Wize High School Algebra I Textbook (Common Core) > Radical Equations and Functions

Solving Radical Equations

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Solving Radical Equations
A radical equation is
f(x)=g(x)\boxed{\sqrt{f(x)}=g(x)}f(x)​=g(x)​
where f(x) & g(x) f(x)~\&~g(x)~f(x) & g(x)  are continuous polynomial functions. 

Solutions to Radical Equations
Solve 2x+1=32\sqrt{x}+1=32x​+1=3. State the restrictions for 'x'.

Let f(x)=2x+1f(x)=2\sqrt{x}+1f(x)=2x​+1 and let g(x)=3g(x)=3g(x)=3. 

Restrictions: x≥0x\geq0x≥0

2x+1=32\sqrt{x}+1=32x​+1=3
  ⟹  2x=2\implies 2\sqrt{x}=2⟹2x​=2
  ⟹  x=1\implies \sqrt{x}=1⟹x​=1
  ⟹  x=1\implies x=1⟹x=1

Graph f(x) f(x)~f(x)  and g(x)g(x)g(x):


f(x) & g(x) f(x)~\&~g(x)~f(x) & g(x)  intersect at (1, 3).

Let's verify the answer:
2x+1=321+1=33=3✓\begin{array}{cccc}
2\sqrt{x}+1&=&3\\\\
2\sqrt{1}+1&=&3\\\\
3&=&3&\checkmark
\end{array}2x​+121​+13​===​333​✓​
Therefore, (1, 3) is the solution.

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Algebraic Solutions to Radical Equations
Solve 5x+6−2=x\sqrt{5x+6}-2=x5x+6​−2=x. State the domain and range

Isolate for 'x'

Restriction:  Since the argument of a square root must be 0 or larger, 5x+6≥05x+6\ge05x+6≥0 which means that our domain can be given by x≥−65x\ge-\frac65x≥−56​, or [−65,∞)[-\frac65,\infin)[−56​,∞) in interval notation.  The range can be found be recognizing that the square root has an output of 0 or higher, and so subtracting 2 from this will give us a possible range of [−2,∞)[-2,\infin)[−2,∞).

5x+6−2=x(5x+6)2=(x+2)25x+6=x2+4x+40=x2−x−20=(x−2)(x+1)∴  x=−1,2\begin{array}{ccc}
\sqrt{5x+6}-2&=&x\\\\
(\sqrt{5x+6})^2&=&(x+2)^2\\\\
5x+6&=&x^2+4x+4\\\\
0&=&x^2-x-2\\\\
0&=&(x-2)(x+1)\\\\
\therefore~~x&=&-1, 2

\end{array}5x+6​−2(5x+6​)25x+600∴  x​======​x(x+2)2x2+4x+4x2−x−2(x−2)(x+1)−1,2​
Verify:
5(−1)+6−2=−1−5+6−2=−11−2=−1−1==−1✓                     5(2)+6−2=210+6−2=216−2=22=2✓\begin{array}{rccc}
\sqrt{5(-1)+6}-2&=&-1\\\\
\sqrt{-5+6}-2&=&-1\\\\
\sqrt{1}-2&=&-1\\\\
-1&=&=-1&\checkmark
\end{array}
~~~~~~~~~~~~~~~~~~~~~
\begin{array}{rccc}
\sqrt{5(2)+6}-2&=&2\\\\
\sqrt{10+6}-2&=&2\\\\
\sqrt{16}-2&=&2\\\\
2&=&2&\checkmark
\end{array}
5(−1)+6​−2−5+6​−21​−2−1​====​−1−1−1=−1​✓​                     5(2)+6​−210+6​−216​−22​====​2222​✓​

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Example: Solving Radical Equations
Solve graphically & algebraically, stating the restrictions on xxx:
x−4+x=4\sqrt{x-4}+x=4x−4​+x=4
Graphically: 

Let f(x)=x−4+xf(x)=\sqrt{x-4}+xf(x)=x−4​+x and g(x)=4g(x)=4g(x)=4. Then, 

The solution is (4, 4)

Restrictions on 'x': x≥4x\geq4x≥4

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Algebraically:
x−4+x=4\sqrt{x-4}+x=4x−4​+x=4

(x−4)2=(−x+4)2x−4=x2−8x+160=x2−9x+200=(x−5)(x−4)∴  x=4,5\begin{array}{rcl}
(\sqrt{x-4})^2&=&(-x+4)^2\\\\
x-4&=&x^2-8x+16\\\\
0&=&x^2-9x+20\\\\
0&=&(x-5)(x-4)\\\\
\therefore~~x&=&4, 5


\end{array}(x−4​)2x−400∴  x​=====​(−x+4)2x2−8x+16x2−9x+20(x−5)(x−4)4,5​
Verify:
x−4+x=44−4+4=44=4✓                  5−4+5=45−4+5=46≠4   Extraneous Solution\begin{array}{rcl}
\sqrt{x-4}+x&=&4\\\\
\sqrt{4-4}+4&=&4\\\\
4&=&4&\checkmark
\end{array}
~~~~~~~~~~~~~~~~~~
\begin{array}{rcl}
\sqrt{5-4}+5&=&4\\\\
\sqrt{5-4}+5&=&4\\\\
6&\neq&4&&~~~\footnotesize{\text{Extraneous Solution}}
\end{array}x−4​+x4−4​+44​===​444​✓​                  5−4​+55−4​+56​===​444​​   Extraneous Solution​

