Question

If a & b are the quadratic equation x^2+5x-1=0 find a^2+b^2

Answer 1

Answer:

27

Step-by-step explanation:

x^2 + 5x -1

sum of roots = α+β = -5

product of roots = αβ = -1

(α+β)² = α² + β² + 2αβ

(-5)² = α² + β² + 2(-1)

α² + β² = 25 + 2 = 27

Answer 2

Answer :

a² + b² = 27

Note:

★ The possible values of the variable which satisfy the equation are called its roots or solutions .

★ A quadratic equation can have atmost two roots .

★ The general form of a quadratic equation is given as ; Ax² + Bx + C = 0

★ If a and b are the roots of the quadratic equation Ax² + Bx + C = 0 , then ;

• Sum of roots , (a + b) = -B/A

• Product of roots , (ab) = C/A

★ If a and b are the roots of a quadratic equation , then that quadratic equation is given as : k•[ x² - (a + b)x + ab ] = 0 , k ≠ 0.

★ The discriminant , D of the quadratic equation Ax² + Bx + C = 0 is given by ;

D = B² - 4AC

★ If D = 0 , then the roots are real and equal .

★ If D > 0 , then the roots are real and distinct .

★ If D < 0 , then the roots are unreal (imaginary) .

Solution :

Here ,

The given quadratic equation is ;

x² + 5x - 1 = 0

Now ,

Comparing the given quadratic equation with the general quadratic equation Ax² + Bx + C = 0 ,

We have ;

A = 1

B = 5

C = -1

Also ,

It is given that , a and b are the roots of the given quadratic equation .

This ,

=> Sum of roots = -B/A

=> a + b = -5/1

=> a + b = -5

Also ,

=> Product of zeros = C/A

=> ab = -1/1

=> ab = -1

Now ,

=> (a + b)² = a² + b² + 2ab

=> (-5)² = a² + b² + 2•(-1)

=> 25 = a² + b² - 2

=> a² + b² = 25 + 2

=> a² + b² = 27

Hence ,

a² + b² = 27