\((x+2)(x-1)=(x+1)(x-3)\)
\((x+1)^3=x^3-2\)
Step 1: Expand Each Given Equation
Option 1: \((x+1)^3 = x^3 - 2\)
Expanding the left-hand side:
\((x+1)^3 = x^3 + 3x^2 + 3x + 1\)
So the equation becomes:
\(x^3 + 3x^2 + 3x + 1 = x^3 - 2\)
Subtracting \(x^3\) from both sides:
\(3x^2 + 3x + 1 = -2\)
\(3x^2 + 3x + 3 = 0\)
Since this contains \(x^3\) initially, it is not a quadratic equation.
Option 2: \((x+1)^2 = 3(x-2)\)
Expanding both sides:
\(x^2 + 2x + 1 = 3x - 6\)
Rearranging:
\(x^2 - x + 7 = 0\)
This is a quadratic equation.
Option 3: \((x+2)^2 + 3 = x - 1\)
Expanding:
\(x^2 + 4x + 4 + 3 = x - 1\)
Rearranging:
\(x^2 + 4x + 7 - x + 1 = 0\)
\(x^2 + 3x + 8 = 0\)
This is a quadratic equation.
Option 4: \((x+2)(x-1) = (x+1)(x-3)\)
Expanding both sides:
\(x^2 - x + 2x - 2 = x^2 - 3x + x - 3\)
\(x^2 + x - 2 = x^2 - 2x - 3\)
Cancel \(x^2\) from both sides:
\(x - 2 = -2x - 3\)
Rearrange:
\(x + 2x = -3 + 2\)
\(3x = -1\)
\(x = -\frac{1}{3}\)
This is a linear equation, not a quadratic equation.
Final Answer: \((x+1)^3 = x^3 - 2\)
To determine which equation is not a quadratic equation, we must recognize a quadratic equation typically takes the form \(ax^2+bx+c=0\), where \(a \neq 0\).
Let's analyze each option:
\((x+2)(x-1)=(x+1)(x-3)\)
Expanding both sides, \(x^2+x-2\) and \(x^2-2x-3\), respectively. Setting them equal gives \(x^2+x-2=x^2-2x-3\).
Simplifying, we get a common \(x^2\) term, leading to \(3x=1\), a linear equation. The expression equals zero, confirming it's a quadratic equation setup.
\((x+1)^2=3(x-2)\)
Expanding gives \(x^2+2x+1=3x-6\).
Rearranging all terms to one side, \(x^2-x+7=0\), shows it is in standard quadratic form.
\((x+2)^2+3=x-1\)
Expanding, we have \(x^2+4x+7=x-1\).
Rearranging, \(x^2+3x+8=0\), confirms a quadratic form.
\((x+1)^3=x^3-2\)
Expanding, we get \(x^3+3x^2+3x+1=x^3-2\).
Upon simplification, \(3x^2+3x+3=0\), clearly representing a cubic term.
This implies it is not a quadratic equation.
Therefore, the non-quadratic equation is \((x+1)^3=x^3-2\).
If the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are real and equal, then: