Multiple choice

lf $\sin\theta, \cos\theta$ are the roots of the equation $ax^{2}+bx+c=0$ then

  1. $ a^{2}-b^{2}+2ac=0$
  2. $ a^{2}+b^{2}+2ac=0$
  3. $ a-b+2ac=0$
  4. $a+b+2c=0$
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A Correct answer
Explanation

The roots are sin(theta) and cos(theta), so their sum is -b/a and product is c/a. Since (sin(theta) + cos(theta))^2 = 1 + 2sin(theta)cos(theta), we obtain b^2/a^2 = 1 + 2c/a. Rearranging gives a^2 - b^2 + 2ac = 0.

AI explanation

Using the sum and product of roots for ax^2+bx+c=0, we have sin x + cos x = -b/a and (sin x)(cos x) = c/a. Squaring the sum gives (sin x + cos x)^2 = sin^2 x + cos^2 x + 2(sin x)(cos x), which translates to (-b/a)^2 = 1 + 2(c/a). Simplifying b^2/a^2 = 1 + 2ac/a^2 yields b^2 = a^2 + 2ac, which rearranges to a^2 - b^2 + 2ac = 0.

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