Back to homeMathematics · Algebra 2 · Change subject
Start for free
Algebra 2 · Mathematics
Equations · Equations

Quadratic Equations and the Quadratic Formula

Before you start, make sure you know: Quadratic equations (Algebra 1)
Before you start, make sure you know: Problems with quadratic equations (Algebra 1)

Marcus punts a football. Its height in feet, $t$ seconds after the kick, is $h = -16t^2 + 48t + 3$. At what times is the ball exactly $35$ ft high? You will find that there are two answers, and you will see why before the end of this page.

In Algebra 1 you solved quadratics by factoring and by the quadratic formula. In Algebra 2 the quadratic equation becomes a tool. You will find out how many solutions it has without solving it, get information from the coefficients alone, and turn complicated-looking equations into quadratics.

Standard Form and the Quadratic Formula

Graphically, the solutions are the $x$-intercepts of the parabola $y = ax^2 + bx + c$. Move the sliders and watch where the parabola crosses the $x$-axis.

Which method should you use? Look at the equation first:

  • If it factors easily, factor and use the zero product property.

  • If there is no $x$ term, isolate $x^2$ and take square roots (next section).

  • The quadratic formula works on every quadratic equation, so when in doubt, use it.

Equations that need to be simplified first

Sometimes the equation is not in standard form yet. Expand, move every term to one side, combine like terms, and then identify $a$, $b$, $c$.

Practice★☆☆Not started
Solve the following complete quadratic equation by applying the quadratic formula: $x^2 - 7x + 10 = 0$

Solve the following complete quadratic equation by applying the quadratic formula:

$x^2 - 7x + 10 = 0$

Solve
Practice★☆☆Not started
Expand the expressions, simplify and solve the resulting quadratic equation: $(x - 2)(x + 3) = 2x - 6$

Expand the expressions, simplify and solve the resulting quadratic equation:

$(x - 2)(x + 3) = 2x - 6$

Solve

Equations Without a Middle or Constant Term

Some problems call these incomplete quadratic equations. You can solve them faster than with the formula.

Flowchart: a quadratic with no x term is solved by isolating x squared; a quadratic with no constant term is solved by factoring out x. no x term:ax² + c = 0 isolate x² x = ±√(−c/a) no constant term:ax² + bx = 0 factor out x x(ax + b) = 0x = 0 or x = −b/a

What if the number under the root is negative? Then $x^2 = -16$ has no real solution. In the complex numbers, $x = \pm 4i$.

Practice★☆☆Not started
Find the solutions to the following incomplete quadratic equation by extracting a common factor: $3x^2 - 12x = 0$

Find the solutions to the following incomplete quadratic equation by extracting a common factor:

$3x^2 - 12x = 0$

Solve
Practice★☆☆Not started
Solve the following incomplete quadratic equation by directly isolating the unknown: $2x^2 - 18 = 0$

Solve the following incomplete quadratic equation by directly isolating the unknown:

$2x^2 - 18 = 0$

Solve

The Discriminant

The expression under the square root, $b^2 - 4ac$, tells you what kind of solutions to expect before you solve anything.

Three parabolas: with a positive discriminant the parabola crosses the x-axis twice, with a zero discriminant it touches the axis once, with a negative discriminant it never meets the axis. Δ > 0two real roots Δ = 0one double root Δ < 0no real roots
Practice★☆☆Not started
Without solving the equation, determine the number of real solutions it has by calculating the value of its discriminant: $5x^2 - 3x + 2 = 0$

Without solving the equation, determine the number of real solutions it has by calculating the value of its discriminant:

$5x^2 - 3x + 2 = 0$

Solve
Practice★★☆Not started
Determine the value (or values) of the parameter $k$ so that the equation $x^2 - kx + 9 = 0$ has a single double real solution.

Determine the value (or values) of the parameter $k$ so that the equation $x^2 - kx + 9 = 0$ has a single double real solution.

