given-a-quadratic-equation-3x-2-x-t-2-4t-3-0-has-roots-sin-and-cos-find-the-value-t-2-4t-5-

Question Number 77180 by jagoll last updated on 04/Jan/20

given a quadratic equation

{3 x}^{2} - x + (t^{2} - 4 t + 3) = 0 has

roots \sin α and \cos α . find the

value \sqrt{t^{2} - 4 t + 5} .

Answered by mr W last updated on 04/Jan/20

\sin α + \cos α = \frac{1}{3} \dots (i)

\sin α \cos α = \frac{t^{2} - 4 t + 3}{3} \dots (i i)

{(i)}^{2} :

1 + 2 \sin α \cos α = \frac{1}{9}

1 + 2 \times \frac{t^{2} - 4 t + 3}{3} = \frac{1}{9}

t^{2} - 4 t + 3 = - \frac{4}{3}

t^{2} - 4 t + 5 = 2 - \frac{4}{3} = \frac{2}{3}

\Rightarrow \sqrt{t^{2} - 4 t + 5} = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}

Commented by jagoll last updated on 04/Jan/20

thanks sir

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