If the roots of the equation ax^2 + bx + c = | vedclass

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(D) Given the roots are $\sin \alpha$ and $\cos \alpha$ for the equation $ax^2 + bx + c = 0$.From the sum of roots: $\sin \alpha + \cos \alpha = -\fra

Vedclass · Accepted Answer

Both $(b)$ and $(c)$

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(D) Given the roots are $\sin \alpha$ and $\cos \alpha$ for the equation $ax^2 + bx + c = 0$.From the sum of roots: $\sin \alpha + \cos \alpha = -\frac{b}{a} \quad (i)$From the product of roots: $\sin \alpha \cos \alpha = \frac{c}{a} \quad (ii)$Squaring equation $(i)$:$(\sin \alpha + \cos \alpha)^2 = \left(-\frac{b}{a}\right)^2$$\sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha = \frac{b^2}{a^2}$Since $\sin^2 \alpha + \cos^2 \alpha = 1$,we have:$1 + 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2}$$1 + \frac{2c}{a} = \frac{b^2}{a^2}$Multiplying by $a^2$: $a^2 + 2ac = b^2 \Rightarrow a^2 - b^2 + 2ac = 0$. This matches option $(b)$.Now,consider $(a+c)^2 = a^2 + 2ac + c^2$. Since $a^2 + 2ac = b^2$,we get $(a+c)^2 = b^2 + c^2$. This matches option $(c)$.Therefore,both $(b)$ and $(c)$ are correct.

If the roots of the equation $ax^2 + bx + c = 0$ are $\sin \alpha$ and $\cos \alpha$,then:

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