quadratic equations problems with solutions

Quadratic equations are second-degree polynomial equations of the form $ax^2 + bx + c = 0$ , where $a \neq 0$ . Mastering them requires fluency across multiple solving methods: factoring, the square root property, completing the square, and the quadratic formula. Understanding the discriminant $\Delta = b^2 - 4ac$ allows you to predict the nature of the roots, whether they are real and distinct, real and equal, or complex. These concepts underpin countless real-world applications, from projectile motion and geometric area problems to number theory and engineering models. The problems below progress from easy to hard within each section, each accompanied by a hint and a full step-by-step solution.

Contents6 sections

Solving Quadratic Equations by Factoring

Factoring is often the fastest method for solving quadratic equations when the roots are rational. The goal is to rewrite $ax^2 + bx + c$ as a product of two linear factors, then apply the Zero Product Property: if $(x - r_1)(x - r_2) = 0$ , then $x = r_1$ or $x = r_2$ . This section builds from simple monic trinomials to equations that must first be rearranged into standard form.

Problem 1: Factoring a Simple Monic Trinomial

Easy

Consider the quadratic equation $x^2 - 7x + 10 = 0$ .

Factor the left-hand side completely.
Find all real solutions of the equation.

Hint

Look for two integers whose product is

+10

and whose sum is

-7

. Once you find them, write the factored form and set each factor equal to zero.

View Solution

Solution to question 1:

We need two integers with product $10$ and sum $-7$ . Those integers are $-2$ and $-5$ .

x^2 - 7x + 10 = (x - 2)(x - 5)

Solution to question 2:

Applying the Zero Product Property:

(x - 2)(x - 5) = 0 \implies x - 2 = 0 \quad \text{or} \quad x - 5 = 0

\boxed{x = 2 \quad \text{or} \quad x = 5}

Problem 2: Factoring with a Leading Coefficient

Medium

Solve the quadratic equation $6x^2 + x - 12 = 0$ by factoring.

Identify the coefficients $a$ , $b$ , and $c$ and compute the product $ac$ .
Use the ac-method to factor and find all solutions.

Hint

Compute

ac = 6 \times (-12) = -72

. Find two integers whose product is

-72

and whose sum is

+1

. Use them to split the middle term and factor by grouping.

View Solution

Solution to question 1:

Here $a = 6$ , $b = 1$ , $c = -12$ , so $ac = -72$ . We need two integers with product $-72$ and sum $1$ : those are $9$ and $-8$ .

Solution to question 2:

Split the middle term and factor by grouping:

6x^2 + 9x - 8x - 12 = 3x(2x + 3) - 4(2x + 3) = (3x - 4)(2x + 3)

3x - 4 = 0 \implies x = \frac{4}{3} \qquad \text{or} \qquad 2x + 3 = 0 \implies x = -\frac{3}{2}

\boxed{x = \frac{4}{3} \quad \text{or} \quad x = -\frac{3}{2}}

Problem 3: Rearranging Before Factoring

Hard

A quadratic equation is given in non-standard form: $3x^2 = 11x - 6$ .

Rewrite the equation in standard form $ax^2 + bx + c = 0$ .
Solve by factoring and verify each solution by substitution.

Hint

Move all terms to one side first. After writing the equation in standard form, compute

ac

and look for factor pairs. Verification means substituting each root back into the original equation.

View Solution

Solution to question 1:

3x^2 - 11x + 6 = 0

Here $a = 3$ , $b = -11$ , $c = 6$ , and $ac = 18$ . Two integers with product $18$ and sum $-11$ : $-2$ and $-9$ .

Solution to question 2:

3x^2 - 9x - 2x + 6 = 3x(x - 3) - 2(x - 3) = (3x - 2)(x - 3)

\boxed{x = \frac{2}{3} \quad \text{or} \quad x = 3}

Verification for $x = 3$ : $3(9) = 11(3) - 6 \Rightarrow 27 = 27 \checkmark$ . Verification for $x = \tfrac{2}{3}$ : $3 \cdot \tfrac{4}{9} = 11 \cdot \tfrac{2}{3} - 6 \Rightarrow \tfrac{4}{3} = \tfrac{4}{3} \checkmark$ .

