Math 2–3 Honors · Lesson Review · October 6, 2026

Solving Quadratics by Completing the Square

Turn a quadratic into a squared binomial, keep both sides balanced, and use the square root principle to find exact solutions, including fractional coefficients and equations with no real solutions.

37-minute edited lesson10 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captions SRT captions

0:0001Review factoring a squared binomial 6:0102Use the square root principle directly 8:1103A second squared-binomial example 9:5104Complete the square in a factorable equation 13:2905Why we add half the coefficient squared 15:2806Irrational solutions and real factors 20:2407Simplify the radical in the answer 22:4608Normalize the leading coefficient 27:4309Complete a square with fractions 34:1510Recognize when there are no real solutions

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02 · Understand

Learning targets

Use the square root principle on an entire squared expression.
Complete a square by adding half the normalized linear coefficient, squared, to both sides.
Handle nonunit leading coefficients and fractions without rounding.
Interpret positive, zero, and negative squared values over the real numbers.

03 · Connect

Key ideas

Completing the square works whether the original quadratic factors easily or has irrational roots.

01The whole expression can be squared: $(x-h)^2=k\quad\Longrightarrow\quad x=h\pm\sqrt{k}\quad(k\ge0)$ The square root principle applies to whatever is squared. You do not need to expand a squared binomial first. For k=0 there is one real solution; for k<0 there are no real solutions.
02Normalize before completing: $ax^2+bx+c=0\quad\Longrightarrow\quad x^2+\frac ba x=-\frac ca\quad(a\ne0)$ Divide every term by the leading coefficient. In the normalized equation, the linear coefficient is b/a. Move the constant to the other side.
03Half the linear coefficient, squared: $x^2+Bx=K\quad\Longrightarrow\quad\left(x+\frac B2\right)^2=K+\left(\frac B2\right)^2$ Use the normalized linear coefficient B, including its sign. Add the same quantity to both sides. The binomial uses B/2, rather than its square.
04Keep exact roots: $\sqrt{28}=2\sqrt7,\qquad\sqrt{\frac{169}{144}}=\frac{13}{12}$ Simplify radicals and fractions without replacing exact solutions by rounded decimals. The ± produces two equations when the right side is positive.
05Integer factoring is not the only factorization: $x^2+8x-5=(x+4-\sqrt{21})(x+4+\sqrt{21})$ A quadratic that does not factor using integers can still have real irrational factors. Completing the square finds the roots and reveals those factors.

04 · Apply

Worked examples

These examples follow the methods developed in the recorded lesson.

Squared binomial

Use the square root principle immediately

(x-3)^2=25\quad\Longrightarrow\quad x-3=\pm5

Treat x−3 as the entire squared quantity.

x=3\pm5\quad\Longrightarrow\quad x=8\text{ or }x=-2

Expanding and factoring also works, but takes more steps.

(x+5)^2=9\quad\Longrightarrow\quad x=-5\pm3=-2\text{ or }-8

Subtract 5 when isolating x. This uses the sign correction made during the recording.

Same equation, another method

Complete a square in a factorable quadratic

x^2+6x+5=0\quad\Longrightarrow\quad x^2+6x=-5

Move the constant. Half of 6 is 3, and 3² is 9.

x^2+6x+9=-5+9\quad\Longrightarrow\quad(x+3)^2=4

Add 9 to both sides, then factor the perfect square.

x=-3\pm2\quad\Longrightarrow\quad x=-1\text{ or }-5

These are the same roots obtained from (x+5)(x+1)=0.

Irrational roots

Solve when integer factoring does not work

x^2+8x-5=0\quad\Longrightarrow\quad x^2+8x=5

Half of 8 is 4, so add 16 to both sides.

(x+4)^2=21\quad\Longrightarrow\quad x=-4\pm\sqrt{21}

The radicand has no perfect-square factor, so this is simplified.

