Question

form, quadratic polynomial , if zeros are (1/2 , -3) ​

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\sf\underline{\purple{Given:-}}$

$\sf\bullet Zero's\:of\: polynomial=\dfrac{1}{2}\:\:and\:\:(-3)\\ \\ \sf\:\:\:\: Where\::\\ \\ \sf\implies let\:\alpha=\dfrac{1}{2}\\ \\ \sf\implies and\:\beta=(-3)$

Now

$\boxed{\sf{\alpha x{}^{2}-(\alpha+\beta)x+\alpha\beta}}$

$\sf\bullet sum\:of\:zero's=\dfrac{1}{2}+(-3)\\ \\ \sf\implies (\alpha+\beta)=\dfrac{1-6}{2}=\dfrac{-5}{2}\\ \\ \sf\bullet Product\:of\: zero's=\dfrac{1}{2}\times (-3)\\ \\ \sf\implies (\alpha\beta)=\dfrac{-3}{2}$

$\sf\underline{\red{A.T.Q:-}}$

$\implies k( x{}^{2}-\dfrac{(-5x)}{2}+\dfrac{(-3)}{2})=0\\ \\ \sf\implies k(\dfrac{x{}^{2}-5x-3}{2})=0\\ \\ \sf\:\:\:taking\:L.C.M\\ \\ \sf\implies k(2x{}^{2}-5x-3)=2\times 0\\ \\ \sf\:\:\:let\:k=1\\ \\ \sf\implies 2x{}^{2}-5x-3=0$

The required polynomial is

$\boxed{\sf{2x{}^{2}-5x-3}}$

Answer 2

|| ☆.QUESTION.☆ ||

Find quadratic polynomial , if zeros are 1/2 and -3 .

|| ☆.SOLUTION.☆ ||

Let, p and q are zeros of polynomial .

So,

• p = 1/2
• q = -3

Now, Calculate

Sum of Zeros

✏ ( p + q ) = (1/ 2) + (-3)

✏ ( p + q ) = (1-6)/2

✏ ( p + q) = (-5/2) .....(1)

Product of Zeros

✏ ( pq ) = (1/)2 × (-3)

✏( pq ) = (-3/)2 .........(2)

Formula Of Quadratic Equation

[ x² - x ( p + q ) + pq = 0 ]

[ Keep value by equation (1) and (2) ]

↪ x² - x { (-5/2) } + ( -3 /2 ) = 0

[ Taking, L.C.M. is 2 ]

↪ [ 2x² - x (-5) - 3 ]/2 = 0

↪ 2x² + 5x - 3 = 0

Thus:-

Required Quadratic Equation

• 2 x² + 5x - 3 = 0

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