Question

find a quadratic polynomial whose zeroes are -3/2 and 3/2.​

Answer 1

Answer:

4x² - 9

Step-by-step explanation:

Let α and β be the zeroes of the required polynomial.

A.T.Q.

α = - 3/2

β = 3/2

Sum of zeroes = α + β = - 3/2 + 3/2 = 0

Product of zeroes = αβ = (- 3/2)(3/2) = - 9/4

The required polynomial is :

$\succ$ p(x) = k [x² - (α + β)x + αβ]

$\succ$ p(x) = k [x² - (0)x + (- 9/4)]

$\succ$ p(x) = k [x² - 9/4]

$\succ$ p(x) = k [(4x² - 9)/4]

$\succ$ p(x) = k/4 [4x² - 9]

Put k = 4, we get

→ p(x) = 4x² - 9

Answer 2

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➡Let, Zeros are

↪P = -3/2.

P = -3/2. ↪q = 3/2.

➡We know,

↪Sum of Roots (p+q)= (-3/2+3/2)

Sum of Roots (p+q)= (-3/2+3/2) = 0.

Sum of Roots (p+q)= (-3/2+3/2) = 0. ↪Product of Roots pq = (-3/2)(3/2)

Sum of Roots (p+q)= (-3/2+3/2) = 0. ↪Product of Roots pq = (-3/2)(3/2) = -9/4

➡Formula of Quadratic equations :-

X² + X(Sum of roots )-(product of roots )=0

=> X² +X(0)-(-9/4)=0

X² +X(0)-(-9/4)=0=>X² + 9/4 =0

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