Quadratic formula — derivation by completing the square: Exercises

Level 1 — Recognition

Goal: read a quadratic and correctly name its parts. No solving yet.

Exercise 1.1

Write $3 x^{2} - 7 x + 2 = 0$ in standard form (it already is) and state $a$ , $b$ , $c$ .

Recall Solution 1.1

Standard form means the equation looks like $a x^{2} + b x + c = 0$ with everything on the left and $0$ on the right. Match term by term:

The $x^{2}$ term is $3 x^{2}$ , so $a = 3$ .
The $x$ term is $- 7 x$ , so $b = - 7$ (the sign travels with the number).
The constant is $+ 2$ , so $c = 2$ .

Answer: $a = 3, b = - 7, c = 2$ .

Exercise 1.2

Rewrite $5 x - 2 = - x^{2} + 4$ in standard form and identify $a$ , $b$ , $c$ .

Recall Solution 1.2

Nothing is in standard form yet — we must gather everything on one side so the right side is $0$ .

Why? The formula and discriminant only work once the equation reads $a x^{2} + b x + c = 0$ .

Bring all terms to the left (add $x^{2}$ to both sides, subtract $4$ ): $x^{2} + 5 x - 2 - 4 = 0 \Rightarrow x^{2} + 5 x - 6 = 0.$ Now read off: $a = 1$ , $b = 5$ , $c = - 6$ .

Answer: $a = 1, b = 5, c = - 6$ .

Exercise 1.3

For $2 x^{2} + 7 x - 4 = 0$ , compute $- b$ and $2 a$ (just these two ingredients).

Recall Solution 1.3

Here $a = 2$ , $b = 7$ , $c = - 4$ .

$- b = - (7) = - 7$ . The formula always uses the opposite of $b$ , never $b$ itself.
$2 a = 2 \cdot 2 = 4$ .

Answer: $- b = - 7, 2 a = 4$ .

Level 2 — Application

Goal: plug into the formula and finish the arithmetic cleanly.

Exercise 2.1

Solve $x^{2} - 5 x + 6 = 0$ .

Recall Solution 2.1

Coefficients: $a = 1, b = - 5, c = 6$ .

Substitute into $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ : $x = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( 1 ) ( 6 )}{2 ( 1 )} = \frac{5 \pm 25 - 24}{2} = \frac{5 \pm 1}{2} .$ Split the $\pm$ : $x = \frac{5 + 1}{2} = 3 or x = \frac{5 - 1}{2} = 2.$ Answer: $x = 2$ or $x = 3$ .

Exercise 2.2

Solve $2 x^{2} + 7 x - 4 = 0$ .

Recall Solution 2.2

Coefficients: $a = 2, b = 7, c = - 4$ .

Discriminant first (keeps the arithmetic honest): $Δ = 7^{2} - 4 (2) (- 4) = 49 - (- 32) = 49 + 32 = 81.$ Notice $- 4 a c = - 4 \cdot 2 \cdot (- 4) = + 32$ : multiplying by the negative $c$ flips the sign back to positive.

Formula: $x = \frac{- 7 \pm 81}{4} = \frac{- 7 \pm 9}{4} .$ $x = \frac{- 7 + 9}{4} = \frac{1}{2} or x = \frac{- 7 - 9}{4} = - 4.$ Answer: $x = \frac{1}{2}$ or $x = - 4$ .

Exercise 2.3

Solve $3 x^{2} - 2 x - 8 = 0$ .

Recall Solution 2.3

Coefficients: $a = 3, b = - 2, c = - 8$ . $Δ = (- 2)^{2} - 4 (3) (- 8) = 4 + 96 = 100.$ $x = \frac{- ( - 2 ) \pm 100}{2 ( 3 )} = \frac{2 \pm 10}{6} .$ $x = \frac{2 + 10}{6} = 2 or x = \frac{2 - 10}{6} = - \frac{8}{6} = - \frac{4}{3} .$ Answer: $x = 2$ or $x = - \frac{4}{3}$ .

Exercise 2.4

Solve $x^{2} - 4 x + 1 = 0$ and leave the answer in exact surd form.

Recall Solution 2.4

Coefficients: $a = 1, b = - 4, c = 1$ . $Δ = (- 4)^{2} - 4 (1) (1) = 16 - 4 = 12.$ $x = \frac{4 \pm 12}{2} .$ Simplify $12 = 4 \cdot 3 = 23$ : $x = \frac{4 \pm 2 3}{2} = 2 \pm 3 .$ Why simplify the surd? $12$ hides a perfect square factor ( $4$ ); pulling it out gives the cleanest exact form.

Answer: $x = 2 + 3$ or $x = 2 - 3$ .

Level 3 — Analysis

Goal: use $Δ$ to reason about the roots, sometimes before computing them.

Exercise 3.1

Without solving, state how many real roots $x^{2} - 6 x + 9 = 0$ has, and why.

