Solving by completing the square

A perfect square trinomial such as $x^2+6x+9$ factors as $(x+3)^2$ because its constant term, $9$, is the square of half the coefficient of $x$ — here, $\left(\frac{6}{2}\right)^2=9$. This pattern always holds: expanding $(x+p)^2$ gives $x^2+2px+p^2$, so the constant term is always the square of half the middle coefficient. Completing the square exploits this pattern in reverse: given $x^2+6x$, add the missing constant $9$ to make it a perfect square. Since an equation must stay balanced, whatever is added to the left must also be added to the right — this is the same balancing rule used whenever a term is moved across an equation, applied here to a term that is added rather than simply moved. For the general equation this pattern needs one extra step first. If $a\neq1$, every term is divided by $a$ so the equation starts in the form $x^2+(\text{something})x=(\text{something})$ before the "add half the middle coefficient, squared" trick can be applied:

where: $a$, $b$, $c$ are the coefficients of the quadratic ($a\neq0$), and $\left(\dfrac{b}{2a}\right)^2$ is the term added to both sides — half the coefficient of $x$ (after dividing by $a$), squared. The left side is now guaranteed to be a perfect square, $\left(x+\dfrac{b}{2a}\right)^2$, no matter what $a$, $b$, $c$ are. Once the equation reads $(x+k)^2=m$, two cases follow. If $m\geq0$, take square roots of both sides — remembering both the positive and negative root — to get $x+k=\pm\sqrt{m}$, then solve for $x$. If $m<0$, no real number squares to a negative value, so the equation has no real roots; this is stated as a conclusion, not worked around. Carrying the same steps through with letters instead of numbers, as the derivation below shows, produces the quadratic formula met in the next topic — nothing new is introduced there; it is exactly this method, done once in general. The quantity $b^2-4ac$ that appears under the root is called the discriminant: it is the same $m$ that must stay non-negative for real roots to exist.

Solve $x^2+6x-7=0$ by completing the square, and identify the value of $k$ and $m$ in $(x+k)^2=m$. This example needs a full walk-through because it must show, in one continuous chain, how moving the constant across, adding the completing term to both sides, and finally taking a signed square root each contribute a necessary piece to reaching both roots correctly — skipping or mishandling any single step below produces a wrong or incomplete answer.

Step 1. The coefficient of $x^2$ is already $1$, so no division by $a$ is needed. Move the constant term to the right side: $x^2+6x=7$.

Step 2. Half the coefficient of $x$ is $\frac{6}{2}=3$, so the completing term is $3^2=9$. Add it to both sides: $x^2+6x+9=7+9$, which simplifies to $x^2+6x+9=16$.

Step 3. The left side is now the perfect square $(x+3)^2$, so the equation reads $(x+3)^2=16$ — here $k=3$ and $m=16$.

Step 4. Take square roots of both sides, keeping the $\pm$: $x+3=\pm4$.

Step 5. Solve the two cases separately: $x=-3+4=1$ or $x=-3-4=-7$.

The roots are $x=1$ and $x=-7$. Checking both in the original equation: $1^2+6(1)-7=0$ and $(-7)^2+6(-7)-7=49-42-7=0$, so both satisfy the equation.

Competitive stretch

For $2x^2-8x+3=0$, the coefficient of $x^2$ is $2$, not $1$, so the division step from Step 1 above is essential this time. Dividing every term by $a=2$ first gives $x^2-4x+\frac{3}{2}=0$, so $x^2-4x=-\frac{3}{2}$. Half of $-4$ is $-2$, squared is $4$, so $x^2-4x+4=-\frac{3}{2}+4=\frac{5}{2}$, giving $(x-2)^2=\frac{5}{2}$. Taking roots, $x=2\pm\sqrt{2.5}$ — two irrational roots left exactly in surd form rather than rounded, since $2.5$ is not a perfect square. Notice that skipping the division and instead adding $\left(\frac{-8}{2}\right)^2=16$ directly to $2x^2-8x+3=0$ would have produced a wrong, non-perfect-square left side — a direct instance of the first common trap below.

• Adding $\left(\dfrac{b}{2}\right)^2$ instead of $\left(\dfrac{b}{2a}\right)^2$ when $a\neq1$ — the equation must be divided by $a$ first, or the completing term is wrong.
• Adding the completing term to only one side. It must go on both sides of the equals sign to keep the equation balanced.
• Dropping the $\pm$ when taking a square root, which silently discards one of the two valid roots.
• Trying to take the square root of a negative $m$. If $(x+k)^2=m$ has $m<0$, the correct conclusion is "no real roots," not an invented answer.
• Rounding a surd answer too early. When $m$ is not a perfect square, leave the roots as $x=-k\pm\sqrt{m}$ in exact form unless the question explicitly asks for a decimal approximation.

Ask the tutor: Stuck on which constant to add for a particular equation, or want completing the square worked line by line for one you are practising? Ask the AI tutor to walk through it with you.