Question

Find the quadratic polynomia whose zeroes are -2over root 3 and root3 over4

Answer 1

||✪✪ QUESTION ✪✪||

Find the quadratic polynomia whose zeroes are (-2/√3) and (√3/4) ?

|| ✰✰ ANSWER ✰✰ ||

☙☙ Method ❶ ☙☙ :-

Given that, Roots are (-2/√3) and (√3/4) ...

So,

→ Sum of roots = (-2/√3) + (√3/4)

→ Sum of roots = [(-8+3) /(4√3) ]

→ Sum of roots = (-5)/(4√3)

And,

→ Product of roots = (-2/√3) * (√3/4)

→ Product of roots = (-1/2)

So,

→ Required equation : x^2 - ( sum of roots )x + product of roots = 0

= > x² - (-5)/(4√3)x + (-1/2) = 0

= > x² + (5/4√3)x - (1/2) = 0

Taking LCM,

= > 4√3x² + 5x - 2√3 = 0

Hence, Required equation is 4√3x² + 5x - 2√3 = 0.

_____________________________

☙☙ Method ❷☙☙ :-

We know, Any quadratic equation can also be written in this form : ( x - a )( x - b ) = 0, where a and b are the roots of the equation.

Thus, here,

= > ( x - (-2/√3) )( x - (√3/4)) = 0

= > ( x + 2/√3) ( x - √3/4)

= > x² - (√3/4)x + (2/√3)x - (1/2) = 0

= > x² [ (-3+8)/4√3)x ] - (1/2) = 0

Taking LCM again ,

= > 4√3x² + 5x - 2√3 = 0

Hence, Required equation is 4√3x² + 5x - 2√3 = 0.

_____________________________

Answer 2

$\huge{\underline{\underline{\mathfrak\green{Question\::}}}}$

$\sf{Find \:the\: quadratic \:polynomial\: whose \:zeroes \:are\: -2/V3\: and \:V3/4.}$

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

$\implies$ $\sf{Let \:,the \:two \:zeroes \:be \:α\: and \: β.}$

$\implies$ .°. $\sf{α = -2/V3 \: and \: β = V3/4}$

$\implies$ $\sf{sum \:of\: zeroes\: = α + β}$

$\implies$ $\sf{(-2/V3) \:+ ( V3/4)}$

$\implies$ ( -8 + 3)/(4V3)

$\implies$ $\sf{(-5)/(4V3)}$

$\implies$ .°. $\sf{sum \:of\: product = α × β}$

$\implies$ $\sf{( -2/V3) \:× ( 3/V4)}$

$\implies$ $\sf{(-1/2)}$

WE KNOW THAT,

$\pink\rightarrow$ $\sf{p(x) = x^2-(α+β)x +(α×β)}$

NOW,

$\implies$ $\sf{p(x) = x^2 - (-5)/(4/V3)x+ (-1/2)}$

$\implies$ $\sf{p(x) = x^2 + ( 5/4V3x)-1/2]}$

$\sf{[By \: taking \:LCM]}$

$\implies$ $\large\sf{\boxed{p(x) = 4/3x^2+5x -2V3}}$