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  1. Home
  2. 12th grade
  3. Precalculus
  4. Exercise : Solve inequalities with radical functions

Solve inequalities with radical functions Precalculus

Solve the following inequalities.

\left(2x-3\right)^{-1/4} \lt \sqrt2

The function \left(\left(2x-3\right)^{-\frac{1}{4}}\right) is defined for x\geqslant\dfrac{3}{2}.

Applying the function x^{-4} (which is decreasing) on both sides of the inequality:

\left( \left(2x-3\right)^{-\frac{1}{4}} \right)^{-4} \gt \left(\sqrt{2}\right)^{-4}

Which is equivalent to:

2x-3 \gt \dfrac{1}{\left(\sqrt{2}\right)^{4}}

2x-3 \gt \dfrac{1}{4}

2x \gt \dfrac{13}{8}

x \gt \dfrac{13}{8}

The solution is included in the definition domain.

The solution of the inequality is \left( \dfrac{13}{8},\infty \right).

\left(4x+1\right)^{-1/2} \lt 3

The function \left(\left(4x+1\right)^{-\frac{1}{2}}\right) is defined for x\geqslant-\dfrac{1}{4}.

Raise both sides to the power:

\left( \left(4x+1\right)^{-\frac{1}{2}} \right)^{-2} \gt 3^{-2}

Which is equivalent to:

4x+1 \gt \dfrac{1}{9}

4x \gt \dfrac{1}{9}-1

4x \gt -\dfrac{8}{9}

x \gt -\dfrac{2}{9}

The solution is included in the definition domain.

The solution of the inequality is \left( -\dfrac{2}{9},\infty \right).

\left(3x-11\right)^{3/2} \leqslant 8

The function \left(3x-11\right)^{\frac{3}{2}} is defined for x\geqslant\dfrac{11}{3}.

Raise both sides to the \left( \dfrac{2}{3} \right) power:

\left( \left(3x-11\right)^{\frac{3}{2}} \right)^{\frac{2}{3}} \leqslant \left(8\right)^{\frac{2}{3}}

Which is equivalent to:

3x-11 \leqslant 8^{\frac{2}{3}}

3x-11 \leqslant4

3x \leqslant15

x \leqslant5

The intersection of the solution and the definition domain is:

\dfrac{11}{3}\leqslant x \leqslant 5

The solution of the inequality is \left[ \dfrac{11}{3},5 \right].

\left(2x+5\right)^{-1/3} \gt 2

The function \left(2x+5\right)^{-\frac{1}{3}} is defined for x \geq -\dfrac{5}{2}.

Raise both sides to the \left(-3\right) power:

\left( \left(2x+5\right)^{-\frac{1}{3}} \right)^{-3} \lt 2^{-3}

Which is equivalent to:

2x+5 \lt \dfrac{1}{8}

2x \lt \dfrac{1}{8}-5

2x \lt -\dfrac{39}{8}

x \lt -\dfrac{39}{16}

The intersection of the solution and the definition domain is:

\dfrac{-5}{2}\leqslant x \leqslant -\dfrac{39}{16}

The solution of the inequality is \left[-\dfrac{5}{2} ,-\dfrac{39}{16}\right).

\sqrt{x-2} \lt 3

The function \sqrt{x-2} is defined for x \geq 2.

Raise both sides to the 2nd power:

\left( \sqrt{x-2} \right)^{2} \lt 3^{2}

Which is equivalent to:

x-2 \lt 9

x \lt 11

The intersection of the solution and the definition domain is:

2\leqslant x \lt 11

The solution of the inequality is \left[2{,}11 \right).

\sqrt{2x+5} \geqslant 4

The function \sqrt{2x+5} is defined for x\geqslant-\dfrac{5}{2}.

Raise both sides to the 2nd power:

\left( \sqrt{2x+5} \right)^{2} \geqslant \left(4\right)^{2}

Which is equivalent to:

2x+5 \geqslant 16

2x \geqslant 11

x\geqslant\dfrac{11}{2}

Because the solution is included in the definition domain:

x\geqslant\dfrac{11}{2}

The solution of the inequality is \left[\dfrac{11}{2},\infty \right).

\sqrt{3x-8} \leqslant 5

The function \sqrt{3x-8} is defined for x\geqslant\dfrac{8}{3}.

Raise both sides to the 2nd power:

\left( \sqrt{3x-8} \right)^{2} \leqslant\left(5\right)^{2}

Which is equivalent to:

3x-8 \leqslant 25

3x \leqslant 33

x\leqslant11

The intersection of the solution and the definition domain is:

\dfrac{8}{3}\leqslant x\leqslant11

The solution of the inequality is \left[\dfrac{8}{3},11 \right].

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See also
  • Course : Radical functions, roots and rational exponents
  • Exercise : Simplify expressions involving exponents
  • Exercise : Determine the domain of radical functions or functions with rational exponents
  • Exercise : Determine the growth or decay of a function with a rational exponent
  • Exercise : Solve equations with radical functions
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