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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.
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.
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.
Solve each equation below.
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).
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:
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.
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.
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.
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.
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.
In general, you can solve $x^2+2mx=n$ by adding $m^2$ to both sides. To see this, notice that $(x+m)^2=x^2+2mx+m^2$:
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
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.
In the last lesson, you learned how to take an expression in standard form $ax^2+bx+c$ and rewrite it in vertex form $a(x-h)^2+k$:
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.
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:
We’ll check algebraically that $a(x-h)^2+k = ax^2+bx+c$. First,
Therefore, whenever $$h=-b/{2a}$$ and $k=-ah^2+c$:
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.
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 $2x^2-5x+3=0$.