Question

Write a quadratic polynomial ,sum of whose zeroes is -3 & product is 2 .​

Answer 1

Answer:

6

is the answer

hope it may be correct

Answer 2

Given:-

Sum of zeroes = 2

Product of zeroes = -8

To find:-

A quadratic Polynomial

Assumption:-

Let sum of zeroes be \alpha+\betaα+β = 2

Let the product of zeroes be \alpha\betaαβ = -8

Solution:-

We know,

Quadratic Equation is always in the form of:-

\sf{x^2 + (\alpha+\beta)x + \alpha\beta}x2+(α+β)x+αβ

Substituting the values,

\sf{x^2 + 2\times x + (-8)}x2+2×x+(−8)

= \sf{x^2 + 2x -8}x2+2x−8

Therefore the quadratic Polynomial is\sf{ x^2 + 2x - 8}x2+2x−8

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Verification:-

Let us find out the zeroes of polynomial

x² + 2x - 8

= \sf{x^2 + 2x - 8}x2+2x−8

By splitting the middle term,

\sf{x^2 + 4x - 2x - 8}x2+4x−2x−8

= \sf{x(x+4) -2(x+4)}x(x+4)−2(x+4)

= \sf{(x+4)(x-2)}(x+4)(x−2)

Either,

\sf{x+4 = 0}x+4=0

= \sf{x = -4}x=−4

Or,

\sf{x-2 = 0}x−2=0

= \sf{x = 2}x=2

Now Let us find the sum of product of zeroes,

\sf{Sum\:of\:zeroes = \dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^2}}Sumofzeroes=Coefficientofx2−Coefficientofx

= \sf{-4+2 = \dfrac{-2}{1}}−4+2=1−2

= \sf{-2 = -2}−2=−2 [Verified]

\sf{Product\:of\:zeroes = \dfrac{Constant\:Term}{Coefficient\:of\:x^2}}Productofzeroes=Coefficientofx2ConstantTerm

= \sf{4\times (-2) = \dfrac{-8}{1}}4×(−2)=1−8

= \sf{-8 = -8}−8=−8 [Verified]

Therefore the Quadratic Equation is \sf{\boxed{\sf{x^2 + 2x - 8}}}x2+2x−8 Hence Verified.

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