Question

If the roots of the quadratic equation x^2+px+q=0 are tan 30 degrees and tan 15 degrees then find the value of 2+q-p.

Answer 1

Answer:

The answer is 3.

Step-by-step explanation:

Given,

tan 30° and tan 15° are the roots of quadratic equation x² + px + q,

Thus, we can write,

$tan 30^{\circ}+tan15^{\circ} = -\frac{p}{1}=-p------(1)$

$tan 30^{\circ}.tan 15^{\circ} = \frac{q}{1}=q-------(2)$

Now, we know that,

tan (30° + 15°) = tan 45°

$\frac{tan 30^{\circ}+tan 15^{\circ}}{1-tan 30^{\circ}.tan 15^{\circ}}=1$

$tan 30^{\circ}+tan 15^{\circ} = 1-tan 30^{\circ}.tan 15^{\circ}$

From Equation (1) and (2),

-p = 1 - q

⇒ q - p = 1

Adding 2 on both sides,

We get,

2 + q - p = 3

Answer 2

Answer:

Value of

$q-p+2=3$

Explanation:

Given

Roots of quadratic equation x²+px+q=0 are tan30 and tan15.

Compare this with

ax²+bx+c=0, we get

a = 1, b = p , c = q

i) Sum of the zeroes = $\frac{-b}{a}$

$tan30+tan15=\frac{-p}{1}$ ---(1)

ii) Product of the zeroes = $\frac{c}{a}$

$tan30tan15=\frac{q}{1}$ ----(2)

_______________________

We know that,

$\boxed {tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}}$

_______________________

Here,

$tan(30+15)=\frac{tan30+tan15}{1-tan30tan15}$

$\implies tan45=\frac{tan30+tan15}{1-tan30tan15}$

$\implies 1=\frac{-p}{1-q}$ /* From (1)&(2) */

$\implies 1-q = -p$

$\implies 1 = q-p$

Add 2 both sides of the equation, we get

$\implies 1+ 2 = q-p+2$

$\implies 3 = q-p+2$

Therefore,

Value of

$q-p+2=3$

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