Multiple choice

lf $\alpha,\ \beta$ are the roots of the equation $x^{2}+mx+l=0$ and $\gamma,\ \delta$ that of $x^{2}+nx+l=0$, then the value of $(\alpha-\gamma)(\beta-\gamma)(\alpha-\delta)(\beta-\delta)=$

  1. $l(m+n)^{2}$
  2. $-l(m+n)^{2}$
  3. $l(m-n)^{2}$
  4. $1 (\mathrm{m}+\mathrm{n})$
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C Correct answer
Explanation

The expression is (alpha - gamma)(beta - gamma)(alpha - delta)(beta - delta). This is equivalent to f(gamma) * f(delta) where f(x) = x^2 + mx + l. Since gamma and delta are roots of x^2 + nx + l = 0, we use the property of roots to evaluate the product, resulting in l(m-n)^2.

AI explanation

We can pair the terms to rewrite the expression as ((alpha - gamma)(beta - gamma))((alpha - delta)(beta - delta)). Using polynomial evaluation, the first part is the quadratic x^2 + mx + l evaluated at x = gamma, yielding gamma^2 + m*gamma + l. Because gamma is a root of x^2 + nx + l = 0, we replace gamma^2 with -n*gamma - l to get (m - n)*gamma. Similarly, the second part evaluates at x = delta to become (m - n)*delta. Multiplying these gives (m - n)^2 * gamma * delta, and since gamma * delta = l from the second equation, the final result is l(m - n)^2.

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