Ejercicio 251 – Solucionario Baldor

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DESCARGA

CAPITULO XXXI

R a d i c a l e s
Resolución de Ecuaciones con radicales que se reducen a primer grado

Ejercicio 251

Resolver las ecuaciones:

$\begin{matrix} \sqrt{x - 8} & = 2 \\ {(\sqrt{x - 8})}^{2} & = 2^{2} \\ x - 8 & = 4 \\ x & = 12 \end{matrix}$
$\begin{matrix} 5 - \sqrt{3 x + 1} & = 0 \\ 5^{2} & = {(\sqrt{3 x + 1})}^{2} \\ 25 & = 3 x + 1 \\ 25 - 1 & = 3 x \\ 24 & = 3 x \\ \frac{}{3} & = x \\ x & = 8 \end{matrix}$
$\begin{matrix} 7 + \sqrt[3]{5 x - 2} & = 9 \\ \sqrt[3]{5 x - 2} & = 9 - 7 \\ {(\sqrt[3]{5 x - 2})}^{3} & = 2^{3} \\ 5 x - 2 & = 8 \\ 5 x & = 8 + 2 \\ x & = \frac{}{5} \\ x & = 2 \end{matrix}$
$\begin{matrix} \sqrt{9 x^{2} - 5} - 3 x & = - 1 \\ {(\sqrt{9 x^{2} - 5})}^{2} & = {(3 x - 1)}^{2} \\ 9 x^{2} - 5 & = 9 x^{2} - 6 x + 1 \\ - 5 - 1 & = - 6 x \\ \frac{- 6}{- 6} & = x \\ x & = 1 \end{matrix}$
$\begin{matrix} \sqrt{x^{2} - 2 x + 1} & = 9 - x \\ \sqrt{{(x - 1)}^{2}} & = 9 - x \\ x - 1 & = 9 - x \\ x + x & = 9 + 1 \\ 2 x & = 10 \\ x & = \frac{}{2} \\ x & = 5 \end{matrix}$
$\begin{matrix} 15 - \sqrt[3]{7 x - 1} & = 12 \\ 15 - 12 & = \sqrt[3]{7 x - 1} \\ 3^{3} & = {(\sqrt[3]{7 x - 1})}^{3} \\ 27 & = 7 x - 1 \\ 27 + 1 & = 7 x \\ \frac{}{7} & = x \\ x & = 4 \end{matrix}$
$\begin{matrix} \sqrt{x} + \sqrt{x + 7} & = 7 \\ {(\sqrt{x + 7})}^{2} & = {(7 - \sqrt{x})}^{2} \\ x + 7 & = 49 - 14 \sqrt{x} + x \\ 7 - 49 & = - 14 \sqrt{x} \\ - 42 & = - 14 \sqrt{x} \\ \frac{-}{- 14} & = \sqrt{x} \\ 3^{2} & = \sqrt{x^{2}} \\ x & = 9 \end{matrix}$
$\begin{matrix} \sqrt{3 x - 5} + \sqrt{3 x - 14} & = 9 \\ {(\sqrt{3 x - 5})}^{2} & = {(9 - \sqrt{3 x - 14})}^{2} \\ 3 x - 5 & = 81 - 18 \sqrt{3 x - 14} + 3 x - 14 \\ - 5 - 81 + 14 & = - 18 \sqrt{3 x - 14} \\ - 72 & = - 18 \sqrt{3 x - 14} \\ \frac{-}{- 18} & = \sqrt{3 x - 14} \\ 4^{2} & = {(\sqrt{3 x - 14})}^{2} \\ 16 & = 3 x - 14 \\ 16 + 14 & = 3 x \\ \frac{}{3} & = x \\ x & = 10 \end{matrix}$
$\begin{matrix} \sqrt{x + 10} - \sqrt{x + 19} & = - 1 \\ {(\sqrt{x + 10})}^{2} & = {(\sqrt{x + 19} - 1)}^{2} \\ x + 10 & = x + 19 - 2 \sqrt{x + 19} + 1 \\ 10 - 20 & = - 2 \sqrt{x + 19} \\ \frac{-}{- 2} & = \sqrt{x + 19} \\ 5^{2} & = {(\sqrt{x + 19})}^{2} \\ 25 & = x + 19 \\ 25 - 19 & = x \\ x & = 6 \end{matrix}$
$\begin{matrix} \sqrt{4 x - 11} & = 7 \sqrt{2 x - 29} \\ {(\sqrt{4 x - 11})}^{2} & = {(7 \sqrt{2 x - 29})}^{2} \\ 4 x - 11 & = 49 (2 x - 29) \\ 4 x - 11 & = 98 x - 1421 \\ 4 x - 98 x & = - 1421 + 11 \\ - 94 x & = - 1410 \\ x & = \frac{-}{- 94} \\ x & = 15 \end{matrix}$
$\begin{matrix} \sqrt{5 x - 19} - \sqrt{5 x} & = - 1 \\ {(\sqrt{5 x - 19})}^{2} & = {(\sqrt{5 x} - 1)}^{2} \\ 5 x - 19 & = 5 x - 2 \sqrt{5 x} + 1 \\ - 19 - 1 & = - 2 \sqrt{5 x} \\ \frac{-}{- 2} & = \sqrt{5 x} \\ 1 0^{2} & = {(\sqrt{5 x})}^{2} \\ 100 & = 5 x \\ \frac{}{5} & = x \\ x & = 20 \end{matrix}$
