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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 A rewrite and a check

Rewrite $x^2-14x+30$ in completed-square form, then expand your form to check that it returns the original expression.
1. Hint 1
  The constant that builds a perfect square is the square of half the linear coefficient.
2. Hint 2
  An expression has no second side to balance against, so add that constant and subtract it again in the same line, then collect the loose constants.
Answer

$(x-7)^2-19$ .

Full solution

Half of the linear coefficient $-14$ is $-7$ , and squaring that half gives the constant $49$ .

Add $49$ and subtract it again, which leaves the expression unchanged.
$x^2-14x+49-49+30$

The first three terms are the perfect square $(x-7)^2$ , and the loose constants combine to $-19$ .
$x^2-14x+30=(x-7)^2-19$

Expanding the answer gives $x^2-14x+49-19$ , which is $x^2-14x+30$ , the original expression.

Answer

$(x-7)^2-19$ .

Key idea

Adding and subtracting the square of half the linear coefficient rewrites a quadratic with leading coefficient $1$ so its variable appears only once.
Problem 1: Mark complete
Problem 2 Matching values

The expression $(x+5)^2-3$ takes a certain value when $x=1$ . Find every real $x$ at which it takes that same value.
1. Hint 1
  Work out the value at $x=1$ first, then find every $x$ that produces it.
2. Hint 2
  Isolate the squared expression and take both square roots.
Answer

$x=1$ or $x=-11$ .

Full solution

At $x=1$ the expression equals $(1+5)^2-3$ , which is $33$ .

A value of $x$ giving $33$ satisfies
$(x+5)^2=36$

Thus $x+5=\pm6$ , giving $x=1$ or $x=-11$ .
Both values put either $6$ or $-6$ inside the square, so both give $33$ .

Answer

$x=1$ or $x=-11$ .

Key idea

Equal squared values can come from opposite values inside a binomial.
Problem 2: Mark complete
Problem 3 An identity condition

For some real number $b$ , the identity $x^2+bx+10=(x+s)^2-6$ holds for every $x$ , and $s$ is negative. Find $s$ .
1. Hint 1
  An identity has matching constant terms on both sides.
2. Hint 2
  Expand the right side, then apply the sign condition to its constant equation.
Answer

$s=-4$ .

Full solution

The right side expands to $x^2+2sx+s^2-6$ .
Matching the constants gives
$s^2-6=10$

Therefore $s^2=16$ .
Since $s$ is negative, $s=-4$ .
The corresponding middle coefficient would be $-8$ , so such an identity exists.

Answer

$s=-4$ .

Key idea

Matching constants in a completed-square identity can recover the binomial shift.
Problem 3: Mark complete
Problem 4 Two sides of an equation

Find all solutions of $2x(x+3)=5-x^2$ .
1. Hint 1
  Expand and gather all squared terms before making a squared binomial.
2. Hint 2
  Divide by the leading coefficient, then add the square of half the linear coefficient to both sides.
Answer

$x=-1\pm\dfrac{2\sqrt{6}}{3}$ , or equivalently $x=\dfrac{-3\pm2\sqrt{6}}{3}$ .

Full solution

Expand and collect the terms.
$3x^2+6x-5=0$

Divide by $3$ and move the constant.
$x^2+2x=\frac53$

Add $1$ to both sides.
$(x+1)^2=\frac83$

Take both square roots and simplify.
$x=-1\pm\frac{2\sqrt6}{3}$

For either value, $(x+1)^2=8/3$ , so $3x^2+6x-5=0$ and the original two expressions agree.

Answer

$x=-1\pm\dfrac{2\sqrt{6}}{3}$ , or equivalently $x=\dfrac{-3\pm2\sqrt{6}}{3}$ .

Key idea

Normalize the squared term before completing a square in an equation assembled from two sides.
Problem 4: Mark complete
Problem 5 A specified value

The quadratic $x^2+8x+q$ equals $11$ when $x=-4$ . Find $q$ , then find every complex $x$ at which the quadratic equals zero.
1. Hint 1
  Use the given value at $x=-4$ to recover the missing constant first.
2. Hint 2
  Rewrite the recovered quadratic as a square centered at $-4$ plus a constant.
Answer

$q=27$ ; $x=-4\pm i\sqrt{11}$ .

Full solution

Substitute $x=-4$ and set the result equal to the given value.
$16-32+q=11$

This gives $q=27$ .
Write the resulting quadratic as a square plus a constant.
$x^2+8x+27=(x+4)^2+11$

For the value zero,
$(x+4)^2=-11.$
Therefore
$x=-4\pm i\sqrt{11}$

The squared term at either value of $x$ is $-11$ , giving the required zero.

Answer

$q=27$ ; $x=-4\pm i\sqrt{11}$ .

Key idea

A known value can supply a missing constant before a quadratic is solved.
Problem 5: Mark complete
Problem 6 Two squared terms

Write $(x-1)^2+(x+3)^2$ in the form $2(x+h)^2+k$ , and use that form to find every real input for which the original expression equals $20$ .
1. Hint 1
  Expand both squares and collect their terms.
2. Hint 2
  Factor the leading coefficient out of the variable terms, complete the square inside the parentheses, then solve the resulting equation.
Answer

$2(x+1)^2+8$ ; $x=-1\pm\sqrt6$ .

