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Question: How do you use the quadratic formula to solve \({\tan ^2}\theta + 3\tan \theta + 1 = 0\)for \(0 \leq...

How do you use the quadratic formula to solve tan⁡2θ+3tan⁡θ+1=0{\tan ^2}\theta + 3\tan \theta + 1 = 0for 0⩽θ⩽360?0 \leqslant \theta \leqslant 360?

Explanation

Solution

Hint : First for all here we will compare the given tangent function as “a” and then Here we will use the standard quadratic equation and will compare the given equation with it and then will find the roots of the given equation by using the formulax=−b±Δ2ax = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}.

Complete step-by-step answer :
Let us assume the tangent function as “g”
i.e. tan⁡θ=g\tan \theta = g
Therefore, the given expression can be written as –
g2+3g+1=0{g^2} + 3g + 1 = 0
Let us consider the general form of the quadratic equation.
ax2+bx+c=0a{x^2} + bx + c = 0
Compare the standard equation with the given equation g2+3g+1=0{g^2} + 3g + 1 = 0 ..... (A)
⇒a=1 ⇒b=3 ⇒c=1  \Rightarrow a = 1 \\\ \Rightarrow b = 3 \\\ \Rightarrow c = 1 \\\
Now, Δ=b2−4ac\Delta = {b^2} - 4ac
Place the values in the above equation –
Δ=32−4(1)(1)\Delta = {3^2} - 4(1)(1)
Simplify the above equation –
Δ=9−4\Delta = 9 - 4
Simplify the above equation –
Δ=5\Delta = 5
Take root on both the sides of the equation –
⇒Δ=5\Rightarrow \sqrt \Delta = \sqrt 5
Now, the roots can be expressed as
g=−b±Δ2ag = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}
Therefore, g=−b+Δ2ag = \dfrac{{ - b + \sqrt \Delta }}{{2a}} or g=−b−Δ2ag = \dfrac{{ - b - \sqrt \Delta }}{{2a}}
Place the known values in the above equation –
g=−(3)+52g = \dfrac{{ - (3) + \sqrt 5 }}{2} or g=−(3)−52g = \dfrac{{ - (3) - \sqrt 5 }}{2}
Replace the value of “g” in the above equation –
tan⁡θ=−(3)+52\tan \theta = \dfrac{{ - (3) + \sqrt 5 }}{2} or tan⁡θ=−(3)−52\tan \theta = \dfrac{{ - (3) - \sqrt 5 }}{2}
Make the required angle the subject –
θ=tan⁡−1(−(3)+52)\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - (3) + \sqrt 5 }}{2}} \right) or θ=tan⁡−1(−(3)−52)\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - (3) - \sqrt 5 }}{2}} \right)
This is the required solution.
So, the correct answer is “θ=tan⁡−1(−(3)+52)\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - (3) + \sqrt 5 }}{2}} \right) or θ=tan⁡−1(−(3)−52)\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - (3) - \sqrt 5 }}{2}} \right)”.

Note : A quadratic equation is an equation of second degree, it means at least one of the terms is squared. Standard equation is ax2+bx+c=0a{x^2} + bx + c = 0 where a, b and c are constant and “a” can never be zero and “x” is unknown. Always remember the standard equation and formula for the sum and product of the roots properly. Follow the given data and conditions carefully and simplify. Every quadratic polynomial has almost two roots. Remember, the quadratic equations having coefficients as the rational numbers has the irrational roots. Also, the quadratic equations whose coefficients are all the distinct irrationals but both the roots are the rational.