Question

form the quadratic equation if the roots are -1/2 and 1/3​

Answer 1

Answer:

The sum if the zeroes (α+β) = $\dfrac{-1}{2}$

Product of the zeroes (αβ) = $\dfrac{1}{3}$

so,

The quadratic equation

= k [x² - (α+β)x + αβ] {where k is constant}

= k [x² - ( $\dfrac{-1}{2}$x) + $\dfrac{1}{3}$ ]

= k [ x² + $\dfrac{1}{2}$x + $\dfrac{1}{3}$ ]

Lcm of 2 and 3 is 6

= k [ x² + $\dfrac{3x + 2}{6}$ ]

= $\dfrac{1}{6}$ [x² + 3x + 2 ]

so , the polynomial is x² + 3x + 2

Answer 2

Answer:

hey mate here is your answer.

Step-by-step explanation:

The two zeroes are given -1/2 and 1/3

first,we find their sum = -1/2+1/3

= -1/6

second, we find their product = -1/2×1/3

= -1/6

Now, we using quadratic polynomial

=k( x²-(a+b)x+ab)

here, a+b is sum of zeroes or roots.

and ab is product of zeroes or roots.

So,on putting values

= k (x²-(-1/6)x+(-1/6))

= k(x² +1/6x-1/6)

here ,k=6

so,equation is 6x²+x-1

hope it helps you.

plz mark it as brainliest.