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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