Question

Y=2(x+1)^2 has how many real roots

Answer 1

The equation y= 2$(x+1)^2$ has one real root and that is x=-1.

What is real roots of the equation?

    We are aware that when we resolve a linear or quadratic equation, we always arrive at the value variable of the equation, or, to put it another way, we always locate the equation's solution. This "solution" is what we refer to as the real roots. For instance, when the equation $X^2$-7x+12=0 is solved, the actual roots are 3 and 4.

Here given,

=> y = 2$(x+1)^2$

Take y=0 then,

=> 2$(x+1)^2$=0

=> $(x+1)^2$=0

=>(x+1)=0

=> x=-1

Hence the given equation has one real root and that is x=-1.

To learn more about real roots refer the below link

brainly.com/question/24147137

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Answer 2

The answer will be:

The equation has one real root. And the root is -1/2.