Question

Does (x-1)²+2(x+1)=0 have a real root. Justify

Answer 1

Hey there!!!
------------------
NO there are no real roots
By opening the brackets we get
x^2+1-2x+2x+2=0
x^2+3=0
x^2=-3
Not possible and also by quadratic formula
We get D as negative indicating it to have no real roots
----------------------
Hope this helps.

Answer 2

Given,

A polynomial p(x) = (x-1)²+2(x+1)

To find,

If p(x) has any real roots.

Solution,

We can simply solve this mathematical problem using the following process:

Mathematically,

If (x+a) and (x+b) are two factors of a quadratic equation p(x), then the condition p(x) = (x+a)(x+b) = 0 must be satisfied.

On simplifying the given polynomial, we get;

p(x) = (x-1)²+2(x+1)

= x²+1-2x+2x+2

= x²+3

Now,

Let us assume that p(x) has real roots.

p(x)=0

=> x²+3 = 0

=> x² = -3

=> x = √(-3)

But, mathematically, root over of a negative integer is not possible. That is, no real value of x exists that can become a factor of p(x).

Hence, p(x) has no real roots.