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Completing the Square: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 An unfinished identity

Find $p$ and $q$ so that $x^2-3x=(x+p)^2+q$ for every real $x$ .
1. Hint 1
  Expand the square and compare the coefficient of $x$ .
2. Hint 2
  After matching the linear term, make the constant terms match as well.
Answer

$p=-\frac32$ , $q=-\frac94$ .

Full solution

Expansion gives $x^2+2px+p^2+q$ .
Matching terms yields
$\begin{aligned} 2p&=-3 \\ p&=-\frac32 \\ p^2+q&=0 \\ q&=-\frac94 \end{aligned}$

Expanding with these values returns $x^2-3x$ .

Answer

$p=-\frac32$ , $q=-\frac94$ .

Key idea

Matching the linear and constant coefficients determines the completed square.
Problem 1: Mark complete
Problem 2 A negative multiplier

Write $-4x^2+8x+1$ in vertex form.
1. Hint 1
  Take the leading coefficient out of the variable terms before completing the square.
2. Hint 2
  Any constant added inside that factor also receives its multiplier.
Answer

$-4(x-1)^2+5$ .

Full solution

Factor $-4$ from the variable terms, then complete the square: half of $-2$ is $-1$ , and $(-1)^2=1$ .
$\begin{aligned} f(x)&=-4(x^2-2x)+1 \\ f(x)&=-4((x-1)^2-1)+1 \\ f(x)&=-4(x-1)^2+5 \end{aligned}$

The $-1$ inside the bracket becomes $+4$ outside.
Expanding the final form gives $-4x^2+8x+1$ .

Answer

$-4(x-1)^2+5$ .

Key idea

Distribute a negative leading coefficient to the square and the compensating constant.
Problem 2: Mark complete
Problem 3 One isolated square

Solve $5-(x+\frac12)^2=2$ over the real numbers.
1. Hint 1
  First leave the squared expression alone on one side.
2. Hint 2
  A positive square value permits two opposite square roots.
Answer

$x=-\frac12\pm\sqrt3$ .

Full solution

Isolate the square by subtracting $5$ from both sides, then multiplying by $-1$ .
$\begin{aligned} (x+\tfrac12)^2&=3 \\ x+\tfrac12&=\pm\sqrt3 \\ x&=-\tfrac12\pm\sqrt3 \end{aligned}$

For either value, the square is $3$ , so the original left side is $5-3=2$ .

Answer

$x=-\frac12\pm\sqrt3$ .

Key idea

Isolating a positive square requires keeping both signs when recovering its base.
Problem 3: Mark complete
Problem 4 Matching two outputs

Find all real $t$ for which $3t^2+6t+2$ equals $5$ , by completing the square.
1. Hint 1
  Move the target output before rewriting the variable terms.
2. Hint 2
  Divide by the leading coefficient so the square is easy to complete.
Answer

$t=-1\pm\sqrt2$ .

Full solution

The equation becomes
$3t^2+6t-3=0.$
Divide by $3$ to get $t^2+2t=1$ , then complete the square: half of $2$ is $1$ , and $1^2=1$ , so add $1$ to both sides.
$\begin{aligned} t^2+2t&=1 \\ (t+1)^2&=2 \\ t&=-1\pm\sqrt2 \end{aligned}$

The original expression is $3(t+1)^2-1$ , which equals $5$ for either candidate.

Answer

$t=-1\pm\sqrt2$ .

Key idea

An output-matching problem becomes a completed square after subtracting the target.
Problem 4: Mark complete
Problem 5 A model’s lowest cost

A model assigns cost $C(q)=2q^2-6q+11$ dollars to a real setting $q\ge0$ . Find the smallest cost and the setting that attains it.
1. Hint 1
  Rewrite the rule so its varying part is a positive multiple of a square.
2. Hint 2
  Check that the setting making the square zero belongs to the domain.
Answer

Minimum cost $\frac{13}{2}$ dollars at $q=\frac32$ .

Full solution

Factor $2$ from the variable terms, then add $\tfrac94$ inside to complete the square, compensating by subtracting $2\cdot\tfrac94=\tfrac92$ outside.
$\begin{aligned} C(q)&=2(q^2-3q)+11 \\ &=2\left(q^2-3q+\tfrac94\right)-\tfrac92+11 \\ &=2\left(q-\tfrac32\right)^2+\tfrac{13}{2} \end{aligned}$

The square is at least zero and is zero when $q=3/2$ , an allowed setting.
Thus the minimum is $13/2$ dollars.
Direct substitution gives
$9/2-9+11=13/2.$

Answer

Minimum cost $\frac{13}{2}$ dollars at $q=\frac32$ .

Key idea

A positive square identifies an unconstrained minimum that must still satisfy the model’s domain.
Problem 5: Mark complete
Problem 6 Restoring an old rule

A notebook shows $f(x)=3[(x+5)^2-25]+2$ . Write it in vertex form, then give the vertex.
1. Hint 1
  The outside multiplier acts on both entries inside the bracket.
2. Hint 2
  Combine the resulting constants outside the square.
Answer

$f(x)=3(x+5)^2-73$ ; vertex $(-5,-73)$ .

Full solution

Distribute the $3$ across both terms inside the bracket, then combine the constants outside the square.
$\begin{aligned} f(x)&=3[(x+5)^2-25]+2 \\ &=3(x+5)^2-75+2 \\ &=3(x+5)^2-73 \end{aligned}$
This is vertex form, with vertex $(-5,-73)$ .