Solving Quadratic Equations Using the Square Root Property

The square root property applies directly when a quadratic equation has no linear term ( $b = 0$ ), taking the form $ax^2 = k$ or $a(x - h)^2 = k$ . In these cases, isolate the squared expression and take the square root of both sides, remembering to include both the positive and negative roots. This method is the foundation for the technique of completing the square.

Problem 4: Pure Quadratic Equation

Easy

Solve the equation $4x^2 - 49 = 0$ using the square root property.

Isolate $x^2$ on one side of the equation.
Apply the square root property to find all solutions.

Hint

Add

49

to both sides, then divide by

4

. When you take the square root, do not forget the

\pm

symbol; there are two solutions.

View Solution

Solution to question 1:

4x^2 = 49 \implies x^2 = \frac{49}{4}

Solution to question 2:

x = \pm \sqrt{\frac{49}{4}} = \pm \frac{7}{2}

\boxed{x = \frac{7}{2} \quad \text{or} \quad x = -\frac{7}{2}}

Problem 5: Binomial Squared Form

Medium

Solve $2(x - 3)^2 - 18 = 0$ .

Isolate the squared binomial $(x-3)^2$ .
Apply the square root property and solve for $x$ .

Hint

Add

18

to both sides, divide by

2

, then take the square root of both sides. Remember that

\sqrt{9} = 3

, and account for both

+

and

-

roots.

View Solution

Solution to question 1:

2(x-3)^2 = 18 \implies (x-3)^2 = 9

Solution to question 2:

x - 3 = \pm 3

x = 3 + 3 = 6 \quad \text{or} \quad x = 3 - 3 = 0

\boxed{x = 6 \quad \text{or} \quad x = 0}

Problem 6: Square Root Property Yielding Irrational Roots

Hard

Solve $3(2x + 1)^2 - 15 = 0$ . Express your answers in simplified radical form.

Isolate $(2x+1)^2$ and apply the square root property.
Solve for $x$ and simplify the radical expressions fully.

Hint

After isolating

(2x+1)^2 = 5

, take the square root to get

2x + 1 = \pm\sqrt{5}

. Then solve for

x

by subtracting

1

and dividing by

2

. The final answers will contain

\sqrt{5}

View Solution

Solution to question 1:

3(2x+1)^2 = 15 \implies (2x+1)^2 = 5 \implies 2x + 1 = \pm\sqrt{5}

Solution to question 2:

2x = -1 \pm \sqrt{5} \implies x = \frac{-1 \pm \sqrt{5}}{2}

\boxed{x = \frac{-1 + \sqrt{5}}{2} \quad \text{or} \quad x = \frac{-1 - \sqrt{5}}{2}}

Decimal approximations: $x \approx 0.618$ or $x \approx -1.618$ .

Solving Quadratic Equations by Completing the Square

Completing the square is a powerful algebraic technique that transforms any quadratic into the form $(x - h)^2 = k$ , making it solvable by the square root property. It is the method from which the quadratic formula is derived. Mastering this technique also reveals the vertex form of a parabola and is essential for advanced topics such as conic sections and calculus optimization.

Problem 7: Completing the Square with Monic Equation

Easy

Solve $x^2 + 8x + 7 = 0$ by completing the square.

Move the constant term to the right side and complete the square on the left side.
Solve for $x$ using the square root property.

Hint

Take half of the coefficient of

x

, which is

\tfrac{8}{2} = 4

, square it to get

16

, and add

16

to both sides. The left side becomes a perfect square trinomial.

