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The number of values aa can take such that x4+ax3+(3a−4)x2+2(a−1)x−4x^4 + ax^3 + (3a - 4)x^2 + 2(a - 1)x - 4 can be expressed as a product of two quadratic polynomials, x2+px+2x^2 + px + 2 and x2+qx−2x^2 + qx - 2, where pp and qq are real, is

Solution

✅ Correct Option: 2

Expand the product (x2+px+2)(x2+qx−2)(x^2 + px + 2)(x^2 + qx - 2):

x4+(p+q)x3+pq⋅x2+2(q−p)x−4x^4 + (p+q)x^3 + pq \cdot x^2 + 2(q-p)x - 4


Match coefficients with x4+ax3+(3a−4)x2+2(a−1)x−4x^4 + ax^3 + (3a-4)x^2 + 2(a-1)x - 4:

p+q=ap + q = a

pq=3a−4pq = 3a - 4

q−p=a−1q - p = a - 1


Add the first and third equations:

2q=2a−12q = 2a - 1, so q=a−12q = a - \dfrac{1}{2}

Subtract to get 2p=12p = 1, so p=12p = \dfrac{1}{2}


Substitute into pq=3a−4pq = 3a - 4:

12(a−12)=3a−4\dfrac{1}{2}\left(a - \dfrac{1}{2}\right) = 3a - 4

Multiply by 44: 2a−1=12a−162a - 1 = 12a - 16, giving a=32a = \dfrac{3}{2}


Only one value of aa works.

Answer =1= 1

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