Solving Quadratic Equations

In earlier lessons, you learned to solve certain quadratic equations by factoring them. In this lesson, you will learn a more general method for solving quadratic equations, by taking square roots.

Solving $(x-h)^2=d$

To solve an equation in which one entire side is squared, you take the square root of both sides and remember the rule that $√{A^2}={|A|}$ (the absolute value of $A$).

Let’s begin by solving the equation $x^2=9$. The two sides of this equation are graphed to the left. Remember that the solutions to the equation are given by the $x$-coordinates of the graphs’ intersection points.

We’ll now look at what happens when you take the square root of both sides of this equation. Use the button to switch between that version of the equation and the original. Do the $x$-coordinates of the points where the two graphs intersect change?
Does the equation $√{x^2}=√9$ have the same solutions as the original equation?
Simplify the equation $√{x^2}=√9$ by applying the rule $√{A^2}={|A|}$ to the left-hand side, and computing the right-hand side.
What are the solutions to this simpler equation?

If $B$ is positive then the solutions to $|A|=B$, where $A$ is some unknown expression, are always given by $A=B$ and $A=-B$. A shorthand for this is to write $A=±B$ (read as “$A$ equals plus or minus $B$”). For example, in basicEx, we found that if $x^2=9$ then $x=±3$.

We want to solve the equation $(x-3)^2=4$. Take the square root of both sides of this equation, and then simplify the resulting equation as in basicEx.

Give the two solutions to this equation, by breaking up the $±$ in the last line into its two possibilities.

Solve each equation below.

Click to see the two sides of the equation $(x+2)^2=-3$ graphed on the left. Does that equation look like it has any solutions?
Can the square of a real number ever be negative?

As you can see:

When $d ≥ 0$, the solutions to the equation $(x-h)^2=d$ are the numbers $h ± √d$. When $d < 0$, the equation has no real solutions.

If you have an equation where the right-hand side isn’t a perfect square, like $(x-3)^2=2$, the only exact way to write the solutions is with the square root symbol left in. That is, you should leave the solutions to $(x-3)^2=2$ as $x=3 ± √2$ unless someone asks you to approximate them (with a calculator).

Click to see a graph of both sides of the equation $(x+4)^2=7$. What are the solutions to that equation?
Now click to see a graph of both sides of the equation $(x+2)^2=6$. What are the solutions to that equation?
What are the approximate decimal values of the solutions to the equation $(x+2)^2=6$ (to 3 decimal places)? Use a calculator.

Solving $a(x-h)^2+k=0$

To solve equations in the form $a(x-h)^2+k=0$, rearrange them so that they are in the form $(x-h)^2=d$, by first subtracting $k$ from both sides and then dividing by $a$.

Solve each equation in the table below, as shown in the example row. The two sides of the equations will be graphed to the left.

If you start with any equation of the form $a(x-h)^2+k=0$ with $a≠0$ and perform these steps, you get:

$$ \cl"tight"{\table a(x-h)^2, +, k, , , =, 0; a(x-h)^2, +, k, -, k, =, -k; a(x-h)^2, , , , , =, -k; {a(x-h)^2}/a, , , , , =, -k/a; (x-h)^2, , , , , =, -k/a; x, , , , , =, h ± √{-k/a}} $$

This gives us a formula for solutions to such equations:

The solutions to the equation $a(x-h)^2+k=0$ with $a≠0$ are $$x= h ± √{-k/a}$$ whenever $$-k/a ≥ 0$$. When $$-k/a > 0$$, the equation has 2 solutions; when $$-k/a = 0$$, it has 1 solution; and when $$-k/a < 0$$, it has no real solutions.

Completing the square — solving $x^2+2mx=n$

The polynomial $x^2+4x$ is shown on the grid to the left, as $x^2+2x+2x$ with one copy of $2x$ to the right of the $x^2$ block and the other above it.

Notice that this picture of $x^2+2x+2x$ looks like a large square with a piece removed. How many grid squares are in that piece? (That is, how many small green squares would you have to add to the grid in order to “complete” the large square $(x+2)^2$?)

Click to get an $e$ slider that fills in the missing piece of the square. If you set $e$ equal to your previous answer, there should be a large, complete, green square on the grid to the left.

When you slide the slider for $x$, does your answer to the previous question change? (That is, does the $e$ that makes $x^2+2x+2x+e = (x+2)^2$ change as $x$ changes?)

The equation $x^2+4x=5$ is illustrated on the grids to the left, with $x^2+4x$ written as $x^2+2x+2x$ as in compSquareQn. The $e$ slider now adds $e$ to both sides of that equation.

We want to solve this equation for $x$ by using the grids to the left. What value of $x$ makes the two grids contain the same number of small green squares?
Does your previous answer change when you slide $e$?

Set $e$ to $4$, as you did in the previous question. There is now a complete large square on the top grid to the left.

