The equations 5x^2 + 12x + 13 = 0 and ax^2 + | vedclass

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(C) The roots of the quadratic equation $5x^2 + 12x + 13 = 0$ are imaginary because the discriminant $D = 12^2 - 4(5)(13) = 144 - 260 = -116 < 0$.Sinc

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) The roots of the quadratic equation $5x^2 + 12x + 13 = 0$ are imaginary because the discriminant $D = 12^2 - 4(5)(13) = 144 - 260 = -116 Since the coefficients $a, b, c$ are real,if one root is complex,the other must be its conjugate. Thus,both roots are common.This implies the ratios of the coefficients are equal: $\frac{a}{5} = \frac{b}{12} = \frac{c}{13} = k$ (where $k > 0$).Thus,$a = 5k$,$b = 12k$,and $c = 13k$.Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the values: $\cos C = \frac{(5k)^2 + (12k)^2 - (13k)^2}{2(5k)(12k)} = \frac{25k^2 + 144k^2 - 169k^2}{120k^2} = \frac{0}{120k^2} = 0$.Since $\cos C = 0$,we have $\angle C = 90^\circ$.

The equations $5x^2 + 12x + 13 = 0$ and $ax^2 + bx + c = 0$ have a common root,where $a, b, c$ are the sides of $\Delta ABC$. Find $\angle C$ in degrees.

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