If $\sin (\alpha)$ and $\cos (\alpha)$ are the roots of the equation $a x^{2}+b x+c=0$, then $b^{2}$ is

What we're given: $\sin(\alpha)$ and $\cos(\alpha)$ are roots of the quadratic equation $ax^2 + bx + c = 0$

What we need to find: The value of $b^2$

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Using the basic properties of quadratic equations:

For any quadratic equation $ax^2 + bx + c = 0$ with roots $p$ and $q$:

• Sum of roots $= p + q = -\dfrac{b}{a}$
• Product of roots $= p \times q = \dfrac{c}{a}$

Since our roots are $\sin(\alpha)$ and $\cos(\alpha)$:

$\sin(\alpha) + \cos(\alpha) = -\dfrac{b}{a}$ ... equation (1)

$\sin(\alpha) \times \cos(\alpha) = \dfrac{c}{a}$ ... equation (2)

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Using the trigonometric identity $\sin^2(\alpha) + \cos^2(\alpha) = 1$

Squaring both sides of equation (1):

$[\sin(\alpha) + \cos(\alpha)]^2 = \left(-\dfrac{b}{a}\right)^2$

Expanding the left side:

$\sin^2(\alpha) + 2\sin(\alpha)\cos(\alpha) + \cos^2(\alpha) = \dfrac{b^2}{a^2}$

Since $\sin^2(\alpha) + \cos^2(\alpha) = 1$:

$1 + 2\sin(\alpha)\cos(\alpha) = \dfrac{b^2}{a^2}$

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From equation (2), we know $\sin(\alpha)\cos(\alpha) = \dfrac{c}{a}$

Substituting:

$1 + 2\left(\dfrac{c}{a}\right) = \dfrac{b^2}{a^2}$

$1 + \dfrac{2c}{a} = \dfrac{b^2}{a^2}$

Multiplying everything by $a^2$:

$a^2 + 2ac = b^2$

Therefore, $b^2 = a^2 + 2ac$