero.

So let's solve the given polynomial by splitting middle term.

$\tt\:4 \sqrt{3} x^{2} +5x-2 \sqrt{3}=0$

$\tt\implies\:4 \sqrt{3} x^{2} +(8-3)x-2 \sqrt{3}=0$

$\tt\implies\:4 \sqrt{3} x^{2} +8x-3x-2 \sqrt{3}=0$

Taking 4x as common from first two terms and √3 from last two terms

$\tt\implies\:4x( \sqrt{3} x +2)-\sqrt{3} ( \sqrt{3}x+2)=0$

Taking ( √3x+2 ) common from whole expression

$\tt\implies\:(4x-\sqrt{3}) ( \sqrt{3} x +2)=0$

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Now zeroes of the polynomial :

Either - $\tt\longrightarrow\:(4x-\sqrt{3})=0$

$\tt\longrightarrow\:4x=\sqrt{3}$

$\tt\color{navy}{\longrightarrow\:x =\frac{ \sqrt{3} }{4}}$

Or -‎ $\tt\longrightarrow\:(\sqrt{3}x+2)=0$

$\tt\longrightarrow\:\sqrt{3}x=-2$

$\tt\color{navy}{\longrightarrow\:x =\frac{-2}{\sqrt{3}}}$

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Therefore :

Two zeroes/roots of the polynomial are $\large{\mathfrak\purple{\frac{\sqrt{3}}{4}}}$ and $\large{\mathfrak\purple{\frac{-2}{\sqrt{3}}}}$

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Now let's verify the relations between the ‎zeroes and the coefficient using the above mentioned formulas :

Given polynomial - $\tt\:4 \sqrt{3} x^{2} +5x-2 \sqrt{3}=0$.

Here :

• a = 4√3 , b = 5 and c = -2√3

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$\tt\:Sum~of~the~roots~=~\frac{-b}{a}$

Plugging the values respectively

$\tt\leadsto\:\frac{\sqrt{3}}{4} + (\frac{-2}{\sqrt{3}})~=~\frac{-5}{4\sqrt{3}}$

$\tt\leadsto\:\frac{\sqrt{3}}{4} - \frac{2}{\sqrt{3}}~=~\frac{-5}{4\sqrt{3}}$

Solving by LCM

$\tt\leadsto\:\frac{(\sqrt{3})^{2}-(4 \times 2)}{4\sqrt{3}} ~=~\frac{-5}{4\sqrt{3}}$

$\tt\leadsto\:\frac{3-8}{4\sqrt{3}} ~=~\frac{-5}{4\sqrt{3}}$

$\tt\leadsto\:\frac{-5}{4\sqrt{3}} ~=~\frac{-5}{4\sqrt{3}}$

Cross multiplying

$\tt\leadsto\:(-5) \times 4\sqrt{3}~=~(-5) \times 4\sqrt{3}$

$\tt\leadsto\:-20\sqrt{3}~=~-20\sqrt{3}$

$\bf\leadsto\:LHS ~=~RHS$

Hence Verified !~

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Again :

$\tt\:Products~of~the~roots~=~\frac{c}{a}$

Plugging the values respectively

$\tt\leadsto\:\frac{\sqrt{3}}{4} \times \frac{-2}{\sqrt{3}}~=~\frac{-2\sqrt{3}}{4\sqrt{3}}$

$\tt\leadsto\:\frac{-2\sqrt{3}}{4\sqrt{3}}~=~\frac{-2\sqrt{3}}{4\sqrt{3}}$

Cross multiplying

$\tt\leadsto\:(-2\sqrt{3}) \times 4\sqrt{3} ~=~(-2\sqrt{3}) \times 4\sqrt{3}$

$\tt\leadsto\:-8\sqrt{9} ~=~-8\sqrt{9}$

$\tt\leadsto\:-8 \times 3 ~=~-8 \times 3$

$\tt\leadsto\:-24 ~=~-24$

$\bf\leadsto\:LHS ~=~RHS$

Hence Verified !~

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Hence we are done with everything :D

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