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Wize High School Grade 12 Pre-Calculus Textbook > Radical 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)}
where f(x) & g(x) f(x)~\&~g(x)~ are continuous polynomial functions.

Solutions to Radical Equations

Solve 2x+1=32\sqrt{x}+1=3. State the restrictions for 'x'.


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

Restrictions: x≥0x\geq0

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

Graph f(x) f(x)~ and g(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}
Therefore, (1, 3) is the solution.
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Algebraic Solutions to Radical Equations

Solve 5x+6−2=x\sqrt{5x+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\ge0 which means that our domain can be given by x≥−65x\ge-\frac65, or [−65,∞)[-\frac65,\infin) 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).

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}
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}

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Example: Solving Radical Equations

Solve graphically & algebraically, stating the restrictions on xx:
x−4+x=4\sqrt{x-4}+x=4

Graphically:

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


The solution is (4, 4)

Restrictions on 'x': x≥4x\geq4



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Algebraically:

x−4+x=4\sqrt{x-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}
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}

Therefore, only x = 4 is a solution.

Practice: Transformations

Let y=f(x)y=\sqrt{f(x)}.

Match the appropriate transformation with its transformed function.
A.
y=33(x+4)y=3\sqrt{3(x+4)}
B.
y=1313x−4y=\frac{1}{3}\sqrt{\frac{1}{3}x}-4
C.
y=3−(x−4)y=3\sqrt{-(x-4)}
D.
y=−3x+4y=-\sqrt{3x}+4
Reflection about the x-axis, horizontal compression by 13\frac{1}{3}, and translation 4 units up.
Reflection about the y-axis, vertical expansion by a factor of 3, and 4 units right
Horizontal expansion by a factor of 3, vertical compression by a factor of 13\frac{1}{3}, and 4 units down
Horizontal compression by a factor of 13\frac{1}{3} , vertical expansion by a factor of 3, and 4 units left

Practice: Solving Radical Equations

Solve algebraically, stating any restrictions on xx:
x−5=2x+3\begin{array}{rcl} x-5&=&2\sqrt{x+3} \end{array}

Practice: Solving Radical Equations

Solve algebraically, stating any restrictions:
x+7+2=3−x\sqrt{x+7}+2=\sqrt{3-x}

Extra Practice
Radical Functions
Solve for xx:
3x3−3x2+1=x^3\sqrt{x^{3}-3x^2}+1=x
Solving Radical Equations
Solve for a:
2a+4−3=3a−22\sqrt{a+4}-3=\sqrt{3a-2}
Solving Radical Equations
Solve for xx:
2x−1=x\sqrt{2x-1}=\sqrt{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}
Solving Radical Equations
Practice: Solving Radical Equations
Solve algebraically, stating any restrictions:
x2−2=x+4\sqrt{x^2-2}=\sqrt{x+4}
Solving Radical Equations
Practice: Solving Radical Equations
Solve algebraically, stating any restrictions on xx:
10+10x−1=13\begin{array}{rcl} 10+\sqrt{10x-1}&=&13 \end{array}
Solving Radical Equations
Practice: Solving Radical Equations
Solve algebraically, stating any restrictions on xx:
x−3=3x−4\begin{array}{rcl} x-3&=&3\sqrt{x-4} \end{array}