View Solution

Solution to question 1:

x^2 + 8x = -7

x^2 + 8x + 16 = -7 + 16 = 9

(x + 4)^2 = 9

Solution to question 2:

x + 4 = \pm 3

x = -4 + 3 = -1 \quad \text{or} \quad x = -4 - 3 = -7

\boxed{x = -1 \quad \text{or} \quad x = -7}

Problem 8: Completing the Square with Non-Unit Leading Coefficient

Medium

Solve $2x^2 - 12x + 10 = 0$ by completing the square.

Divide through by the leading coefficient to obtain a monic equation.
Complete the square and find all solutions, expressing them in exact form.

Hint

Divide every term by

2

first to get

x^2 - 6x + 5 = 0

. Then proceed with the standard completing-the-square algorithm: half of

-6

-3

, and

(-3)^2 = 9

View Solution

Solution to question 1:

x^2 - 6x + 5 = 0 \implies x^2 - 6x = -5

Solution to question 2:

x^2 - 6x + 9 = -5 + 9 = 4

(x - 3)^2 = 4 \implies x - 3 = \pm 2

x = 3 + 2 = 5 \quad \text{or} \quad x = 3 - 2 = 1

\boxed{x = 5 \quad \text{or} \quad x = 1}

Problem 9: Completing the Square with Irrational and Fractional Steps

Hard

Solve $3x^2 + 5x - 2 = 0$ by completing the square. Express the solutions in exact simplified form.

Divide by the leading coefficient and isolate the quadratic and linear terms.
Complete the square, then apply the square root property to obtain the two roots.

Hint

After dividing by

3

, you get

x^2 + \tfrac{5}{3}x = \tfrac{2}{3}

. Take half of

\tfrac{5}{3}

, which is

\tfrac{5}{6}

, then square it:

\left(\tfrac{5}{6}\right)^2 = \tfrac{25}{36}

. Add

\tfrac{25}{36}

to both sides before taking the square root.

View Solution

Solution to question 1:

x^2 + \frac{5}{3}x = \frac{2}{3}

Solution to question 2:

x^2 + \frac{5}{3}x + \frac{25}{36} = \frac{2}{3} + \frac{25}{36} = \frac{24}{36} + \frac{25}{36} = \frac{49}{36}

\left(x + \frac{5}{6}\right)^2 = \frac{49}{36} \implies x + \frac{5}{6} = \pm \frac{7}{6}

x = -\frac{5}{6} + \frac{7}{6} = \frac{2}{6} = \frac{1}{3} \quad \text{or} \quad x = -\frac{5}{6} - \frac{7}{6} = -\frac{12}{6} = -2

\boxed{x = \frac{1}{3} \quad \text{or} \quad x = -2}

Solving Quadratic Equations with the Quadratic Formula

The quadratic formula is the universal method for solving any equation of the form $ax^2 + bx + c = 0$ :

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

It is derived by completing the square on the general form. The expression $b^2 - 4ac$ is the discriminant, and its sign determines the nature of the solutions. This section practices applying the formula across equations with rational, irrational, and complex roots.

Problem 10: Two Rational Roots via the Quadratic Formula

Easy

Solve $x^2 - 5x - 14 = 0$ using the quadratic formula.

Identify $a$ , $b$ , and $c$ , then substitute into the quadratic formula.
Simplify the discriminant and state both solutions.

Hint

With

a=1

b=-5

c=-14

, compute

\Delta = (-5)^2 - 4(1)(-14)

. If

\Delta

is a perfect square, the roots will be rational.

View Solution

Solution to question 1:

a = 1,\quad b = -5,\quad c = -14

x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-14)}}{2(1)} = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2}

Solution to question 2:

x = \frac{5 + 9}{2} = 7 \quad \text{or} \quad x = \frac{5 - 9}{2} = -2

\boxed{x = 7 \quad \text{or} \quad x = -2}

Problem 11: Two Irrational Roots via the Quadratic Formula

Medium

Solve $2x^2 - 6x + 1 = 0$ using the quadratic formula. Leave answers in exact radical form.

Compute the discriminant and determine the nature of the roots before solving.
Apply the quadratic formula and simplify the radical completely.

