# 1.4e: Exercises - Radical Equations

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Exercise $$\PageIndex{A}$$

$$\bigstar$$ Solve.

 $$\sqrt { x } = 7$$ $$\sqrt { x } = 4$$ $$\sqrt { x } + 8 = 9$$ $$\sqrt { x } - 4 = 5$$ $$\sqrt { x } + 7 = 4$$ $$\sqrt { x } + 3 = 1$$ $$5 \sqrt { x } - 1 = 0$$ $$3 \sqrt { x } - 2 = 0$$ $$\sqrt { 3 x + 1 } = 2$$ $$\sqrt { 5 x - 4 } = 4$$ $$\sqrt { 7 x + 4 } + 6 = 11$$ $$\sqrt { 3 x - 5 } + 9 = 14$$ $$2 \sqrt { x - 1 } - 3 = 0$$ $$3 \sqrt { x + 1 } - 2 = 0$$ $$\sqrt { x + 1 } = \sqrt { x } + 1$$ $$\sqrt { 2 x - 1 } = \sqrt { 2 x } - 1$$ $$\sqrt { 4 x - 1 } = 2 \sqrt { x } - 1$$ $$\sqrt { 4 x - 11 } = 2 \sqrt { x } - 1$$ $$\sqrt { x + 8 } = \sqrt { x } - 4$$ $$\sqrt { 25 x - 1 } = 5 \sqrt { x } + 1$$ $$\sqrt [ 3 ] { x } = 3$$ $$\sqrt [ 3 ] { x } = - 4$$ $$\sqrt [ 3 ] { 2 x + 9 } = 3$$ $$\sqrt [ 3 ] { 4 x - 11 } = 1$$ $$\sqrt [ 3 ] { 5 x + 7 } + 3 = 1$$ $$\sqrt [ 3 ] { 3 x - 6 } + 5 = 2$$ $$4 - 2 \sqrt [ 3 ] { x + 2 } = 0$$ $$6 - 3 \sqrt [ 3 ] { 2 x - 3 } = 0$$ $$\sqrt [ 5 ] { 3 ( x + 10 ) } = 2$$ $$\sqrt [ 5 ] { 4 x + 3 } + 5 = 4$$

1. $$49$$    3. $$1$$    5. $$Ø$$    7. $$\frac{1}{25}$$     9. $$1$$     11. $$3$$     13. $$\frac{13}{4}$$     15. $$0$$     17. $$\frac{1}{4}$$     19. $$Ø$$     21. $$27$$     23. $$9$$     25. $$−3$$     27. $$6$$     29. $$\frac{2}{3}$$.

Exercise $$\PageIndex{B}$$

$$\bigstar$$ Solve.

 $$\sqrt { 8 x + 11 } = 3 \sqrt { x + 1 }$$ $$2 \sqrt { 3 x - 4 } = \sqrt { 2 ( 3 x + 1 ) }$$ $$\sqrt { 2 ( x + 10 ) } = \sqrt { 7 x - 15 }$$ $$\sqrt { 5 ( x - 4 ) } = \sqrt { x + 4 }$$ $$\sqrt [ 3 ] { 5 x - 2 } = \sqrt [ 3 ] { 4 x }$$ $$\sqrt [ 3 ] { 9 ( x - 1 ) } = \sqrt [ 3 ] { 3 ( x + 7 ) }$$ $$\sqrt [ 3 ] { 3 x + 1 } = \sqrt [ 3 ] { 2 ( x - 1 ) }$$ $$\sqrt [ 3 ] { 9 x } = \sqrt [ 3 ] { 3 ( x - 6 ) }$$ $$\sqrt [ 5 ] { 3 x - 5 } = \sqrt [ 5 ] { 2 x + 8 }$$ $$\sqrt [ 5 ] { x + 3 } = \sqrt [ 5 ] { 2 x + 5 }$$

31. $$2$$     33. $$7$$     35. $$2$$     37. $$−3$$     39. $$13$$.

