\(2^{x+1} -32 = 0\)
1.3 Solving surd equations
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1.2 Rational exponents and surds
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1.4 Applications of exponentials
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1.3 Solving surd equations (EMBFB)
We also need to be able to solve equations that involve surds.
Worked example 13: Surd equations
Solve for x: 53√x4=405
Write in exponential notation
5(x4)13=4055x43=405
Divide both sides of the equation by 5 and simplify
5x435=4055x43=81x43=34
Simplify the exponents
(x43)34=(34)34x=33x=27
Check the solution by substituting the answer back into the original equation
LHS=53√x4=5(27)43=5(33)43=5(34)=405=RHS
Worked example 14: Surd equations
Solve for z: z−4√z+3=0
Factorise
z−4√z+3=0z−4z12+3=0(z12−3)(z12−1)=0
Solve for both factors
The zero law states: if a×b=0, then a=0 or b=0.
∴(z12−3)=0 or (z12−1)=0
Therefore
z12−3=0z12=3(z12)2=32z=9
or
z12−1=0z12=1(z12)2=12z=1
Check the solution by substituting both answers back into the original equation
If z=9:
LHS=z−4√z+3=9−4√9+3=12−12=0=RHS
If z=1:
LHS=z−4√z+3=1−4√1+3=4−4=0=RHS
Write the final answer
The solution to z−4√z+3=0 is z=9 or z=1.
Worked example 15: Surd equations
Solve for p: √p−2−3=0
Write the equation with only the square root on the left hand side
Use the additive inverse to get all other terms on the right hand side and only the square root on the left hand side.
√p−2=3Square both sides of the equation
(√p−2)2=32p−2=9p=11
Check the solution by substituting the answer back into the original equation
If p=11:
LHS=√p−2−3=√11−2−3=√9−3=3−3=0=RHS
Write the final answer
The solution to √p−2−3=0 is p=11.
Solving surd equations
Solve for the unknown variable (remember to check that the solution is valid):
\(\text{125} \left ( 3^p \right ) = 27 \left ( 5^p \right )\)
\(2y^{\frac{1}{2}} - 3y^{\frac{1}{4}} + 1 = 0\)
\(t-1 = \sqrt{7-t}\)
\(2z - 7\sqrt{z} + 3 = 0\)
\(x^{\frac{1}{3}}(x^{\frac{1}{3}} + 1) = 6\)
\(2^{4n} - \dfrac{1}{\sqrt[4]{16}} = 0\)
\(\sqrt{31 -10d} = 4 - d\)
\(y - 10\sqrt{y} + 9 = 0\)
\(f = 2 + \sqrt{19 - 2f}\)
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1.2 Rational exponents and surds
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1.4 Applications of exponentials
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