Full solution

Expanding each square gives
$x^2-2x+1+x^2+6x+9$

Combining like terms gives
$2x^2+4x+10$

This is an expression, not an equation, so there is no other side to divide; factor $2$ from the variable terms instead.
$2x^2+4x+10=2(x^2+2x)+10$

Complete the square inside the parentheses by adding and subtracting $1$ .
$x^2+2x=(x+1)^2-1$

Substituting that back gives
$2\left((x+1)^2-1\right)+10$

Distributing the $2$ and combining the constants gives
$2(x+1)^2+8$

Set this equal to $20$ , subtract $8$ , and divide by $2$ .
$(x+1)^2=6$

The inputs are
$x=-1\pm\sqrt6.$
Substitution into the completed form gives $2(6)+8=20$ for both.

Answer

$2(x+1)^2+8$ ; $x=-1\pm\sqrt6$ .

Key idea

Combining several squares may produce one completed square that is easier to solve.
Problem 6: Mark complete
Problem 7 A product condition

Find all complex numbers $x$ for which $(x-1)(x+6)=-16$ , and explain whether either solution is real.
1. Hint 1
  Expand the product so its linear term is visible.
2. Hint 2
  Half of an odd linear coefficient is a fraction; complete the square anyway, then inspect the sign of the number the square must equal.
Answer

$x=\dfrac{-5\pm i\sqrt{15}}{2}$ ; neither solution is real.

Full solution

Expand and gather every term on the left.
$x^2+5x+10=0$

Half of $5$ is $\frac{5}{2}$ , whose square is $\frac{25}{4}$ .
Move $10$ across and add that square to both sides.
$\left(x+\frac{5}{2}\right)^2=\frac{25}{4}-10$

The right side is negative.
$\left(x+\frac{5}{2}\right)^2=-\frac{15}{4}$

No real square is negative, so neither solution is real.
Since $\sqrt{-15}=i\sqrt{15}$ , taking both roots gives
$x+\frac{5}{2}=\pm\frac{i\sqrt{15}}{2}$

Subtract $\frac{5}{2}$ and write the pair over one denominator.
$x=\frac{-5\pm i\sqrt{15}}{2}$

Each value makes $x^2+5x+10=0$ , so the original product equals $-16$ .

Answer

$x=\dfrac{-5\pm i\sqrt{15}}{2}$ ; neither solution is real.

Key idea

The sign on the other side of a completed square distinguishes real from nonreal solutions.
Problem 7: Mark complete
Problem 8 A family of equations

For a real number $p$ , consider $x^2+2px+p^2+7=0$ . A student says that no choice of $p$ gives a real solution for $x$ . Is the claim correct? Explain.

How many distinct real solutions would the equation have if its final constant $7$ were replaced by $0$ ?
1. Hint 1
  The first three terms form a square.
2. Hint 2
  Compare the value required of that square with the possible values of a real square.
3. Hint 3
  Replacing the final constant changes the number on the other side of the completed square.
Answer

The claim is correct. With $7$ replaced by $0$ , there is one distinct real solution, $x=-p$ .

Full solution

The equation is equivalent to
$(x+p)^2=-7$

For real $x$ and real $p$ , the square on the left is zero or positive.
It cannot equal $-7$ , regardless of the chosen real $p$ .

After replacing $7$ by $0$ , the equation becomes
$(x+p)^2=0$

A square equals zero exactly when its base is zero.
Thus $x=-p$ is the only distinct real solution, a repeated root.

Answer

The claim is correct. With $7$ replaced by $0$ , there is one distinct real solution, $x=-p$ .

Key idea

The sign of the value required of a completed square determines the number of distinct real roots.
Problem 8: Mark complete
Problem 9 A replacement of the input

A student says that $x^2+6x+10=0$ and $(2x)^2+6(2x)+10=0$ have the same solutions. Decide whether the claim is correct, and give the solution set of each equation.
1. Hint 1
  Complete the square in the expression that plays the role of the input.
2. Hint 2
  In the second equation, isolate $2x$ before dividing to obtain $x$ .
Answer

False. First: $x=-3\pm i$ . Second: $x=\dfrac{-3\pm i}{2}$ .

Full solution

The first equation becomes
$(x+3)^2=-1$

Its solutions are $x=-3\pm i$ .
In the second equation, the same square is built from $2x$ .
$(2x+3)^2=-1$

Thus $2x=-3\pm i$ , so
$x=\frac{-3\pm i}{2}$

These are half the first pair, not the same pair.
Substitution gives squared binomial value $-1$ in each respective equation.

Answer

False. First: $x=-3\pm i$ . Second: $x=\dfrac{-3\pm i}{2}$ .

Key idea

Replacing $x$ by $2x$ in an equation halves each of its solutions.
Problem 9: Mark complete
Problem 10 Two pairs of roots

The equations $x^2+6x+1=0$ and $x^2+6x-3=0$ each have two real roots. A student says that the average of the two roots is the same for both equations. Decide whether the claim is correct, and find that average.
1. Hint 1
  Complete the square in each equation.
2. Hint 2
  The plus and minus square-root terms cancel when each pair of roots is added.
Answer

The claim is correct; the common average is $-3$ .

Full solution

The first equation becomes
$(x+3)^2=8$

Its roots are $-3\pm2\sqrt2$ , whose sum is $-6$ .
The second equation becomes
$(x+3)^2=12$

Its roots are $-3\pm2\sqrt3$ , whose sum is also $-6$ .
Each average is
$\frac{-6}{2}=-3$

The different distances from $-3$ do not change the shared average.

Answer

The claim is correct; the common average is $-3$ .

Key idea

A pair of roots written as a fixed number plus or minus a real amount has that fixed number as its average.
Problem 10: Mark complete

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