Math · College algebra · Worked example
Solve a quadratic by completing the square
Solve 2x² + 8x − 10 = 0 by completing the square, then write the quadratic in vertex form.
Make the leading coefficient 1
Completing the square is simplest when x² has coefficient 1. Divide every term by 2; dividing both sides by a nonzero number keeps the same solutions.
Move the constant term
Add 5 to both sides so the x terms stand alone on the left.
Add the square of half the x coefficient
Half of 4 is 2, and 2² = 4. Adding 4 turns x² + 4x into the perfect square x² + 4x + 4. Add the same 4 to the right side to keep the equation balanced.
Write the left side as a square
Factor the perfect square. The equation now says that x + 2 is a number whose square is 9.
Take both square roots
A positive number has two square roots, so x + 2 = 3 or x + 2 = −3. Writing only +3 would lose a solution.
Solve each case and check
Subtract 2 in each case. Both values check in the original equation: 2(1)² + 8(1) − 10 = 0 and 2(−5)² + 8(−5) − 10 = 50 − 40 − 10 = 0.
Read the vertex form
The same steps rewrite y = 2x² + 8x − 10 in vertex form. Its vertex is (−2, −18), the lowest point of the parabola, halfway between the x-intercepts −5 and 1.
Result
x = 1 or x = −5. In vertex form, 2x² + 8x − 10 = 2(x + 2)² − 18, with vertex (−2, −18).
Your turn
Solve x² − 8x + 3 = 0 by completing the square.
Show the answer and explanation
x = 4 ± √13.
Move the constant: x² − 8x = −3. Half of −8 is −4 and (−4)² = 16, so add 16 to both sides: (x − 4)² = 13. Then x − 4 = ±√13 and x = 4 ± √13, about 7.606 and 0.394.
Keep exploring
Open the steps in Math and replace the 10 with 16: completing the square then gives (x + 2)² = 12, and the roots become irrational.
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