Questions: Solve the quadratic equation. Check the answers. x^2 + x - 3 = 3

Solution Steps

To solve the quadratic equation \(x^2 + x - 3 = 3\), we first need to bring it to the standard form \(ax^2 + bx + c = 0\). Then, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the solutions. Finally, we will check the solutions by substituting them back into the original equation.

1. Rewrite the equation in standard form.
2. Identify coefficients \(a\), \(b\), and \(c\).
3. Use the quadratic formula to find the solutions.
4. Check the solutions by substituting them back into the original equation.

The given equation is: \[ x^2 + x - 3 = 3 \] Subtract 3 from both sides to bring it to the standard form \(ax^2 + bx + c = 0\): \[ x^2 + x - 6 = 0 \]

From the standard form \(x^2 + x - 6 = 0\), we identify the coefficients: \[ a = 1, \quad b = 1, \quad c = -6 \]

The discriminant \(\Delta\) is given by: \[ \Delta = b^2 - 4ac \] Substituting the values of \(a\), \(b\), and \(c\): \[ \Delta = 1^2 - 4 \cdot 1 \cdot (-6) = 1 + 24 = 25 \]

The solutions to the quadratic equation are given by: \[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \] Substituting the values of \(a\), \(b\), and \(\Delta\): \[ x = \frac{-1 \pm \sqrt{25}}{2 \cdot 1} = \frac{-1 \pm 5}{2} \] This gives us two solutions: \[ x_1 = \frac{-1 + 5}{2} = \frac{4}{2} = 2 \] \[ x_2 = \frac{-1 - 5}{2} = \frac{-6}{2} = -3 \]

Substitute \(x_1 = 2\) back into the original equation: \[ 2^2 + 2 - 3 = 4 + 2 - 3 = 3 \] Substitute \(x_2 = -3\) back into the original equation: \[ (-3)^2 + (-3) - 3 = 9 - 3 - 3 = 3 \] Both solutions satisfy the original equation.

Final Answer