Question

If \( \alpha \) is the common root of the quadratic equations \( x^2 - 5x + 4a = 0 \) and \( x^2 - 2ax - 8 = 0 \), where \( a \in \mathbb{R} \), then the value of \( \alpha^4 - \alpha^3 + 68 \) is:

• A. 260
• B. 250
• C. 0
• D. 240

Hint:

For common root problems in quadratics, try solving by equating the expressions and verifying roots through substitution. Pick integer roots to simplify calculations.

Answer 1

The Correct Option is A

Given that \( \alpha \) is a common root of the equations: \[ x^2 - 5x + 4a = 0
\text{and}
x^2 - 2ax - 8 = 0 \]
Since \( \alpha \) satisfies both equations: \[ \alpha^2 - 5\alpha + 4a = 0
\text{(1)} \] \[ \alpha^2 - 2a\alpha - 8 = 0
\text{(2)} \]
Subtract (2) from (1): \[ (\alpha^2 - 5\alpha + 4a) - (\alpha^2 - 2a\alpha - 8) = 0 \] \[ -5\alpha + 4a + 2a\alpha + 8 = 0 \Rightarrow (2a - 5)\alpha + 4a + 8 = 0 \]
Solve for \( \alpha \): \[ (2a - 5)\alpha = -4a - 8 \Rightarrow \alpha = \frac{-4a - 8}{2a - 5}
\text{(3)} \]
Substitute (3) in one of the original equations, say (1): \[ \alpha^2 - 5\alpha + 4a = 0 \Rightarrow \text{Compute } \alpha^4 - \alpha^3 + 68 \]
Use the fact from (1): \( \alpha^2 = 5\alpha - 4a \)
Square both sides: \[ \alpha^4 = (\alpha^2)^2 = (5\alpha - 4a)^2 = 25\alpha^2 - 40a\alpha + 16a^2 \]
But \( \alpha^2 = 5\alpha - 4a \), so: \[ \alpha^4 = 25(5\alpha - 4a) - 40a\alpha + 16a^2 = 125\alpha - 100a - 40a\alpha + 16a^2 \]
Now compute \( \alpha^3 \). Multiply both sides of (1) by \( \alpha \): \[ \alpha^3 = 5\alpha^2 - 4a\alpha = 5(5\alpha - 4a) - 4a\alpha = 25\alpha - 20a - 4a\alpha \]
Now find \( \alpha^4 - \alpha^3 + 68 \): \[ \alpha^4 - \alpha^3 + 68 = (125\alpha - 100a - 40a\alpha + 16a^2) - (25\alpha - 20a - 4a\alpha) + 68 \]
Simplify: \[ = (125\alpha - 25\alpha) + (-40a\alpha + 4a\alpha) + (-100a + 20a) + 16a^2 + 68 \] \[ = 100\alpha - 36a\alpha - 80a + 16a^2 + 68 \]
Now substitute \( \alpha = 2 \), which satisfies both equations when \( a = 1 \) \[ \alpha = 2,
a = 1 \] \[ \Rightarrow \alpha^4 = 2^4 = 16,
\alpha^3 = 8 \Rightarrow \alpha^4 - \alpha^3 + 68 = 16 - 8 + 68 = 76 \]
Wait — there's a contradiction, so check which value of \( \alpha \) satisfies both: Try \( a = 2 \) check if a common root exists For \( a = 2 \), first equation: \( x^2 - 5x + 8 = 0 \), roots are complex. Try \( a = 1 \), then: \[ x^2 - 5x + 4 = 0 \Rightarrow x = 1, 4
\text{and} \] \[ x^2 - 2x - 8 = 0 \Rightarrow x = 4, -2 \Rightarrow \alpha = 4 \]
So finally: \[ \alpha = 4 \Rightarrow \alpha^4 = 256,
\alpha^3 = 64 \Rightarrow \alpha^4 - \alpha^3 + 68 = 256 - 64 + 68 = 260 \] \[ \boxed{\text{Correct Option: (1)}} \] % Tip