Answer

$f(x)=3(x+5)^2-73$ ; vertex $(-5,-73)$ .

Key idea

Distributing a leading coefficient across a completed square must reach every term inside it, including the subtracted constant.
Problem 6: Mark complete
Problem 7 A family with a parameter

For real $b$ , write $f(x)=2x^2+bx+1$ in vertex form and express the vertex coordinates in terms of $b$ .
1. Hint 1
  Factor 2 out of the variable terms.
2. Hint 2
  Half of the inside coefficient is $b/4$ ; add and subtract its square, $b^2/16$ .
Answer

$f(x)=2(x+\frac b4)^2+1-\frac{b^2}{8}$ ; vertex $(-\frac b4,1-\frac{b^2}{8})$ .

Full solution

Inside the leading factor, the linear coefficient is $b/2$ , so half of it is $b/4$ , and its square is $b^2/16$ .
Add and subtract $b^2/16$ inside the parentheses.
$\begin{aligned} f(x)&=2\left(x^2+\tfrac b2x+\tfrac{b^2}{16}-\tfrac{b^2}{16}\right)+1 \\ f(x)&=2\left(\left(x+\tfrac b4\right)^2-\tfrac{b^2}{16}\right)+1 \end{aligned}$

Distribute the $2$ across both terms inside the outer parentheses; the subtracted term becomes $b^2/8$ .
$f(x)=2\left(x+\tfrac b4\right)^2+1-\tfrac{b^2}{8}$
This is vertex form, with horizontal coordinate $-b/4$ and height $1-b^2/8$ .

Answer

$f(x)=2(x+\frac b4)^2+1-\frac{b^2}{8}$ ; vertex $(-\frac b4,1-\frac{b^2}{8})$ .

Key idea

Completing the square works with parameter coefficients as well as numbers.
Problem 7: Mark complete
Problem 8 A proposed square form

A student rewrites $3x^2-6x+8$ as $3(x-1)^2+7$ , saying the added $1$ was subtracted from $8$ . Find the error and correct the vertex form.
1. Hint 1
  The added completing constant sits inside a factor of 3.
2. Hint 2
  Expand the proposed form to see how much its constant differs from the original.
Answer

Correct form: $3(x-1)^2+5$ ; the added inside $1$ contributes $3$ outside.

Full solution

The variable terms are $3(x^2-2x)$ .
Adding $1$ inside increases the whole expression by $3$ , so subtract $3$ outside.
$\begin{aligned} f(x)&=3(x^2-2x+1)+8-3 \\ f(x)&=3(x-1)^2+5 \end{aligned}$

The proposed expression expands to $3x^2-6x+10$ , so it was too large by $2$ .

Answer

Correct form: $3(x-1)^2+5$ ; the added inside $1$ contributes $3$ outside.

Key idea

The amount compensated outside equals the inside completing constant times the leading coefficient.
Problem 8: Mark complete
Problem 9 A parameterized minimum

For real $b$ , the expression $x^2+bx+b^2/4+2$ is claimed to have minimum $2$ . Is this true for every $b$ ? Explain.
1. Hint 1
  The first three terms may make a complete square already.
2. Hint 2
  Check whether its base can equal zero for each real parameter.
Answer

Yes; its minimum is $2$ at $x=-b/2$ .

Full solution

The expression rewrites exactly as
$(x+\tfrac b2)^2+2$
The square is at least zero for every real input.
It reaches zero at $x=-b/2$ , which is real for every real $b$ .
The claimed minimum is therefore attained in every case.

Answer

Yes; its minimum is $2$ at $x=-b/2$ .

Key idea

A real shift of a square changes the minimizing input but not its added constant minimum.
Problem 9: Mark complete
Problem 10 Counting solutions in a family

For which real $c$ does $2x^2-10x+c=0$ have two real solutions, exactly one, or none? Give the answer for each case.
1. Hint 1
  Divide by the leading coefficient before completing the square.
2. Hint 2
  The number of solutions depends on the sign of the constant once the square is isolated.
Answer

Two solutions when $c<\frac{25}{2}$ ; exactly one when $c=\frac{25}{2}$ ; none when $c>\frac{25}{2}$ .

Full solution

Divide by the leading coefficient, then complete the square: half of $-5$ is $-\tfrac52$ , and $\left(-\tfrac52\right)^2=\tfrac{25}{4}$ .
$\begin{aligned} x^2-5x+\tfrac c2&=0 \\ \left(x-\tfrac52\right)^2&=\tfrac{25}{4}-\tfrac c2 \end{aligned}$

If $\tfrac{25}{4}-\tfrac c2>0$ , that is $c<\tfrac{25}{2}$ , there are two real solutions.
If $\tfrac{25}{4}-\tfrac c2=0$ , that is $c=\tfrac{25}{2}$ , there is exactly one.
If $\tfrac{25}{4}-\tfrac c2<0$ , that is $c>\tfrac{25}{2}$ , no real number squares to a negative value, so there is no real solution.

Answer

Two solutions when $c<\frac{25}{2}$ ; exactly one when $c=\frac{25}{2}$ ; none when $c>\frac{25}{2}$ .

Key idea

The number of real solutions of a quadratic with a nonzero leading coefficient turns on the sign of the constant left over after completing the square.
Problem 10: Mark complete

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