Hint

The discriminant

\Delta = 36 - 8 = 28

. Since

28 = 4 \times 7

, simplify

\sqrt{28} = 2\sqrt{7}

before dividing by

2a = 4

View Solution

Solution to question 1:

$\Delta = (-6)^2 - 4(2)(1) = 36 - 8 = 28 > 0$ . Since $\Delta > 0$ and not a perfect square, there are two distinct irrational real roots.

Solution to question 2:

x = \frac{6 \pm \sqrt{28}}{2 \cdot 2} = \frac{6 \pm 2\sqrt{7}}{4} = \frac{3 \pm \sqrt{7}}{2}

\boxed{x = \frac{3 + \sqrt{7}}{2} \quad \text{or} \quad x = \frac{3 - \sqrt{7}}{2}}

Problem 12: Non-Standard Form and Complex Roots

Hard

Solve $(2x - 1)(x + 3) = -8$ using the quadratic formula.

Expand the left side, move all terms to one side, and write the equation in standard form $ax^2 + bx + c = 0$ .
Apply the quadratic formula. If the discriminant is negative, express the solutions as complex numbers in the form $a \pm bi$ .

Hint

After expanding:

2x^2 + 5x - 3 = -8

, so

2x^2 + 5x + 5 = 0

. Compute

\Delta = 25 - 40

. A negative discriminant means the roots are complex: use

\sqrt{-k} = i\sqrt{k}

View Solution

Solution to question 1:

2x^2 + 6x - x - 3 = -8 \implies 2x^2 + 5x - 3 + 8 = 0 \implies 2x^2 + 5x + 5 = 0

Solution to question 2:

\Delta = 5^2 - 4(2)(5) = 25 - 40 = -15

Since $\Delta < 0$ , the roots are complex (non-real).

x = \frac{-5 \pm \sqrt{-15}}{4} = \frac{-5 \pm i\sqrt{15}}{4}

\boxed{x = \frac{-5 + i\sqrt{15}}{4} \quad \text{or} \quad x = \frac{-5 - i\sqrt{15}}{4}}

The Discriminant and the Nature of Roots

The discriminant $\Delta = b^2 - 4ac$ of a quadratic equation $ax^2 + bx + c = 0$ classifies the roots without requiring a full solution. The three cases are: $\Delta > 0$ (two distinct real roots), $\Delta = 0$ (one repeated real root, also called a double root), and $\Delta < 0$ (two complex conjugate roots). This section develops the skill of analyzing the discriminant and working backwards from root conditions to determine unknown parameters.

Discriminant $\Delta = b^2 - 4ac$	Nature of Roots	Number of Real Solutions
$\Delta > 0$ and a perfect square	Two distinct rational roots	2
$\Delta > 0$ and not a perfect square	Two distinct irrational roots	2
$\Delta = 0$	One repeated (double) real root	1
$\Delta < 0$	Two complex conjugate roots	0

Problem 13: Classifying Roots Using the Discriminant

Easy

Without solving, classify the roots of each equation using the discriminant.

$x^2 - 6x + 9 = 0$
$2x^2 + 3x - 5 = 0$
$x^2 + x + 4 = 0$

Hint

For each equation, identify

a

b

c

and compute

\Delta = b^2 - 4ac

. Then compare

\Delta

to zero and check whether it is a perfect square.

View Solution

Solution to question 1:

\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0

One repeated real root (double root).

Solution to question 2:

\Delta = 3^2 - 4(2)(-5) = 9 + 40 = 49 > 0 \text{ (perfect square)}

Two distinct rational roots.

Solution to question 3:

\Delta = 1^2 - 4(1)(4) = 1 - 16 = -15 < 0

Two complex conjugate roots (no real solutions).

Problem 14: Finding a Parameter for a Double Root

Medium

The quadratic equation $x^2 - kx + 16 = 0$ has exactly one real solution (a double root).

Set up the condition on the discriminant for a double root and solve for all possible values of $k$ .
For each value of $k$ , state the corresponding double root.

