Question

Find a quadratic polynomial whose sum and product are 2/3, -1.​

Answer 1

Answer :

3x² - 2x - 3

Solution :

Given,

Sum and product of roots = 2/3 and -1

⇝ α + β = 2/3

⇝ αβ = -1

p(x) = x² - (α + β)x + αβ

= x² - (2/3)x + (-1)

= 3x² - 2x - 3

Answer 2

Answer:

Here, The quadratic equation is 3x^2+x-2=0.

Step-by-step explanation:

Here, As per our given question,

=Let the zeroes be alpha and beta respectively,

So, sum of zeroes =(alpha +beta)=(2/3)+(-1)=

=After taking Lcm and solving it, we get,

=[2+(-3)]/3=(-1)/3

Product of zeroes=(alpha×beta)=(-1)×2/3=(-2)/3

Now,As we have the formula to find out a quadratic equation, is: (x^2-(alpha+beta)x+alpha beta=0)

So, By putting value, we get,

=x^2-[(-1)/3]x+[(-2)/3)

After taking Lcm and solving it, we get,

=[3x^2-(-1)x+(-2)]/3=0

=3x^2+x-2=0 (On transposing 3 to another side, it gets multiplied with 0 and becomes 0).

So, Our quadratic equation is =3x^2+x-2=0

Now, Proof: (For proof we have to dug out the given zeroes from this equation),

=3x^2+x-2=0

By solving it with middle-term split method, we get,

=3x^2+3x-2x-2=0

=3x(x+1) -2(x+1) (On taking common)

=(3x-2) ,(x+1) (Again by taking common)

=So, 3x-2=0, = x+1=0

=3x=2 =x=(-1)

=x=2/3

So, We have the zeroes as=(-1) and 2/3 which were given in our question,

Hence, Proved.

Thank you.