Question

Solve the quadratic equation \(ax^2 + bx + c = 0\), \((a \neq 0)\) by the method of completing square.

Hint:

In completing the square, always make the coefficient of \(x^2\) equal to \(1\) first. Then add the square of half the coefficient of \(x\).

Answer 1

Step 1: Write the given quadratic equation.}
We are given the quadratic equation:
\[ ax^2 + bx + c = 0, \qquad a \neq 0 \]
Step 2: Divide the whole equation by \(a\).}
To make the coefficient of \(x^2\) equal to \(1\), divide every term by \(a\):
\[ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \]
Step 3: Shift the constant term to the right side.}
Subtract \(\frac{c}{a}\) from both sides:
\[ x^2 + \frac{b}{a}x = -\frac{c}{a} \]
Step 4: Complete the square.}
Take half of the coefficient of \(x\), which is \(\frac{b}{2a}\), and square it:
\[ \left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} \] Now add \(\frac{b^2}{4a^2}\) to both sides:
\[ x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2} \]
Step 5: Write the left side as a perfect square.}
The left-hand side becomes:
\[ \left(x + \frac{b}{2a}\right)^2 \] So, we get:
\[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \]
Step 6: Take square root on both sides.}
Taking square roots, we get:
\[ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \]
Step 7: Find the value of \(x\).}
Subtract \(\frac{b}{2a}\) from both sides:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Step 8: State the final result.}
Hence, the solution of the quadratic equation \(ax^2 + bx + c = 0\) is:
\[ \boxed{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}} \]