Recall Solution 3.1

$a = 1, b = - 6, c = 9$ . $Δ = (- 6)^{2} - 4 (1) (9) = 36 - 36 = 0.$ Since $Δ = 0$ , the $\pm Δ$ adds/subtracts zero, so the two roots collapse into one.

Answer: exactly one repeated real root (it equals $x = - \frac{b}{2 a} = 3$ ). Geometrically the parabola just touches the $x$ -axis — see the middle curve below.

Figure — Quadratic formula — derivation by completing the square

Exercise 3.2

For $x^{2} + 2 x + 5 = 0$ , show there are no real roots, then give the complex roots.

Recall Solution 3.2

$a = 1, b = 2, c = 5$ . $Δ = 2^{2} - 4 (1) (5) = 4 - 20 = - 16.$ $Δ < 0$ , so $Δ$ is not a real number — no real roots.

In the complex numbers we use $- 1 = i$ , so $- 16 = 4 i$ : $x = \frac{- 2 \pm 4 i}{2} = - 1 \pm 2 i .$ Answer: $x = - 1 + 2 i$ or $x = - 1 - 2 i$ (a conjugate pair). See Complex numbers and quadratic equations.

Exercise 3.3

Find the value(s) of $k$ for which $x^{2} + k x + 9 = 0$ has exactly one repeated real root.

Recall Solution 3.3

"Exactly one repeated real root" is the signature of $Δ = 0$ .

Here $a = 1, b = k, c = 9$ : $Δ = k^{2} - 4 (1) (9) = k^{2} - 36.$ Set it to zero: $k^{2} - 36 = 0 \Rightarrow k^{2} = 36 \Rightarrow k = \pm 6.$ Why two answers? A perfect-square trinomial can be $(x + 3)^{2}$ (giving $k = 6$ ) or $(x - 3)^{2}$ (giving $k = - 6$ ); both touch the axis once.

Answer: $k = 6$ or $k = - 6$ .

Exercise 3.4

For which values of $m$ does $2 x^{2} + 4 x + m = 0$ have two distinct real roots?

Recall Solution 3.4

"Two distinct real roots" means $Δ > 0$ (strictly positive, so $\pm Δ$ genuinely splits).

$a = 2, b = 4, c = m$ : $Δ = 4^{2} - 4 (2) (m) = 16 - 8 m .$ Require $16 - 8 m > 0$ : $16 > 8 m \Rightarrow m < 2.$ Answer: $m < 2$ . (See Discriminant and nature of roots and Quadratic inequalities for handling the inequality carefully.)

Level 4 — Synthesis

Goal: combine the formula with factoring, Vieta's relations, or reverse-engineering.

Exercise 4.1

The equation $x^{2} + p x + q = 0$ has roots $2$ and $- 5$ . Find $p$ and $q$ .

Recall Solution 4.1

Two routes; both use Vieta's formulas for sum and product of roots. For $x^{2} + p x + q = 0$ :

sum of roots $= - p$ ,
product of roots $= q$ .

Sum: $2 + (- 5) = - 3 = - p \Rightarrow p = 3$ . Product: $2 \cdot (- 5) = - 10 = q$ .

Check by rebuilding: $(x - 2) (x + 5) = x^{2} + 3 x - 10$ , matching $p = 3, q = - 10$ . ✓

Answer: $p = 3, q = - 10$ .

Exercise 4.2

Solve $\frac{1}{x} + \frac{1}{x + 2} = \frac{3}{4}$ for $x$ .

Recall Solution 4.2

Clear the fractions. Multiply every term by the common denominator $4 x (x + 2)$ (valid provided $x \neq = 0$ and $x \neq = - 2$ ): $4 (x + 2) + 4 x = 3 x (x + 2) .$ Expand: $4 x + 8 + 4 x = 3 x^{2} + 6 x \Rightarrow 8 x + 8 = 3 x^{2} + 6 x .$ Bring to standard form: $0 = 3 x^{2} - 2 x - 8.$ This is exactly Exercise 2.3, whose roots are $x = 2$ and $x = - \frac{4}{3}$ .

Check the forbidden values: neither $2$ nor $- \frac{4}{3}$ equals $0$ or $- 2$ , so both are valid.

Answer: $x = 2$ or $x = - \frac{4}{3}$ .

Exercise 4.3

A rectangle is $3$ cm longer than it is wide and has area $40 cm^{2}$ . Find its width.

Recall Solution 4.3

Let the width be $w$ cm; then the length is $w + 3$ cm.

Model: area $=$ width $\times$ length: $w (w + 3) = 40 \Rightarrow w^{2} + 3 w - 40 = 0.$ $a = 1, b = 3, c = - 40$ : $Δ = 3^{2} - 4 (1) (- 40) = 9 + 160 = 169, 169 = 13.$ $w = \frac{- 3 \pm 13}{2} .$ $w = \frac{10}{2} = 5 or w = \frac{- 16}{2} = - 8.$ Reject the negative root: a width cannot be $- 8$ cm.