Exercise $$\PageIndex{C}$$

$$\bigstar$$ Solve.

 $$\sqrt { 4 x + 21 } = x$$ $$\sqrt { 8 x + 9 } = x$$ $$\sqrt { 4 ( 2 x - 3 ) } = x$$ $$\sqrt { 3 ( 4 x - 9 ) } = x$$ $$2 \sqrt { x - 1 } = x$$ $$3 \sqrt { 2 x - 9 } = x$$ $$\sqrt { 9 x + 9 } = x + 1$$ $$\sqrt { 3 x + 10 } = x + 4$$ $$\sqrt { x - 1 } = x - 3$$ $$\sqrt { 2 x - 5 } = x - 4$$ $$\sqrt { 16 - 3 x } = x - 6$$ $$\sqrt { 7 - 3 x } = x - 3$$ $$3 \sqrt { 2 x + 10 } = x + 9$$ $$2 \sqrt { 2 x + 5 } = x + 4$$ $$3 \sqrt { x - 1 } - 1 = x$$ $$2 \sqrt { 2 x + 2 } - 1 = x$$ $$\sqrt { 10 x + 41 } - 5 = x$$ $$\sqrt { 6 ( x + 3 ) } - 3 = x$$ $$\sqrt { 8 x ^ { 2 } - 4 x + 1 } = 2 x$$ $$\sqrt { 18 x ^ { 2 } - 6 x + 1 } = 3 x$$ $$5 \sqrt { x + 2 } = x + 8$$ $$4 \sqrt { 2 ( x + 1 ) } = x + 7$$ $$\sqrt { x ^ { 2 } - 25 } = x$$ $$\sqrt { x ^ { 2 } + 9 } = x$$ $$3 + \sqrt { 6 x - 11 } = x$$ $$2 + \sqrt { 9 x - 8 } = x$$ $$\sqrt { 4 x + 25 } - x = 7$$ $$\sqrt { 8 x + 73 } - x = 10$$ $$2 \sqrt { 4 x + 3 } - 3 = 2 x$$ $$2 \sqrt { 6 x + 3 } - 3 = 3 x$$ $$2 x - 4 = \sqrt { 14 - 10 x }$$ $$3 x - 6 = \sqrt { 33 - 24 x }$$ $$\sqrt [ 3 ] { x ^ { 2 } - 24 } = 1$$ $$\sqrt [ 3 ] { x ^ { 2 } - 54 } = 3$$ $$\sqrt [ 3 ] { x ^ { 2 } + 6 x } + 1 = 4$$ $$\sqrt [ 3 ] { x ^ { 2 } + 2 x } + 5 = 7$$ $$\sqrt [ 3 ] { 25 x ^ { 2 } - 10 x - 7 } = - 2$$ $$\sqrt [ 3 ] { 9 x ^ { 2 } - 12 x - 23 } = - 3$$ $$\sqrt [ 3 ] { 4 x ^ { 2 } - 1 } - 2 = 0$$ $$4 \sqrt [ 3 ] { x ^ { 2 } } - 1 = 0$$ $$\sqrt [ 5 ] { x ( 2 x + 1 ) } - 1 = 0$$ $$\sqrt [ 5 ] { 3 x ^ { 2 } - 20 x } - 2 = 0$$

41. $$7$$     43. $$2, 6$$     45. $$2$$     47. $$−1, 8$$     49. $$5$$     51. $$Ø$$     53. $$−3, 3$$     55. $$2, 5$$     57. $$−4, 4$$     59. $$\frac{1}{2}$$     61. $$2, 7$$     63. $$Ø$$     65. $$10$$     67. $$−6, −4$$     69. $$−\frac{1}{2}, \frac{3}{2}$$     71. $$Ø$$     73. $$−5, 5$$     75. $$−9, 3$$     77. $$\frac{1}{5}$$     79. $$− \frac{3}{2} ,\frac{ 3}{2}$$     81. $$−1, \frac{1}{2}$$

Exercise $$\PageIndex{D}$$

$$\bigstar$$ Solve.