Solve

Sum and Product of the Roots

You can learn a lot about the solutions without finding them.

Diagram relating the coefficients of x squared plus x minus 20 equals zero to the sum and the product of its two roots. x² + x − 20 = 0 sum of rootsx₁ + x₂ = −1 (= −b) product of rootsx₁ · x₂ = −20 (= c) x₁ = 4, x₂ = −5
Practice★☆☆Not started
Given the equation $4x^2 - 5x - 9 = 0$, calculate the sum and the product of its roots using Vieta's formulas, without needing to solve the equation.

Given the equation $4x^2 - 5x - 9 = 0$, calculate the sum and the product of its roots using Vieta's formulas, without needing to solve the equation.

Solve
Practice★☆☆Not started
Write a quadratic equation in the form $ax^2 + bx + c = 0$ (with $a=1$) knowing that its solutions are $x_1 = 4$ and $x_2 = -3$.

Write a quadratic equation in the form $ax^2 + bx + c = 0$ (with $a=1$) knowing that its solutions are $x_1 = 4$ and $x_2 = -3$.

Solve
Practice★★★Not started
Let $x_1$ and $x_2$ be the roots of the equation $3x^2 - 5x + 1 = 0$. Without explicitly calculating the value of $x_1$ and $x_2$, use the properties of the roots to…

Let $x_1$ and $x_2$ be the roots of the equation $3x^2 - 5x + 1 = 0$. Without explicitly calculating the value of $x_1$ and $x_2$, use the properties of the roots to find the exact value of the expression $\frac{1}{x_1^2} + \frac{1}{x_2^2}$.

Solve

Zeros and Factors

Solving $ax^2 + bx + c = 0$ is the same as finding the zeros of the polynomial $P(x) = ax^2 + bx + c$. The Factor Theorem connects zeros to factors: if $r$ is a zero of $P$, then $(x - r)$ is a factor of $P$.

Drag the sliders: the zeros $a$ and $b$ are exactly where the parabola crosses the $x$-axis.

A zero is a number that makes the equation true. So if a problem tells you one solution, substitute it.

Practice★☆☆Not started
Find the roots of the polynomial $P(x) = 2x^2 + 5x - 3$ and use them to write the factored polynomial.

Find the roots of the polynomial $P(x) = 2x^2 + 5x - 3$ and use them to write the factored polynomial.

Solve
Practice★★☆Not started
In the parametric equation $2x^2 + mx + 15 = 0$, it is known that one of the solutions is $x = 3$. Without needing to use the quadratic formula, calculate the value of…

In the parametric equation $2x^2 + mx + 15 = 0$, it is known that one of the solutions is $x = 3$. Without needing to use the quadratic formula, calculate the value of the parameter $m$ and find the other solution.

Solve

Equations with a Parameter

Sometimes an equation contains an extra letter, a parameter such as $k$ or $m$. Then the question is how the number of solutions changes with the parameter. Try it: the slider $a$ below plays the role of $k$ in $kx^2 + 2x + 1 = 0$.

The discriminant is also the tool for proofs about solutions. Compare $\Delta$ with something whose sign you know.

Practice★★★Not started
Discuss the number of real solutions to the equation $(m-1)x^2 - 2mx + (m+2) = 0$ depending on the different values that the parameter $m$ can take.

Discuss the number of real solutions to the equation $(m-1)x^2 - 2mx + (m+2) = 0$ depending on the different values that the parameter $m$ can take.

Solve
Practice★★★Not started
Prove algebraically that the quadratic equation $ax^2 + bx + c = 0$ (with $a \neq 0$) always has two distinct real solutions if the coefficients $a$ and $c$ have…

Prove algebraically that the quadratic equation $ax^2 + bx + c = 0$ (with $a \neq 0$) always has two distinct real solutions if the coefficients $a$ and $c$ have opposite signs (that is, if $a \cdot c < 0$ is satisfied).

