Question

find the quadritic polynomial whose zeroes are 2/3 and -1/4​

Answer 1

here is ur answer

$\Huge\orange{\boxed{\boxed{\underline{ANSWER}}}}$

$\bf given \: zeroes \:of \: polynomial \: are \: \frac {2}{3} , \frac {-1}{4}$

$\bf for \: quadratic \: equation\: if \: zeroes \: are\: given \: then \:$

$\Huge \bf {\boxed {\underline {quadratic \: equation = x^2-(a+b)x+ab }}}$

if a,b are the zeroes of the polynomial

$\bf sum \:of \:zeroes \\ =a+b =\frac {2}{3}-\frac {1}{4} \\ = \frac {5}{12}$

$\bf product \:of \:zeroes \\=ab=\frac{2}{3}×\frac {-1}{4} \\=\frac {-1}{6}$

then the quadratic equation =

$\bf x^2- \frac {5x}{12}-\frac {1}{6}$

let us equate the equation to 0

then after solving

quadratic equation =>

12x²-5x-2 =0

Some more useful formulae;

→sum of zeroes =-b/a

=coefficient of x/coefficient of x²

→product of zeroes =c/a=constant term/coefficient of x²

hope the answer helps

$\huge \pink{ \boxed{ \boxed{ \mathbb{ \mid \ulcorner THANKS \urcorner \mid }}}}$

Answer 2

Answer:

$\large{\underline{\boxed{\sf 12x^2 - 5x - 2}}}$

Step-by-step explanation:

Given that ;

$\sf \dfrac{2}{3}\:and \: - \dfrac{1}{4}$ are the zeroes of the polynomial.

Sum of zeroes = $\sf \dfrac{2}{3} - \dfrac{1}{4}$

⇒ α + β = $\sf \dfrac{8-3}{12}$

⇒ α + β = $\sf \dfrac{5}{12}$

Product of zeroes = $\sf \dfrac{2}{3}*( - \dfrac{1}{4})$

⇒ αβ = $\sf -\dfrac{2}{12}$

⇒ αβ = $\sf - \dfrac{1}{6}$

We know that ;

A quadratic polynomial is given by ;

⇒ x² - (α + β)x + αβ = 0

⇒ x² - $\sf \dfrac{5}{12}x$ +$\sf - \dfrac{1}{6}$ = 0

⇒ x² - $\sf \dfrac{5}{12} x - \dfrac{1}{6}$ = 0

⇒ $\sf \dfrac{12x {}^{2} - 5x - 2 }{12} = 0$

⇒ 12x² - 5x - 2 = 0

Hence, the required quadratic polynomial is 12x² - 5x - 2.