Question

find the quadratic polynomial whose sum of zeroes is √3/2 and product of zeroes is -4​

Answer 1

Answer:

3x² - 4x + 9 = 0

Step-by-step explanation:

We know that every Quadratic Equation is of form

k [ x² - ( sum of roots) x + (product of roots)], where k is any constant value

Now sum of roots = -4/3 and product of roots = 3

∴ Desired equation will be

k[ x² - (-4/3)x + 3] = 0

Let k = 3 [ LCM of denominators is 3]

3[x² - (4/3)x +3] = 0

3x² - 4x + 9 = 0

Desired Equation is 3x² - 4x + 9 = 0

Answer 2

$\large \red{\bf{given}} \: :$

$\sf \boxed{ \sf \alpha + \beta = \frac{ \sqrt{3} }{2} } \: and \: \boxed{ \sf \alpha \beta = - 4}$

$\green{ \bf to \: \: \: \: find} \: :$

$\pink{ \bf quadratic \: \: \: polynomial \: \: \: \: equation}$

$\large \purple{ \underline{ \underline{ \orange{ \mathfrak{ \: \: ☢ \: \: solution \: \: \: }}}}} \: :$

$\bf general \: \: \: form \: \: :$

$\blue{ \overbrace{ { \underbrace{ \: \: \boxed{ \pink{ \bf {x}^{2} + ( \alpha + \beta )x - ( \alpha \beta ) }} \: \: }}}}$

$\bf hence \: \: \: \: by \: \: \: applying \: \: \: values$

$\sf the \: \: \: equation \: \: is$

$\sf⇒ \: \: {x}^{2} + ( \frac{ \sqrt{3} }{2} )x - ( - 4)$

$\sf⇒ \: \: {x}^{2} + ( \frac{ \sqrt{3}x }{2}) + 4$

by taking LCM ,

$\bf we \: \: \: obtain$

$\sf⇒2 {x}^{2} + \sqrt{3} + 8$

$\sf \ast \: \: the \: \: \: equation \: \: \: is \: \: \boxed{ \orange{\sf 2{x}^{2} + \sqrt{3} + 8}}$