Q. (y^2 = x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1.)

Problem

Solve the equation:

\(y^2 = x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1\)

Step 1: Identify the variable we need to solve for

The equation is:

\(y^2 = x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1\)

The variable \(y\) is squared. That means we need to undo the square in order to solve for \(y\).

The inverse operation of squaring is taking the square root.

Step 2: Take the square root of both sides

Start with:

\(y^2 = x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1\)

Now take the square root of both sides:

\(\sqrt{y^2} = \sqrt{x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1}\)

The left side simplifies, but we must be careful. The square root of \(y^2\) gives the absolute value of \(y\):

\(\sqrt{y^2} = |y|\)

So the equation becomes:

\(|y| = \sqrt{x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1}\)

Step 3: Remove the absolute value

If \(|y|\) equals a nonnegative expression, then \(y\) can be either the positive value or the negative value of that expression.

For example:

\(5^2 = 25\)

\((-5)^2 = 25\)

Both \(5\) and \(-5\) have the same square. That is why every time we solve an equation of the form \(y^2 = A\), we write:

\(y = \pm \sqrt{A}\)

In this problem, the expression \(A\) is:

\(x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1\)

Therefore:

\(y = \pm \sqrt{x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1}\)

Step 4: Check whether the expression under the square root can be simplified

Now we check whether the polynomial under the square root is a perfect square.

The expression is:

\(x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1\)

Since the highest power is \(x^6\), a possible square would come from a cubic expression, because:

\((x^3)^2 = x^6\)

So we can compare the given expression with the square of a general cubic expression.

Suppose the expression were a perfect square:

\((x^3 + ax^2 + bx + c)^2\)

Expand this form:

\((x^3 + ax^2 + bx + c)^2 = x^6 + 2ax^5 + (a^2 + 2b)x^4 + (2ab + 2c)x^3 + (b^2 + 2ac)x^2 + 2bcx + c^2\)

Now compare this with the given polynomial:

\(x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1\)

Match the coefficients one by one.

First, compare the \(x^5\) terms:

\(2a = 4\)

So:

\(a = 2\)

Next, compare the \(x^4\) terms:

\(a^2 + 2b = 6\)

Substitute \(a = 2\):

\(2^2 + 2b = 6\)

\(4 + 2b = 6\)

\(2b = 2\)

\(b = 1\)

Now compare the constant terms:

\(c^2 = 1\)

So:

\(c = 1\) or \(c = -1\)

Next, compare the \(x\) terms:

\(2bc = 2\)

Since \(b = 1\), we get:

\(2(1)c = 2\)

\(2c = 2\)

\(c = 1\)

So the only possible perfect-square form is:

\((x^3 + 2x^2 + x + 1)^2\)

Now expand it to check:

\((x^3 + 2x^2 + x + 1)^2\)

Using the expansion formula, with \(a = 2\), \(b = 1\), and \(c = 1\):

\((x^3 + 2x^2 + x + 1)^2 = x^6 + 4x^5 + 6x^4 + 6x^3 + 5x^2 + 2x + 1\)

This is not the same as the original polynomial, because the original polynomial has:

\(2x^3 + x^2\)

But the square gives:

\(6x^3 + 5x^2\)

Therefore, the expression under the square root is not a simple perfect square.

Step 5: Write the two solution branches

Since the square root cannot be simplified as a perfect square, we keep it under the radical.

The positive solution is:

\(y = \sqrt{x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1}\)

The negative solution is:

\(y = -\sqrt{x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1}\)

Step 6: Consider the real-number restriction

If we want real-number values of \(y\), the expression inside the square root must be greater than or equal to zero.

So we need:

\(x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1 \ge 0\)

This condition tells us which \(x\)-values give real values of \(y\).

For values of \(x\) where the polynomial is negative, the square root is not real. For values of \(x\) where the polynomial is zero, there is one real value:

\(y = 0\)

For values of \(x\) where the polynomial is positive, there are two real values:

\(y = \sqrt{x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1}\)

and

\(y = -\sqrt{x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1}\)

Final answer

The equation solved for \(y\) is:

\(y = \pm \sqrt{x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1}\)

So the two solutions are:

\(y = \sqrt{x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1}\)

and

\(y = -\sqrt{x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1}\)

For real-number solutions, the expression under the square root must satisfy:

\(x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1 \ge 0\)