Does (x-1)²+2(x+1)=0have a real roots ? Justify your answer
Answers
Answer:
Roots of this equation are not real.
Step-by-step explanation:
= > ( x - 1 )^2 + 2( x + 1 ) = 0
Method 1
From the properties of expansion :
- ( a - b )^2 = a^2 + b^2 - 2ab
= > [ ( x )^2 + 1 - 2( x × 1 ) + 2( x + 1 ) = 0
= > [ x^2 + 1 - 2x ] + 2x + 2 = 0
= > x^2 + 1 - 2x + 2x + 2 = 0
= > x^2 + 1 + 2 = 0
= > x^2 + 3 = 0
= > x^2 = - 3
= > x = ±√( - 3 )
Square root of any negative number doesn't exists in real, so the given equation doesn't have real roots.
Method 2
= > [ ( x )^2 + 1 - 2( x × 1 ) + 2( x + 1 ) = 0
= > [ x^2 + 1 - 2x ] + 2x + 2 = 0
= > x^2 + 1 - 2x + 2x + 2 = 0
= > x^2 + 1 + 2 = 0
= > x^2 + 3 = 0
We know, discriminant of an equation shows whether that equation has real roots, two distinct root or no real roots.
Also, we know discriminant of equation ax^2 + bx + c is b^2 - 4ac.
On comparing the given equation with ax^2 + bx + c = 0, we get :
a = 1 ; b = 0 ; c = 3
Thus,
= > Discriminant = ( 0 )^2 - 4( 1 × 3 )
= > Discriminant = 0 - 12
= > Discriminant = - 12
As the discriminant of this equation is less than 0, roots of this equation are not real.
Answer:
no it does not have a real root