(4-\sqrt{21})+(4+\sqrt{21})=8,\qquad(4-\sqrt{21})(4+\sqrt{21})=-5

These irrational constants give the real factors of the original polynomial.

Simplify the radical

Keep the answer exact

x^2+10x-3=0\quad\Longrightarrow\quad x^2+10x=3

Add 25 to both sides.

(x+5)^2=28\quad\Longrightarrow\quad x=-5\pm2\sqrt7

Use √28=2√7 and keep both signs.

Fractional coefficients

Divide first, then complete

2x^2-3x-5=0\quad\Longrightarrow\quad x^2-\frac32x=\frac52

Divide every term by 2. Half the new linear coefficient is −3/4.

\left(x-\frac34\right)^2=\frac52+\frac9{16}=\frac{49}{16}

Add 9/16 to both sides and use a common denominator.

x=\frac34\pm\frac74\quad\Longrightarrow\quad x=\frac52\text{ or }-1

A perfect-square fraction gives rational roots.

Second fractional example

Connect the resulting roots back to factoring

6x^2-7x-5=0\quad\Longrightarrow\quad x^2-\frac76x=\frac56

Half the normalized linear coefficient is −7/12.

\left(x-\frac7{12}\right)^2=\frac56+\frac{49}{144}=\frac{169}{144}

Convert 5/6 to 120/144 before adding.

x=\frac7{12}\pm\frac{13}{12}\quad\Longrightarrow\quad x=\frac53\text{ or }-\frac12

These roots correspond to (3x−5)(2x+1)=0. Preserve the leading coefficient when reconstructing factors.

No real solutions

Interpret a negative squared value

x^2+2x+5=0\quad\Longrightarrow\quad(x+1)^2=-4

Move 5, then add 1 to both sides.

\text{Real solution set: }\varnothing

Every real square is nonnegative. Complex solutions are a later topic, so state no real solutions precisely.

05 · Avoid

Common mistakes

Completing before dividing.

Make the coefficient of x² equal one first. Halve the new linear coefficient after normalization.

Adding to only one side.

The equation stays equivalent only when the same square is added to both sides.

Using the squared value inside the binomial.

Add (B/2)² to complete the square, but write x+B/2 inside the resulting binomial.

Dropping a sign or a solution.

Include ±, then isolate x carefully. A negative right side means no real solutions, rather than two real negative roots.

06 · Check yourself

Try it first

Use these five fresh problems to check your understanding. Open a solution after you try it.

1Solve using the square root principle:

(x-4)^2=18

$x-4=\pm\sqrt{18}=\pm3\sqrt2$

Apply the principle to the entire binomial.

$x=4\pm3\sqrt2$

Add 4 after taking both square roots.

2Complete the square and solve:

x^2-10x+6=0

$x^2-10x=-6,\qquad(x-5)^2=-6+25=19$

Half of −10 is −5; add 25 to both sides.

$x=5\pm\sqrt{19}$

The radical is already simplified.

3Divide first, then complete the square:

2x^2+5x-1=0

$x^2+\frac52x=\frac12,\qquad\left(x+\frac54\right)^2=\frac12+\frac{25}{16}=\frac{33}{16}$

Divide all terms by 2, then add 25/16 to both sides.

$x=\frac{-5\pm\sqrt{33}}4$

Both exact roots solve the original equation.

4Determine the real solution set:

x^2-4x+8=0

$x^2-4x=-8\quad\Longrightarrow\quad(x-2)^2=-4$

Add 4 to both sides.

$\text{Real solution set: }\varnothing$

A real square cannot be negative.

5A student writes (x+2)²=7. Correct the missing step and solve:

x^2+4x=7

$x^2+4x+4=7+4\quad\Longrightarrow\quad(x+2)^2=11$

The added 4 belongs on both sides.

$x=-2\pm\sqrt{11}$

Subtract 2 after taking both square roots.

07 · Revisit

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