Question

if tan theta , see theta are roots of ax^2+bx+c=0 then prove that a^4 =b^2 (4ac-b^2)

Answer 1

Answer:

Given that tanθ and secθ are the roots of quadratic equation $ax^2+bx+c=0$

As we know that for quadratic equation $ax^2+bx+c=0$

sum of roots = $-\frac{b}{a}$

And product of roots = $\frac{c}{a}$

Therefore,

$\tan\theta+\sec\theta = -\frac{b}{a}$

⇒ $\frac{1+\sin\theta}{\cos\theta} =-\frac{b}{a}$      ..... (1)

Also, $\tan\theta \times\sec\theta=\frac{c}{a}$

⇒ $\frac{\sin\theta}{\cos^2\theta} =\frac{c}{a}$        ....... (2)

Squaring eq (1)

$(\frac{1+\sin\theta}{\cos\theta})^2 =(-\frac{b}{a})^2$

or, $\frac{(1+\sin\theta)^2}{\cos^2\theta} =\frac{b^2}{a^2}$      ...... (3)

Dividing (3) by (2) we get

$\frac{(1+\sin\theta)^2}{\sin\theta} =\frac{b^2}{ac}$         ............. (4)

Subtracting both the sides from 4

$4-\frac{(1+\sin\theta)^2}{\sin\theta} =4-\frac{b^2}{ac}$    

$\frac{4\sin\theta-(1+\sin\theta)^2}{\sin\theta} =\frac{4ac-b^2}{ac}$

or, $\frac{(1-\sin\theta)^2}{\sin\theta} =\frac{4ac-b^2}{ac}$              ............. (5)

Dividing eq (5) by eq (4)

$\frac{(1-\sin\theta)^2}{(1+\sin\theta)^2} =\frac{4ac-b^2}{b^2}$         ....... (6)

From equation (1)

$\frac{1+\sin\theta}{\cos\theta} =-\frac{b}{a}$

Squaring both sides

$(\frac{1+\sin\theta}{\cos\theta})^2 =(-\frac{b}{a})^2$

Squaring on both sides again

$(\frac{1+\sin\theta}{\cos\theta})^4 =(-\frac{b}{a})^4$

or, $\frac{(1+\sin\theta)^4}{\cos^4\theta} =\frac{b^4}{a^4}$

or, $\frac{(1+\sin\theta)^4}{\cos^2\theta \times \cos^2\theta} =\frac{b^4}{a^4}$

or, $\frac{(1+\sin\theta)^4}{(1-\sin^2\theta) \times (1-\sin^2\theta)} =\frac{b^4}{a^4}$

or, $\frac{(1+\sin\theta)^4}{(1-\sin\theta)^2 \times (1+\sin\theta)^2} =\frac{b^4}{a^4}$

or, $\frac{(1+\sin\theta)^4}{(1-\sin\theta)^2} =\frac{b^4}{a^4}$        ........... (7)

from equation (6) and (7)

$\frac{4ac-b^2}{b^2}=\frac{a^4}{b^4}$

⇒ $a^4=b^2(4ac-b^2)$       (Hence Proved)

Answer 2

Answer:

a⁴=b²(b²-4ac)

This is a very simple answer, you'll like it!

The key is to know the formula for difference of roots of a quadratic equation.

The modulus of difference of roots is given by √D over a.

If X1 and X2 are roots, then

|X1-X2|=(√D)/4a

Here, D is the DISCRIMINANT or DELTA (∆) equal to b²-4ac.

Step-by-step explanation:

WKT, sec²p-tan²p=1. (take theta in the question as p, I don't have a theta on my keyboard)

=> (sec p-tan p)(sec p+tan p) =1

=> (sec p-tan p)=1/(sec p+tan p)

=> (√D)/a = -a/b

{Sum of roots= -b/a, Difference of roots= (√D)/4a}

=> D/a²=a²/b²

=> b²-4ac = a⁴/b²

=> b⁴-4ab²c=a⁴

=> a⁴-b⁴+4ab²c=0

The question should actually be a⁴=b²(b²-4ac) which is in the 3rd step from the end.

Hope you liked the answer!