Question

Find the zeroes of given quadratic polynomial

p(x)=4y^2-5√2y-7

PLEASE SOLVE THESE QUESTION.

THESE IS OF 100 POINTS.​

Answer 1

$\huge \underline \mathfrak \color{blue}{ || ⫷\:  \: correct \: question \:  \: ⫸ || }$

p(y)=4√3y²+5y-2√3

$\huge \underline \mathfrak \color{red}{ || ⫷\:  \: Answer \:  \: ⫸ || }$

The roots of the given quadratic equation are

$\boxed{\red{\sf\:y\:=\:-\:\frac{2\:\sqrt{3}}{3}}}\:\:\sf\:or\:\:\:\boxed{\red{\sf\:y\:=\:\frac{\sqrt{3}}{4}}}$

The given quadratic equation is

$\sf\:4\:\sqrt{3}\:y^{2}\:+\:5y\:-\:2\:\sqrt{3}\:=\:043y2+5y−23=0</p><p>\begin{gathered}\therefore\sf\:4\:\sqrt{3}\:y^{2}\:+\:5y\:-\:2\:\sqrt{3}\:=\:0\\\\:\implies\sf\:4\:\sqrt{3}\:y^{2}\:+\:8y\:-\:3y\:-\:2\:\sqrt{3}\:=\:0\\\\\implies\sf\:4\:\sqrt{3}\:y^{2}\:+\:8y\:-\:\sqrt{3}\:\times\:\sqrt{3}\:\times\:y\:-\:2\:\sqrt{3}\:=\:0\:\:\:-\:-\:-\:[\:Expressing\:in\:terms\:of\:\sqrt{3}\:]\\\\\implies\sf\:4y\:(\:\sqrt{3}\:y\:+\:2\:)\:-\:\sqrt{3}\:(\:\sqrt{3}\:y\:+\:2\:)\:=\:0\\\\\implies\sf\:(\:\sqrt{3}\:y\:+\:2\:)\:(\:4y\:-\:\sqrt{3}\:)\:=\:0\\\\\implies\sf\:\sqrt{3}\:y\:+\:2\:=\:0\:\:\:or\:\:\:4y\:-\:\sqrt{3}\:=\:0\\\\\implies\sf\:\sqrt{3}\:y\:=\:-\:2\:\:\:or\:\:\:4y\:=\:\sqrt{3}\\\\\implies\sf\:y\:=\:-\:\frac{2}{\sqrt{3}}\:\:\:or\:\:\:y\:=\:\frac{\sqrt{3}}{4}\\\\\implies\boxed{\red{\sf\:y\:=\:-\:\frac{2\:\sqrt{3}}{3}}}\:\:\sf\:or\:\:\:\boxed{\red{\sf\:y\:=\:\frac{\sqrt{3}}{4}}}\:\:\:[\:Rationalising\:the\:denominator\:]\end{gathered}∴43y2+5y−23=0:⟹43y2+8y−3y−23=0⟹43y2+8y−3×3×y−23=0−−−[Expressingintermsof3]⟹4y(3y+2)−3(3y+2)=0⟹(3y+2)(4y−3)=0⟹3y+2=0or4y−3=0⟹3y=−2or4y=3⟹y=−32ory=43⟹y=−323ory=43$

[Rationalisingthedenominator]

Now,

$\begin{gathered}\sf\:Comparing\:4\:\sqrt{3}\:y^{2}\:+\:5y\:-\:2\:\sqrt{3}\:=\:0\:with\:ax^{2}\:+\:bx\:+\:c\:=\:0,\:we\:get,\\\\\sf\bullet\:a\:=\:4\:\sqrt{3}\\\\\sf\bullet\:b\:=\:5\\\\\sf\bullet\:c\:=\:-\:2\:\sqrt{3}\end{gathered}Comparing43y2+$