Answer: width $= 5$ cm (length $8$ cm, area $40$ ✓). See figure below.

Level 5 — Mastery

Goal: a small creative leap — hidden quadratics, parameters, and proof-style reasoning.

Exercise 5.1

Solve $x^{4} - 5 x^{2} + 4 = 0$ .

Recall Solution 5.1

This is degree $4$ , not $2$ — but it is quadratic in $x^{2}$ . Let $u = x^{2}$ (a substitution that makes the hidden quadratic visible): $u^{2} - 5 u + 4 = 0.$ Factor or use the formula: $Δ = 25 - 16 = 9$ , so $u = \frac{5 \pm 3}{2} = 4 or 1.$ Now undo the substitution, $u = x^{2}$ :

$x^{2} = 4 \Rightarrow x = \pm 2$ ,
$x^{2} = 1 \Rightarrow x = \pm 1$ .

Answer: $x = - 2, - 1, 1, 2$ (four real roots).

Exercise 5.2

Find $k$ so that $k x^{2} - 4 x + 1 = 0$ has real roots. Consider all cases, including $k = 0$ .

Recall Solution 5.2

Case $k = 0$ (degenerate). The equation becomes $- 4 x + 1 = 0$ , a linear equation with the single real solution $x = \frac{1}{4}$ . So $k = 0$ does give a real root — but this is not a quadratic anymore.

Case $k \neq = 0$ (genuine quadratic). Real roots require $Δ \geq 0$ : $Δ = (- 4)^{2} - 4 (k) (1) = 16 - 4 k \geq 0 \Rightarrow k \leq 4.$ Combined with $k \neq = 0$ : real quadratic roots for all $k \leq 4$ with $k \neq = 0$ .

Answer: the equation has (at least one) real solution for every $k \leq 4$ — including $k = 0$ , where it degenerates to linear with root $x = \frac{1}{4}$ .

Exercise 5.3

Show that if $x^{2} + b x + c = 0$ has roots $r$ and $s$ , then $r + s = - b$ and $r s = c$ , using the quadratic formula directly.

Recall Solution 5.3

With $a = 1$ the two roots are $r = \frac{- b + Δ}{2}, s = \frac{- b - Δ}{2}, Δ = b^{2} - 4 c .$ Sum (the $Δ$ pieces cancel): $r + s = \frac{( - b + Δ ) + ( - b - Δ )}{2} = \frac{- 2 b}{2} = - b .$ Product (difference of squares: $(A + B) (A - B) = A^{2} - B^{2}$ ): $r s = \frac{( - b ) ^{2} - ( Δ ) ^{2}}{4} = \frac{b ^{2} - ( b ^{2} - 4 c )}{4} = \frac{4 c}{4} = c .$ Answer: $r + s = - b$ and $r s = c$ , exactly Vieta's formulas for sum and product of roots.

Exercise 5.4

The quadratic $x^{2} - (t + 3) x + (2 t + 2) = 0$ has one root equal to $2$ . Find $t$ and the other root.

Recall Solution 5.4

A root satisfies the equation — substitute $x = 2$ : $2^{2} - (t + 3) (2) + (2 t + 2) = 0.$ $4 - 2 t - 6 + 2 t + 2 = 0 \Rightarrow 0 = 0.$ This is an identity — true for every $t$ ! So $x = 2$ is a root no matter what $t$ is. We need another relation.

Use Vieta's product. With $a = 1$ , product of roots $= c = 2 t + 2$ . If the roots are $2$ and $r$ : $2 r = 2 t + 2 \Rightarrow r = t + 1.$ Use Vieta's sum as a cross-check: sum $= t + 3$ , and $2 + r = 2 + (t + 1) = t + 3$ ✓ — consistent for all $t$ .

Answer: the condition holds for every $t$ ; the other root is $r = t + 1$ . (For example, $t = 4$ gives roots $2$ and $5$ ; indeed $x^{2} - 7 x + 10 = (x - 2) (x - 5)$ .)

Connections

Parent: full derivation
Completing the square technique
Perfect square trinomials
Discriminant and nature of roots
Factoring quadratics
Complex numbers and quadratic equations
Parabola and its vertex form
Vieta's formulas for sum and product of roots
Quadratic inequalities

Recall Quick self-quiz

For $2 x^{2} + 4 x + m = 0$ , distinct real roots need $m$ ::: $m < 2$ Roots of $x^{2} - 5 x + 6 = 0$ ::: $x = 2$ and $x = 3$ Sign of $Δ$ for one repeated root ::: $Δ = 0$ $- 16$ in complex form ::: $4 i$ Width from $w^{2} + 3 w - 40 = 0$ (physical) ::: $5$ cm