Hint

A double root occurs when

\Delta = 0

. Here

a = 1

b = -k

c = 16

. Set

(-k)^2 - 4(1)(16) = 0

and solve for

k

. There will be two values.

View Solution

Solution to question 1:

\Delta = k^2 - 64 = 0 \implies k^2 = 64 \implies k = \pm 8

Solution to question 2:

For $k = 8$ : equation is $x^2 - 8x + 16 = (x-4)^2 = 0$ , so the double root is $\boxed{x = 4}$ .

For $k = -8$ : equation is $x^2 + 8x + 16 = (x+4)^2 = 0$ , so the double root is $\boxed{x = -4}$ .

Problem 15: Range of a Parameter for Real Solutions

Hard

Consider the quadratic equation $2x^2 + mx + 8 = 0$ , where $m$ is a real constant.

Determine the values of $m$ for which the equation has two distinct real roots.
Determine the values of $m$ for which the equation has no real roots.

Hint

Compute

\Delta = m^2 - 4(2)(8) = m^2 - 64

. For two distinct real roots, require

\Delta > 0

; for no real roots, require

\Delta < 0

. Solve each inequality.

View Solution

Solution to question 1: Two distinct real roots ( $\Delta > 0$ ):

m^2 - 64 > 0 \implies m^2 > 64 \implies |m| > 8

\boxed{m < -8 \quad \text{or} \quad m > 8}

Solution to question 2: No real roots ( $\Delta < 0$ ):

m^2 - 64 < 0 \implies m^2 < 64 \implies |m| < 8

\boxed{-8 < m < 8}

Quadratic Equations in Real-World Word Problems

Quadratic equations arise naturally in many practical contexts: projectile motion (height as a function of time), geometric area problems (dimensions of rectangles and triangles), and number theory (consecutive integers). The key skill is translating a verbal description into standard form, solving, and then interpreting the solution in context: discarding any physically or geometrically inadmissible roots (e.g., negative lengths or times).

Problem 16: Consecutive Integer Problem

Easy

The product of two consecutive positive integers is 182.

Let the smaller integer be $n$ . Write a quadratic equation in $n$ .
Solve the equation and identify the two integers.

Hint

The two consecutive integers are

n

and

n + 1

. Their product is

n(n+1) = 182

. Expand and rearrange to standard form, then factor or use the quadratic formula.

View Solution

Solution to question 1:

n(n+1) = 182 \implies n^2 + n - 182 = 0

Solution to question 2:

(n - 13)(n + 14) = 0 \implies n = 13 \quad \text{or} \quad n = -14

Since the integers must be positive, $n = 13$ . The two consecutive integers are $\boxed{13 \text{ and } 14}$ .

Verification: $13 \times 14 = 182 \checkmark$ .

Problem 17: Rectangular Area Problem

Medium

A rectangular garden has a length that is 5 meters more than twice its width. The area of the garden is 75 square meters.

Define a variable for the width and write a quadratic equation representing the area.
Solve the equation and state the dimensions of the garden.

Hint

Let

w

be the width in meters. Then the length is

2w + 5

. Area

= w(2w + 5) = 75

. Expand and rearrange, then solve. Discard any negative solution since width must be positive.

View Solution

Solution to question 1:

w(2w + 5) = 75 \implies 2w^2 + 5w - 75 = 0

Solution to question 2:

\Delta = 25 + 4(2)(75) = 25 + 600 = 625

w = \frac{-5 \pm 25}{4}

w = \frac{20}{4} = 5 \quad \text{or} \quad w = \frac{-30}{4} \text{ (rejected, negative)}

Width $= 5$ m, Length $= 2(5) + 5 = 15$ m.

\boxed{\text{Width} = 5 \text{ m}, \quad \text{Length} = 15 \text{ m}}

Verification: $5 \times 15 = 75 \checkmark$ .

