Quadratic Equations Question - None

Question

If $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$ have a common root $\alpha \neq 0$, then $\frac{a^3 + b^3 + c^3}{abc}$ is equal to

Accepted Answer

Step 1: Since α is a common root of both equations, it satisfies:

aα² + bα + c = 0 ... (1)

bα² + cα + a = 0 ... (2)

Step 2: Multiply equation (1) by α:

aα³ + bα² + cα = 0 ... (3)

Step 3: From equation (2): bα² + cα + a = 0, we have:

bα² + cα = -a

Step 4: Substitute this into equation (3):

aα³ - a = 0

aα³ = a

Since α ≠ 0 and a ≠ 0 (for a quadratic), we get: α³ = 1

Step 5: Therefore α is a cube root of unity. Since α ≠ 0 and is real or complex, we have α³ = 1.

Step 6: Now multiply equation (1) by α²:

aα⁴ + bα³ + cα² = 0

Since α³ = 1: aα + b + cα² = 0 ... (4)

Step 7: From equation (1): aα² + bα + c = 0 ... (1)

From equation (2): bα² + cα + a = 0 ... (2)

From Step 4 (rewritten): aα + b + cα² = 0 ... (4)

Step 8: Multiply equation (1) by 1, equation (2) by 1, and equation (4) by 1, then add cyclically:

(aα² + bα + c) + (bα² + cα + a) + (aα + b + cα²) = 0

Collecting terms: (a+b+c)α² + (a+b+c)α + (a+b+c) = 0

(a + b + c)(α² + α + 1) = 0

Step 9: If a + b + c ≠ 0, then α² + α + 1 = 0, which gives α = ω or α = ω² (complex cube roots of unity).

For cube roots of unity: 1 + ω + ω² = 0

Step 10: Since α³ = 1 and α ≠ 1, we have α² + α + 1 = 0, which means:

a + b + c = 0 OR α² + α + 1 = 0

Step 11: Using the identity: when α³ = 1 and a, b, c satisfy the given conditions,

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

Since α² + α + 1 = 0 implies the roots cycle through coefficients, this gives:

a³ + b³ + c³ = 3abc

Step 12: Therefore: (a³ + b³ + c³)/(abc) = 3abc/(abc) = 3

∴ Answer: C

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