Question

If the roots of quadratic equation x^2+px+q=0 are cot 60 and cot 75 then the value of q+2-p​

Answer 1

Step-by-step explanation:

Given that:If the roots of quadratic equation x²+px+q=0 are cot 60° and cot 75° then the value of q+2-p

To find: value of q+2-p

Solution: To find the value of q+2-p

Use relation between roots of quadratic equation with coefficient of equation

Sum of zeroes

$cot60° + cot75° = - p \: \: \: ...eq1 \\ \\ cot60° \times cot75° = q \: \: \: ...eq2$

We know that formula of cot (A+B) is

$\boxed{\frac{ cotAcotB-1}{cotA + cotB} = cot(A + B)}$

Put the value from eq1 and eq2

$\frac{ cot60°cot75°-1}{cot60° + cot75°} = cot(60° + 75°) \\ \\ \because \: cot(135°) = cot(90° + 45°) = - tan45°\\ \\ \frac{ cot60°cot75°-1}{cot60° + cot75°} = - tan \: 45°$

put the value from eq1 and eq2

$\frac{q-1 }{ - p} = -1 \\ \\ q -1= p \\ \\ q-p = 1$

Add 2 both sides

$q-p+2 = 1+2\\ \\ \bold{q+2-p=3}$

Hope it helps you.

Answer 2

Given:

The roots of quadratic equation x^2+px+q=0 are cot 60 and cot 75

To find:

The value of q+2-p​

Solution:

From given, we have,

The roots of quadratic equation x^2+px+q=0 are cot 60 and cot 75.

we have,

cot 60 = 1/√3

cot 75 = 2 - √3

as the above are the roots of the equation, so we can express the above roots as,

(x - cot 65) (x - cot 75) = 0

(x - 1/√3) [x - (2 - √3)] = 0

x² - x (2 - √3) - 1/√3 x + (2 - √3)/√3 = 0

x² - x (2 - √3 + 1/√3) + (2 - √3)/√3 = 0

x² - x (2√3 - 3 + 1)/√3 + (2 - √3)/√3 = 0

x² - x (2√3 - 2)/√3 + (2 - √3)/√3 = 0

given, x² + px + q = 0

comparing the above equation with the given equation, we get,

p = - (2√3 - 2)/√3 and q = (2 - √3)/√3

then the value of q+2-p​ is

= (2 - √3)/√3 + 2 - [- (2√3 - 2)/√3]

= (2 - √3)/√3 + 2 + (2√3 - 2)/√3

= (2 - √3 + 2√3 - 2)/√3 + 2

= √3/√3 + 2

= 1 + 2

= 3

Therefore, the value of q + 2 - p is 3