Problem 18: Projectile Motion (Height and Time)

Medium

A ball is launched vertically upward from the top of a building 20 m tall with an initial velocity of 15 m/s. Its height above the ground at time $t$ seconds is modeled by: $h(t) = -5t^2 + 15t + 20$

Find the time(s) at which the ball is at a height of 30 m above the ground.
Find the time at which the ball hits the ground. Discard any non-physical solution.

Hint

For question 1, set

h(t) = 30

and simplify. For question 2, set

h(t) = 0

. In both cases, a negative value of

t

has no physical meaning and must be discarded.

View Solution

Solution to question 1:

-5t^2 + 15t + 20 = 30 \implies -5t^2 + 15t - 10 = 0 \implies t^2 - 3t + 2 = 0

(t-1)(t-2) = 0 \implies t = 1 \text{ s or } t = 2 \text{ s}

The ball reaches 30 m on the way up at $t = 1$ s and again on the way down at $t = 2$ s.

Solution to question 2:

-5t^2 + 15t + 20 = 0 \implies t^2 - 3t - 4 = 0 \implies (t-4)(t+1) = 0

t = 4 \text{ s} \quad \text{or} \quad t = -1 \text{ s (rejected)}

\boxed{t = 4 \text{ seconds}}

Problem 19: Uniform Border Around a Rectangle

Hard

A photograph measuring 12 cm by 8 cm is mounted on a card with a uniform border of width $x$ cm on all four sides. The total area of the mounted card is 180 cm².

Express the total area of the card in terms of $x$ and set up a quadratic equation.
Solve for $x$ using the quadratic formula, rounding your answer to two decimal places if necessary.

Hint

The card’s total dimensions are

(12 + 2x)

cm by

(8 + 2x)

cm. Set the product equal to 180, expand, and rearrange into standard form. Discard the negative root.

View Solution

Solution to question 1:

(12 + 2x)(8 + 2x) = 180

96 + 24x + 16x + 4x^2 = 180

4x^2 + 40x + 96 = 180 \implies 4x^2 + 40x - 84 = 0 \implies x^2 + 10x - 21 = 0

Solution to question 2:

\Delta = 100 + 84 = 184

x = \frac{-10 \pm \sqrt{184}}{2} = \frac{-10 \pm 2\sqrt{46}}{2} = -5 \pm \sqrt{46}

x = -5 + \sqrt{46} \approx -5 + 6.782 \approx 1.78 \text{ cm} \quad \text{(the negative root is rejected)}

\boxed{x \approx 1.78 \text{ cm}}

Problem 20: Rate and Work Problem Modeled by a Quadratic

Hard

Two pipes together fill a tank in 6 hours. Pipe A alone takes 5 hours longer than Pipe B alone to fill the same tank. Let $t$ represent the time (in hours) for Pipe B to fill the tank alone.

Set up a rational equation expressing the combined rate, then clear fractions to form a quadratic equation in $t$ .
Solve the quadratic equation. State the time for each pipe to fill the tank individually, discarding any inadmissible solution.

Hint

Pipe B fills

\tfrac{1}{t}

of the tank per hour; Pipe A fills

\tfrac{1}{t+5}

per hour. Together they fill

\tfrac{1}{6}

per hour. Multiply through by

6t(t+5)

to eliminate fractions.

View Solution

Solution to question 1:

\frac{1}{t} + \frac{1}{t+5} = \frac{1}{6}

Multiply through by $6t(t+5)$ :

6(t+5) + 6t = t(t+5)

6t + 30 + 6t = t^2 + 5t

t^2 + 5t - 12t - 30 = 0 \implies t^2 - 7t - 30 = 0

Solution to question 2:

(t - 10)(t + 3) = 0 \implies t = 10 \quad \text{or} \quad t = -3 \text{ (rejected)}

\boxed{t = 10 \text{ h (Pipe B)}, \quad t + 5 = 15 \text{ h (Pipe A)}}

Verification: $\tfrac{1}{10} + \tfrac{1}{15} = \tfrac{3}{30} + \tfrac{2}{30} = \tfrac{5}{30} = \tfrac{1}{6} \checkmark$ .