Completing the Square and Equivalent Forms
Rewriting a quadratic in vertex form reveals its vertex instantly — and the SAT often asks you to match equivalent forms.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
Completing the square turns $x^2 + bx + c$ into $(x + h)^2 + k$. Whatever you add to complete the square must be balanced so the expression stays equal.
How to solve one, step by step
Example: rewrite $x^2 + 6x + 5$ in vertex form.
- Half of $6$ is $3$; $3^2 = 9$.
- Write $(x + 3)^2 - 9 + 5 = (x + 3)^2 - 4$.
The mistakes that cost points
- Not balancing the added term. Adding a number to complete the square requires subtracting it back to keep the value the same.
- Confusing factored and vertex form. Factored form shows the zeros; vertex form shows the vertex — they look different on purpose.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
For what value of $m$ does the equation $x^2 + mx + 25 = 0$ have two unequal real solutions?
- A$m = 9$
- B$m = -11$
- C$m = 10$
- D$m = -9$
Show solution
Answer: B, $m = -11$. STEP 1: The equation is already in standard form: $x^2 + mx + 25 = 0$.
STEP 2: Identify the discriminant. For $x^2 + mx + 25 = 0$, the discriminant is $m^2 - 100$.
STEP 3: For two unequal real solutions, set $m^2 - 100 > 0$.
STEP 4: Solve. The boundary values are $m = -10$ and $m = 10$. The solution is $m < -10$ or $m > 10$.
STEP 5: The matching answer is $m = -11$.
Medium
If $m$ is an integer, for which value of $m$ will the equation $-x^2 + mx = 16$ have no real solutions?
- A$m = -9$
- B$m = 8$
- C$m = -2$
- D$m = 9$
Show solution
Answer: C, $m = -2$. STEP 1: Rearrange the equation to standard form: $x^2 - mx + 16 = 0$.
STEP 2: Identify the discriminant. For $x^2 - mx + 16 = 0$, the discriminant is $m^2 - 64$.
STEP 3: For no real solutions, set $m^2 - 64 < 0$.
STEP 4: Solve. The boundary values are $m = -8$ and $m = 8$. The solution is $-8 < m < 8$.
STEP 5: The matching answer is $m = -2$.
Hard
For what values of $p$ does the equation $2x^2 + px + 32 = 0$ have two unequal real solutions?
- A$p = -16$ or $p = 16$
- B$p > 16$
- C$-16 < p < 16$
- D$p < -16$ or $p > 16$
Show solution
Answer: D, $p < -16$ or $p > 16$. STEP 1: The equation $2x^2 + px + 32 = 0$ is already in standard form with $a = 2$, $b = p$, $c = 32$.
STEP 2: Compute the discriminant: $b^2 - 4ac = p^2 - 4(2)(32) = p^2 - 256$.
STEP 3: For two unequal real solutions, set $p^2 - 256 > 0$.
STEP 4: The boundary values are $p = -16$ and $p = 16$ (since $\sqrt{256} = 16$). Test values from each region of the number line.
STEP 5: The solution is $p < -16$ or $p > 16$.
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