Question

If  α, ß are the roots of x^2+x+1=0 then find the value of α-ß


fast answer ​

Answer 1

Answer:

Step-by-step explanation:

α  

2

+β  

2

=(α+β)  

2

−2αβ

=(a−2)  

2

+2(a+1)

=a  

2

−4a+4+2a+2

=a  

2

−2a+6

=(a−1)  

2

+5

Minimum value is at a=1.

Value of expression at a=1 is 5

Answer 2

Given:

x² + x + 1 = 0 is a quadratic equation whose roots are α and ß.

To find:

Value of α - ß

Solution:

Inorder to solve the given problem, firstly we will find the roots of the given equation and then by subtracting the both, we will get our desired result.

The general form of a quadratic equation is :

• ax² + bx + c = 0

On comparing the given equation with general form of quadratic equation, we get :

• a = 1
• b = 1
• c = 1

We will use quadratic formula to find the roots of the given equation.

Quadratic Formula

$\underline{\boxed{ \sf Roots = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}}$

By substituting the given values in this formula, we get :

$\sf Roots = \dfrac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}$

$\sf Roots = \dfrac{-1\pm\sqrt{1-4}}{2}$

$\sf Roots = \dfrac{-1\pm\sqrt{ - 3}}{2}$

$\sf Roots = \dfrac{-1\pm\sqrt{3} i}{2}$

Since negative numbers are not allowed inside square root function, √-3 = 3i because √-1 = i ( iota ) .

$\sf If \: \alpha = \dfrac{-1-\sqrt{3} i}{2} \: then \: \beta=\dfrac{-1+\sqrt{3}i}{2}$

Required answer will be,

$\implies \sf \alpha - \beta$

${ \implies \sf \dfrac{ - 1 - \sqrt{3}i }{2} - \left ( \dfrac{ - 1 + \sqrt{3} i}{2} \right)}$

${ \implies \sf \dfrac{ - 1 - \sqrt{3}i }{2} + \dfrac{ 1 - \sqrt{3} i}{2} }$

${ \implies \sf \dfrac{ - 1 - \sqrt{3}i + 1 - \sqrt{3} i}{2}}$

${ \implies \sf \dfrac{ - 2 \sqrt{3}i}{2}}$

${ \implies \sf - \sqrt{3}i}$

Or,

$\sf If \: \beta = \dfrac{-1-\sqrt{3} i}{2} \: then \: \alpha=\dfrac{-1+\sqrt{3}i}{2}$

Required answer will be,

$\implies \sf \alpha - \beta$

${ \implies \sf\dfrac{ - 1 + \sqrt{3} i}{2} - \left (\dfrac{ - 1 - \sqrt{3}i}{2}\right)}$

${ \implies \sf \dfrac{ - 1 + \sqrt{3} i}{2} + \dfrac{ 1 + \sqrt{3}i}{2}}$

${ \implies \sf \dfrac{ - 1 + \sqrt{3} i + 1 + \sqrt{3}i}{2}}$

${ \implies \sf\dfrac{\sqrt{3} i+ \sqrt{3}i}{2}}$

${ \implies \sf \dfrac{2\sqrt{3} i}{2}}$

${ \implies \sf \sqrt{3} i}$

∴ α - ß = √3i or - √3 i