Question

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

Answer 1

the given polynomial y has 2 real roots

Explanation:

y=2(x+1)^2

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

y=2x^2+2+4x

y=2x^2+2x+2x+2

y=2x(x+1)+2(x+1)

y=(x+1)(2x+2)

Answer 2

The equation Y = 2(x+1)^2 represents a quadratic equation in the form of Y = a(x-h)^2 + k, where a, h, and k are constants.

To determine the number of real roots, we can analyze the discriminant of the quadratic equation. The discriminant is the part of the quadratic formula that lies inside the square root (√).

In this case, the discriminant is given by:

b^2 - 4ac

Since the equation Y = 2(x+1)^2 does not have a linear term (x), the coefficient b is equal to 0.

Therefore, the discriminant simplifies to:

0^2 - 4(2)(1)

Which further simplifies to:

0 - 8

Finally, we have:

-8

The value of the discriminant is negative (-8), which indicates that the quadratic equation Y = 2(x+1)^2 does not have any real roots.

Instead, it has complex roots, which involve imaginary numbers.

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