How long is one side of this square?
When $e$ is set to $4$, what is another way of writing the equation $x^2+4x=5$?
Using the method you learned in the first few questions of this lesson, what are the solutions to that equation?

In general, you can solve $x^2+2mx=n$ by adding $m^2$ to both sides. To see this, notice that $(x+m)^2=x^2+2mx+m^2$:

$$ \cl"tight"{\table , \colspan 7 (x+m)^2; =, \colspan 7 (x+m)(x+m); =, \colspan 3 x(x+m), +, \colspan 3 m(x+m); =, x^2, +, m(x), +, x(m), +, m^2; =, x^2, +, mx, +, mx, +, m^2; =, x^2, +, \colspan 3 2mx, +, m^2; } $$

That is, $x^2+2mx+m^2$ can be factored as $(x+m)^2$. So, when you add $m^2$ to both sides of $x^2+2mx=n$, the equation becomes

$$ \cl"tight"{\table x^2+2mx+m^2, =, n + m^2; \colspan 3 \text"or"; (x+m)^2, =, n + m^2; } $$

which can be solved by the method given in the section on solving $(x-h)^2=d$ above.

This method for solving $x^2+2mx=n$ is referred to as completing the square.

Reducing to vertex form

In the last lesson, you learned how to take an expression in standard form $ax^2+bx+c$ and rewrite it in vertex form $a(x-h)^2+k$:

  • The coefficient $a$ stays the same between the two forms.
  • $h$ is the $x$-coordinate of the axis of symmetry (or the vertex), so it’s given by $$h=-b/{2a}$$.
  • $k$ is the $y$-coordinate of the vertex, so it’s given by plugging $h$ back into the original expression: $k=ah^2+bh+c$.

This is very much like the process of completing the square that you performed in Questions compSquareQn and compSquareQn2, when $a=1$ and $b=2m$.

We’ll start by rewriting $x^2+4x+3$ in vertex form. What are the values of $a, h, k$ for this polynomial? Use scratch paper.

What is the vertex form of $x^2+4x+3$?

Check your answer to the last question, by taking the polynomial in vertex form you just found, expanding it, and simplifying the result. If you’ve done all your algebra correctly, you should get back the original polynomial $x^2+4x+3$.

We can simplify the formula $k=ah^2+bh+c$. Multiplying both sides of $$h=-b/{2a}$$ by $-2a$, we get $-2ah=b$. Plugging this equation for $b$ into the one for $k$, we get:

$$ \cl"tight"{\table k, =, ah^2, +, bh, +, c; , =, ah^2, -, 2ah(h), +, c; , =, ah^2, -, 2ah^2, +, c; , =, \colspan 3 -ah^2, +, c} $$
Use the formula $k=-ah^2+c$ to compute $k$ for the quadratic polynomial $x^2+4x+3$. Remember that you already determined that $$h=-4/{2(1)}=-2$$ for this polynomial.
Does this match the value of $k$ you found for that polynomial in vertForm?

We’ll check algebraically that $a(x-h)^2+k = ax^2+bx+c$. First,

$$ \cl"tight"{\table (x-h)^2, =, \colspan 7 (x-h)(x-h); , =, \colspan 3 x(x-h), -, \colspan 3 h(x-h); , =, x^2, -, hx, -, hx, +, h^2; , =, x^2, -, \colspan 3 2hx, +, h^2; $$

Therefore, whenever $$h=-b/{2a}$$ and $k=-ah^2+c$:

$$ \cl"tight"{\table a(x-h)^2+k, =, {a(x^2}, -, 2hx, +, {h^2)}, +, \colspan 3 k; , =, ax^2, -, 2ahx, +, ah^2, +, \colspan 3 (-ah^2+c); , =, ax^2, -, 2a(-b/{2a})x, +, ah^2, -, ah^2, +, c; , =, ax^2, +, bx, , , , , +, c} $$

Rewrite each polynomial below in vertex form, using scratch paper and the formulas $$h=-b/{2a}$$ and $k=-ah^2+c$. Then check your answer. (Remember to write fractions in reduced form.)

In Questions vertForm through vertFormTableQn, you learned how to rewrite expressions of the form $ax^2+bx+c$ in vertex form (that is, in the form $a(x-h)^2+k$). In solveVert, you learned how to solve equations where one side is in vertex form and the other side is $0$. Putting these two methods together gives you a way to solve equations of the form $ax^2+bx+c=0$.

Use this method to solve the equation $2x^2-5x+3=0$, the left-hand side of which is graphed on the grid to the left.

First, find the values $a, h, k$ for the polynomial $2x^2-5x+3$ in vertex form.

What is an equivalent equation to $2x^2-5x+3=0$, in which the left-hand side is in vertex form? (Write fractions by using the ‘/’ key, or as decimals.)

Using the formula $x=h ± √{-k ∕ a}$, find the solutions to this rewritten equation. (Remember that $$√{a/b}={√a}/{√b}$$.)

Check those solutions in the original equation $2x^2-5x+3=0$.