Question

write nature of roots find if they are real. x2 -(√2+1)x + √2 = 0​

Answer 1

Answer:

answer for the given problem is given

Answer 2

Answer :

Real and distinct

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 α and ß are the roots of the quadratic equation ax² + bx + c = 0 , then ;

• Sum of roots , (α + ß) = -b/a

• Product of roots , (αß) = c/a

★ If α and ß are the roots of a quadratic equation , then that quadratic equation is given as : k•[ x² - (α + ß)x + αß ] = 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 :

The given quadratic equation is ;

x² - (√2 + 1)x + √2 = 0 .

Comparing the given quadratic equation with the general quadratic equation ax² + bx + c = 0 ,

We have ;

a = 1

b = -(√2 + 1)

c = √2

Now ,

The discriminant of the given quadratic equation will be ;

=> D = b² - 4ac

=> D = [-(√2 + 1) ]² - 4×1×√2

=> D = (√2 + 1)² - 4√2

=> D = √2² + 1² + 2√2 - 4√2

=> D = 2 + 1 - 2√2

=> D = 3 - 2√2

=> D = 3 - 2×1.414

=> D = 3 - 2.828

=> D = 0.172

=> D > 0

Since ,

The discriminant of the given quadratic equation is greater than zero .

Hence ,

The roots of the given quadratic equation are real and distinct .