Therefore, only x = 4 is a solution. 

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Example: Solving Radical Equations
Mia was working on building a fence around a triangular portion of their garden.
If she has exactly 12 meters of fencing, what are the side lengths of the garden?

First, we want to set up an equation that represents the fencing being used:

12=3+x−22+x−112=x−19+x−1\begin{aligned}
12 &= 3 + x - 22 + \sqrt{x-1} \\
12 &= x - 19 + \sqrt{x-1}
\end{aligned}1212​=3+x−22+x−1​=x−19+x−1​​
To find the missing sides we have to solve this equation for xxx.

12=x−19+x−131−x=x−1(31−x)2=(x−1)2961−62x+x2=x−1x2−63x+962=0(x−26)(x−37)=0\begin{aligned}
12 &= x - 19 + \sqrt{x - 1} \\
31 - x &= \sqrt{x - 1} \\
(31 - x)^2 &= (\sqrt{x-1})^2 \\
961 - 62x + x^2 &= x - 1 \\
x^2 - 63x + 962 &= 0 \\
(x - 26)(x - 37) & = 0
\end{aligned}1231−x(31−x)2961−62x+x2x2−63x+962(x−26)(x−37)​=x−19+x−1​=x−1​=(x−1​)2=x−1=0=0​

So we have that x=26x = 26x=26 or x=37x = 37x=37.

If x=26x = 26x=26 then the sides of the triangle will be 3, 4, and 5 meters.
This agrees with the fact that Mia has 12 meters of fencing.

If x=37x = 37x=37 then the sides of the triangle will be 3, 15, 6 meters.
The would mean Mia had 24 meters of fencing, which does not agree with the given information.

Therefore, the side lengths are 3, 4, 53,\ 4,\ 53, 4, 5 meters

Practice: Solving Radical Equations
For each of the following radical expressions, determine the correct restriction on the values of aaa.

a) 5−a\sqrt{5-a}5−a​

b) 2a−4\sqrt{2a-4}2a−4​

c) a2−9+a\sqrt{a^2-9}+\sqrt{a}a2−9​+a​

a \le

a \ge

a \ge

I don't know

Practice: Solving Radical Equations
Solve algebraically, stating any restrictions on xxx:
x−5=2x+3\begin{array}{rcl}
x-5&=&2\sqrt{x+3}
\end{array}x−5​=​2x+3​​

Answer

I don't know

Practice: Solving Radical Equations
Solve algebraically, stating any restrictions:
x+7+2=3−x\sqrt{x+7}+2=\sqrt{3-x}x+7​+2=3−x​

Extra Practice

Solving Radical Equations

Solve for a:
2a+4−3=3a−22\sqrt{a+4}-3=\sqrt{3a-2}2a+4​−3=3a−2​

Solving Radical Equations

Solve for xxx: 
2x−1=x\sqrt{2x-1}=\sqrt{x}2x−1​=x​

Solving Radical Equations

Practice: Solving Radical Equations
Solve algebraically, stating any restrictions:
x−6+3=x+9\sqrt{x-6}+3=\sqrt{x+9}x−6​+3=x+9​

Solving Radical Equations

Practice: Solving Radical Equations
Solve algebraically, stating any restrictions:
x2−2=x+4\sqrt{x^2-2}=\sqrt{x+4}x2−2​=x+4​

Solving Radical Equations

Practice: Solving Radical Equations
Solve algebraically, stating any restrictions on xxx:
10+10x−1=13\begin{array}{rcl}
10+\sqrt{10x-1}&=&13
\end{array}10+10x−1​​=​13​

Solving Radical Equations

Practice: Solving Radical Equations
Solve algebraically, stating any restrictions on xxx:
x−3=3x−4\begin{array}{rcl}
x-3&=&3\sqrt{x-4}
\end{array}x−3​=​3x−4​​