\((x^2+1)^2-x^2=0\) has
four real roots
two real roots
no real roots
one real root
Step 1: Start with the given equation:
\((x^2 + 1)^2 - x^2 = 0\)
Step 2: To make it easier, put \(y = x^2\). Then the equation becomes:
\((y + 1)^2 - y = 0\)
Step 3: Expand \((y + 1)^2\):
\(y^2 + 2y + 1\)
Step 4: Substitute back:
\(y^2 + 2y + 1 - y = 0\)
Step 5: Simplify:
\(y^2 + y + 1 = 0\)
Step 6: This is a quadratic equation. The discriminant \(D = b^2 - 4ac\). Here, \(a = 1, b = 1, c = 1\).
So, \(D = 1^2 - 4(1)(1) = 1 - 4 = -3\).
Step 7: Since the discriminant is negative (\(D < 0\)), this quadratic has no real solutions for \(y\).
Step 8: But \(y = x^2\). If \(y\) has no real value, then \(x^2\) also has no real value. That means \(x\) has no real solution.
Final Answer: The equation has no real roots.