Solve

Equations That Reduce to Quadratics

Many equations that do not look quadratic become quadratic after a little work.

Rational equations

Multiply both sides by the least common denominator (LCD). Write down the restrictions first: values that make a denominator zero cannot be solutions.

Watch out: after clearing denominators the $x^2$ terms sometimes cancel. Then you are left with a linear equation and only one solution. That is fine, just solve it.

Substitution (quadratic in form)

If an equation has the form $A u^2 + B u + C = 0$ where $u$ is some expression in $x$, replace that expression with a new letter, solve for $u$, and then go back to $x$.

Flowchart for solving x to the fourth minus 13 x squared plus 36 equals zero by substituting u equals x squared, solving for u, and then undoing the substitution. x⁴ − 13x² + 36 = 0 u = x² u² − 13u + 36 = 0 solve u = 4 or u = 9 x² = u x² = 4 or x² = 9 x = ±2 or x = ±3

The substituted piece can be a whole expression, such as $u = x^2 + 1$ or $u = x^2 - 3x$.

Practice★★☆Not started
Solve the following equation with algebraic fractions by working it out until it is reduced to a quadratic equation: $\frac{x+1}{x-2} + \frac{3x}{x+2} = 4$

Solve the following equation with algebraic fractions by working it out until it is reduced to a quadratic equation:

$\frac{x+1}{x-2} + \frac{3x}{x+2} = 4$

Solve
Practice★★★Not started
Solve the following fourth-degree equation by making a suitable change of variable to transform it into a quadratic equation: $(x^2 - 3x)^2 - 2(x^2 - 3x) - 8 = 0$

Solve the following fourth-degree equation by making a suitable change of variable to transform it into a quadratic equation:

$(x^2 - 3x)^2 - 2(x^2 - 3x) - 8 = 0$

Solve

Projectile and Profit Models

When a quadratic function models a situation, questions like "when?" or "how many?" turn into quadratic equations. Use this plan:

  1. Identify what the variable stands for (and its units).

  2. Write the equation from the situation.

  3. Solve it.

  4. Check that each solution makes sense in context.

Back to the start: here is Marcus's punt together with the line $h = 35$. The times when the ball is $35$ ft high are where the curve crosses the line.

Practice★★☆Not started
In a physics experiment, a water rocket is launched vertically upward. Its height $h$ in feet above the ground, as a function of time $t$ in seconds, is given by the…

In a physics experiment, a water rocket is launched vertically upward. Its height $h$ in feet above the ground, as a function of time $t$ in seconds, is given by the expression $h(t) = -5t^2 + 20t + 2$. Determine at what instant or instants the rocket is exactly $17$ feet high.

Solve
Practice★★☆Not started
A technology company manufactures external hard drives. The monthly profit, in thousands of dollars, when selling $x$ hundreds of units is modeled by the quadratic…

A technology company manufactures external hard drives. The monthly profit, in thousands of dollars, when selling $x$ hundreds of units is modeled by the quadratic function $B(x) = -2x^2 + 24x - 54$. How many units must be sold for the profit to be zero (break-even point)?

Solve

Area Problems

Area is length times width, so two unknown lengths multiplied together give a quadratic. Draw a figure and label every side with an expression.

A rectangular photo 12 inches by 8 inches surrounded by a frame of uniform width x, so the outer rectangle measures 12 plus 2x by 8 plus 2x. photo12 in. × 8 in.xx outer: (12 + 2x) by (8 + 2x)
Practice★★☆Not started
An architect is designing a rectangular solar panel for the roof of an industrial building. Due to design constraints, the base of the panel must measure exactly $3$…

An architect is designing a rectangular solar panel for the roof of an industrial building. Due to design constraints, the base of the panel must measure exactly $3$ feet more than its height. If the total area of the panel must be $28 \text{ ft}^2$, set up a quadratic equation and calculate its dimensions.

x + 3 x Area = 28 ft²
Solve
Practice★★☆Not started
A photograph $20 \text{ cm}$ high by $30 \text{ cm}$ wide is to be framed with a wooden frame of uniform width $x$. If the total area of the frame (excluding the…

A photograph $20 \text{ cm}$ high by $30 \text{ cm}$ wide is to be framed with a wooden frame of uniform width $x$. If the total area of the frame (excluding the photograph) is $216 \text{ cm}^2$, set up an equation and find the width of the frame.

