Question

if sintheta and sectheta (0<theta<90) are the roots of the equation √3x^2+kx+3=0.find k​

Answer 1

Step-by-step explanation:

Given:if sintheta and sectheta (0<theta<90) are the roots of the equation √3x²2+kx+3=0

To find:find k

Solution:

We know that,there is a relation between roots of quadratic equation and coefficient of x²,x and constant term.

So,apply the relation

$\text{sum \: of \: zeros} = \frac{ - b}{a} \\ \\ \text{multiplication \: of \: zeros} = \frac{ c}{a}$

Here,

$\sqrt{3} {x}^{2} + kx + 3 = 0 \\ \\ a = \sqrt{3} \\ b = k \\ c = 3 \\ \\ zeroes \: are \: sin \theta \: and \: sec\theta \\ \\ sin \theta + sec\theta = \frac{ - k}{ \sqrt{3} } \: \: \: ...eq1\\ \\ sin \theta \times sec\theta = \frac{ 3}{ \sqrt{3} }...eq2$

We know that

$sec\theta = \frac{ 1}{ cos\theta }$

So,put this to eq2

$sin \theta \times \frac{1}{ cos\theta} = \sqrt{3} \\ \\ tan\theta = \sqrt{3} \\ \\ tan\theta = tan60° \\ \\ \theta = 60°$

Put this value in eq1

and value of angle 60° for both trigonometric functions

$sin 60°+ sec60° = \frac{ - k}{ \sqrt{3} } \\ \\ \frac{ \sqrt{3} }{2} + 2 = \frac{ - k}{ \sqrt{3} } \\ \\ - k = \frac{ \sqrt{3} (\sqrt{3} + 4)}{2} \\ \\\bold{ k = \frac{ - \sqrt{3} (\sqrt{3} + 4)}{2} } \\ \\ or\\\\\bold{ k = \frac{ - (3 +4\sqrt{3} )}{2} }$

Hope it helps you.

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