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Question: If one root of the equation $x^2 + ax + b = 0$ is two times the square of the other root, then:...

If one root of the equation x2+ax+b=0x^2 + ax + b = 0 is two times the square of the other root, then:

A

2a3−b(3a+1)+4b2=02a^3 - b(3a + 1) + 4b^2 = 0

B

2a3−b(6a−1)+4b2=02a^3 - b(6a - 1) + 4b^2 = 0

C

2a3+b(6a−1)−4b2=02a^3 + b(6a - 1) - 4b^2 = 0

D

2a3+b(3a−1)−4b2=02a^3 + b(3a - 1) - 4b^2 = 0

Answer

2a^3 - b(6a - 1) + 4b^2 = 0

Explanation

Solution

Let the roots be α\alpha and β\beta. Assume β=2α2\beta = 2\alpha^2.

Using Viète’s formulas for the quadratic x2+ax+b=0x^2 + ax + b = 0, we have:

α+β=−a  ⟹  α+2α2=−a  ⟹  a=−α(1+2α)\alpha + \beta = -a \implies \alpha + 2\alpha^2 = -a \implies a = -\alpha(1+2\alpha)

αβ=b  ⟹  α⋅2α2=2α3=b\alpha\beta = b \implies \alpha \cdot 2\alpha^2 = 2\alpha^3 = b

Express aa and bb in terms of α\alpha:

a=−α(1+2α),b=2α3a = -\alpha(1+2\alpha), \quad b = 2\alpha^3

Now check Option 2: 2a3−b(6a−1)+4b2=02a^3 - b(6a-1)+4b^2=0.

Compute:

  • a3=[−α(1+2α)]3=−α3(1+2α)3a^3 = [-\alpha(1+2\alpha)]^3 = -\alpha^3(1+2\alpha)^3 so that 2a3=−2α3(1+2α)32a^3 = -2\alpha^3(1+2\alpha)^3.
  • b(6a−1)=2α3[6(−α(1+2α))−1]=2α3[−6α−12α2−1].b(6a-1) = 2\alpha^3[6(-\alpha(1+2\alpha))-1] = 2\alpha^3[-6\alpha-12\alpha^2-1].
  • 4b2=4(2α3)2=16α6.4b^2 = 4(2\alpha^3)^2 = 16\alpha^6.

Thus the expression becomes:

−2α3(1+2α)3−2α3[−6α−12α2−1]+16α6-2\alpha^3(1+2\alpha)^3 - 2\alpha^3[-6\alpha-12\alpha^2-1] + 16\alpha^6.

Notice that −2α3[−6α−12α2−1]=2α3(6α+12α2+1)-2\alpha^3[-6\alpha-12\alpha^2-1] = 2\alpha^3(6\alpha+12\alpha^2+1).

Also, expanding (1+2α)3=1+6α+12α2+8α3(1+2\alpha)^3 = 1+6\alpha+12\alpha^2+8\alpha^3, we have:

−2α3(1+6α+12α2+8α3)=−2α3−12α4−24α5−16α6-2\alpha^3(1+6\alpha+12\alpha^2+8\alpha^3) = -2\alpha^3 -12\alpha^4 -24\alpha^5 -16\alpha^6.

Adding:

−2α3−12α4−24α5−16α6+2α3+12α4+24α5+16α6=0-2\alpha^3 -12\alpha^4 -24\alpha^5 -16\alpha^6 + 2\alpha^3 +12\alpha^4 +24\alpha^5 +16\alpha^6 = 0.

Since the expression is identically zero, Option 2 is correct.