PHOTOGRAPH 20 × 30 in x x FRAME 216 sq in
Solve
Practice★★★Not started
In a right triangle with legs $6 \text{ cm}$ and $8 \text{ cm}$, a rectangle is inscribed so that two of its sides lie on the legs and one of its vertices lies on the…

In a right triangle with legs $6 \text{ cm}$ and $8 \text{ cm}$, a rectangle is inscribed so that two of its sides lie on the legs and one of its vertices lies on the hypotenuse. If the area of the rectangle must be exactly $12 \text{ cm}^2$, set up a quadratic equation and find the possible dimensions of the rectangle.

8 in 6 in P(x, y) Area = 12
Solve

Number and Sharing Problems

For "consecutive" numbers, use $n$ and $n + 1$ (consecutive integers) or $n$ and $n + 2$ (consecutive odd or consecutive even integers). For sharing problems, use cost per person $= \dfrac{\text{total}}{\text{number of people}}$.

Two bars of the same total 100 dollars: split among 5 friends each pays 20 dollars, split among 4 friends each pays 25 dollars. 5 friends pay $20 each $20$20$20$20$20 4 friends pay $25 each $25$25$25$25 Same total: $100
Practice★★☆Not started
The product of two consecutive odd integers is $323$. Set up a quadratic equation and solve it to find what these two numbers are. Is there more than one possible pair?

The product of two consecutive odd integers is $323$. Set up a quadratic equation and solve it to find what these two numbers are. Is there more than one possible pair?

Solve
Practice★★★Not started
A group of friends decides to rent a sailboat for the weekend for a total cost of $\$360…

A group of friends decides to rent a sailboat for the weekend for a total cost of $\$360$, which they agree to pay in equal parts. At the last minute, two of the friends cannot attend, forcing the rest of the group to pay $\$9$ more each to cover the total cost. Set up an equation that reduces to a quadratic equation to determine how many friends made up the original group.

Solve

Distance and Ratio Problems

When two motions are at right angles (north and east), the distance between the objects is the hypotenuse of a right triangle, so the Pythagorean Theorem gives a quadratic. Proportions between lengths also lead to quadratics after cross-multiplying.

Right triangle: one cyclist rides north a distance v miles, another rides east v plus 7 miles, and they end up 13 miles apart. north: v mi east: v + 7 mi 13 mi start
Practice★★★Not started
Two surveillance drones depart simultaneously from the same base. The first drone flies in a straight line toward the north at a constant speed of $v \text{ km/h}$, and…

Two surveillance drones depart simultaneously from the same base. The first drone flies in a straight line toward the north at a constant speed of $v \text{ km/h}$, and the second flies toward the east at a speed of $v + 5 \text{ km/h}$. After $2$ hours of flight, the straight-line distance separating them is $50 \text{ km}$. Set up a quadratic equation and determine the speed of each drone.

Solve
Practice★★★Not started
The golden ratio is obtained by dividing a segment of length $L$ into two parts $x$ and $y$ (with $x > y$) in such a way that the ratio of the total segment to the…

The golden ratio is obtained by dividing a segment of length $L$ into two parts $x$ and $y$ (with $x > y$) in such a way that the ratio of the total segment to the larger part is equal to the ratio of the larger part to the smaller part. If we set the length of the smaller part as $y = 1$, set up the resulting quadratic equation for $x$ and find the exact value of the golden ratio $\Phi$.

Solve

Summary

See all the exercises in this section