 $$\sqrt { 2 x ^ { 2 } - 15 x + 25 } = \sqrt { ( x + 5 ) ( x - 5 ) }$$ $$\sqrt { x ^ { 2 } - 4 x + 4 } = \sqrt { x ( 5 - x ) }$$ $$\sqrt [ 3 ] { 2 \left( x ^ { 2 } + 3 x - 20 \right) } = \sqrt [ 3 ] { ( x + 3 ) ^ { 2 } }$$ $$\sqrt [ 3 ] { 3 x ^ { 2 } + 3 x + 40 } = \sqrt [ 3 ] { ( x - 5 ) ^ { 2 } }$$ $$\sqrt { 2 x - 5 } + \sqrt { 2 x } = 5$$ $$\sqrt { 4 x + 13 } - 2 \sqrt { x } = 3$$ $$\sqrt { 8 x + 17 } - 2 \sqrt { 2 - x } = 3$$ $$\sqrt { 3 x - 6 } - \sqrt { 2 x - 3 } = 1$$ $$\sqrt { 2 ( x - 2 ) } - \sqrt { x - 1 } = 1$$ $$\sqrt { 2 x + 5 } - \sqrt { x + 3 } = 2$$ $$\sqrt { 2 ( x + 1 ) } - \sqrt { 3 x + 4 } - 1 = 0$$ $$\sqrt { 6 - 5 x } + \sqrt { 3 - 3 x } - 1 = 0$$ $$\sqrt { x - 2 } - 1 = \sqrt { 2 ( x - 3 ) }$$ $$\sqrt { 14 - 11 x } + \sqrt { 7 - 9 x } = 1$$ $$\sqrt { x + 1 } = \sqrt { 3 } - \sqrt { 2 - x }$$ $$\sqrt { 2 x + 9 } - \sqrt { x + 1 } = 2$$

83. $$5, 10$$     85. $$−7, 7$$     87. $$\frac{9}{2}$$     89. $$1$$     91. $$10$$     93. $$Ø$$     95. $$3$$     97. $$-1, 2$$.

Exercise $$\PageIndex{E}$$

$$\bigstar$$ Solve.

 $$x ^ { 1 / 2 } - 10 = 0$$ $$x ^ { 1 / 2 } - 6 = 0$$ $$x ^ { 1 / 3 } + 2 = 0$$ $$x ^ { 1 / 3 } + 4 = 0$$ $$( x - 1 ) ^ { 1 / 2 } - 3 = 0$$ $$( x + 2 ) ^ { 1 / 2 } - 6 = 0$$ $$( 2 x - 1 ) ^ { 1 / 3 } + 3 = 0$$ $$( 3 x - 1 ) ^ { 1 / 3 } - 2 = 0$$ $$( 4 x + 15 ) ^ { 1 / 2 } - 2 x = 0$$ $$( 3 x + 2 ) ^ { 1 / 2 } - 3 x = 0$$ $$( 2 x + 12 ) ^ { 1 / 2 } - x = 6$$ $$( 4 x + 36 ) ^ { 1 / 2 } - x = 9$$ $$2 ( 5 x + 26 ) ^ { 1 / 2 } = x + 10$$ $$3 ( x - 1 ) ^ { 1 / 2 } = x + 1$$ $$x ^ { 1 / 2 } + ( 3 x - 2 ) ^ { 1 / 2 } = 2$$ $$( 6 x + 1 ) ^ { 1 / 2 } - ( 3 x ) ^ { 1 / 2 } = 1$$ $$( 3 x + 7 ) ^ { 1 / 2 } + ( x + 3 ) ^ { 1 / 2 } - 2 = 0$$ $$( 3 x ) ^ { 1 / 2 } + ( x + 1 ) ^ { 1 / 2 } - 5 = 0$$ $$\sqrt{3x+7} +\sqrt{x+1} = 2$$ $$\sqrt{2x+4} - \sqrt{x+3} =1$$ $$\sqrt{2x}-\sqrt{x+1}=1$$

99. $$100$$     101. $$−8$$     103. $$10$$     105. $$−13$$     107. $$\frac{5}{2}$$     109. $$−6, −4$$     111. $$−2, 2$$     113. $$1$$     115. $$−2$$      117. {-1}     119. {8}

$$\star$$

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