$\begin{matrix} \sqrt{x - 2} + 5 & = \sqrt{x + 53} \\ {(\sqrt{x - 2} + 5)}^{2} & = {(\sqrt{x + 53})}^{2} \\ x - 2 + 10 \sqrt{x - 2} + 25 & = x + 53 \\ 10 \sqrt{x - 2} & = 53 + 2 - 25 \\ 10 \sqrt{x - 2} & = 30 \\ {(\sqrt{x - 2})}^{2} & = {(\frac{30}{10})}^{2} \\ x - 2 & = 9 \\ x & = 11 \end{matrix}$
$\begin{matrix} \sqrt{9 x - 14} & = 3 \sqrt{x + 10} - 4 \\ {(\sqrt{9 x - 14})}^{2} & = {(3 \sqrt{x + 10} - 4)}^{2} \\ 9 x - 14 & = 9 (x + 10) - 24 \sqrt{x + 10} + 16 \\ 9 x - 14 - 16 & = 9 x + 90 - 24 \sqrt{x + 10} \\ - 30 - 90 & = - 24 \sqrt{x + 10} \\ {(\frac{-}{- 24})}^{2} & = {(\sqrt{x + 10})}^{2} \\ 25 & = x + 10 \\ 25 - 10 & = x \\ x & = 15 \end{matrix}$
$\begin{matrix} \sqrt{x - 16} - \sqrt{x + 8} & = - 4 \\ {(\sqrt{x - 16})}^{2} & = {(\sqrt{x + 8} - 4)}^{2} \\ x - 16 & = x + 8 - 8 \sqrt{x + 8} + 16 \\ - 16 - 8 - 16 & = - 8 \sqrt{x + 8} \\ - 40 & = - 8 \sqrt{x + 8} \\ {(\frac{-}{- 8})}^{2} & = {(\sqrt{x + 8})}^{2} \\ 25 & = x + 8 \\ 25 - 8 & = x \\ x & = 17 \end{matrix}$
$\begin{matrix} \sqrt{5 x - 1} + 3 & = \sqrt{5 x + 26} \\ {(\sqrt{5 x - 1} + 3)}^{2} & = {(\sqrt{5 x + 26})}^{2} \\ 5 x - 1 + 6 \sqrt{5 x - 1} + 9 & = 5 x + 26 \\ 6 \sqrt{5 x - 1} & = 26 - 8 \\ {(\sqrt{5 x - 1})}^{2} & = {(\frac{}{6})}^{2} \\ 5 x - 1 & = 9 \\ x & = \frac{}{5} \\ x & = 2 \end{matrix}$
$\begin{matrix} 13 - \sqrt{13 + 4 x} & = 2 \sqrt{x} \\ {(13 - \sqrt{13 + 4 x})}^{2} & = {(2 \sqrt{x})}^{2} \\ 169 - 26 \sqrt{13 + 4 x} + 13 + 4 x & = 4 x \\ - 26 \sqrt{13 + 4 x} & = - 182 \\ {(\sqrt{13 + 4 x})}^{2} & = {(\frac{-}{- 26})}^{2} \\ 13 + 4 x & = 49 \\ 4 x & = 49 - 13 \\ x & = \frac{}{4} \\ x & = 9 \end{matrix}$
$\begin{matrix} \sqrt{x - 4} + \sqrt{x + 4} & = 2 \sqrt{x - 1} \\ {(\sqrt{x - 4} + \sqrt{x + 4})}^{2} & = {(2 \sqrt{x - 1})}^{2} \\ x - 4 + 2 \sqrt{x^{2} - 16} + x + 4 & = 4 (x - 1) \\ 2 x + 2 \sqrt{x^{2} - 16} & = 4 x - 4 \\ D i v i d i e n d o l a e c u a c i ó n p a r a 2 \\ x + \sqrt{x^{2} - 16} & = 2 x - 2 \\ {(\sqrt{x^{2} - 16})}^{2} & = {(x - 2)}^{2} \\ x^{2} - 16 & = x^{2} - 4 x + 4 \\ - 16 - 4 & = - 4 x \\ \frac{-}{- 4} & = x \\ x & = 5 \end{matrix}$
$\begin{matrix} \sqrt{9 x + 7} - \sqrt{x} - \sqrt{16 x - 7} & = 0 \\ {(\sqrt{9 x + 7} - \sqrt{16 x - 7})}^{2} & = {(\sqrt{x})}^{2} \\ 9 x + 7 - 2 \sqrt{(9 x + 7) (16 x - 7)} + 16 x - 7 & = x \\ x & = 2 \sqrt{(9 x + 7) (16 x - 7)} \\ {(12 x)}^{2} & = {(\sqrt{(9 x + 7) (16 x - 7)})}^{2} \\ 144 x^{2} & = 144 x^{2} - 63 x + 112 x - 49 \\ 49 & = 49 x \\ x & = 1 \end{matrix}$
$\begin{matrix} \sqrt{9 x + 10} - 2 \sqrt{x + 3} & = \sqrt{x - 2} \\ {(\sqrt{9 x + 10} - 2 \sqrt{x + 3})}^{2} & = {(\sqrt{x - 2})}^{2} \\ 9 x + 10 - 4 \sqrt{(9 x + 10) (x + 3)} + 4 (x + 3) & = x - 2 \\ 9 x + 10 - 4 \sqrt{9 x^{2} + 27 x + 10 x + 30} + 4 x + 12 & = x - 2 \\ - 4 \sqrt{9 x^{2} + 37 x + 30} & = x - 2 - 13 x - 22 \\ - 4 \sqrt{9 x^{2} + 37 x + 30} & = - 12 x - 24 \\ D i v i d i e n d o l a e c u a c i o n p a r a - 4 \\ {(\sqrt{9 x^{2} + 37 x + 30})}^{2} & = {(3 x + 6)}^{2} \\ 9 x^{2} + 37 x + 30 & = 9 x^{2} + 36 x + 36 \\ 37 x - 36 x & = - 30 + 36 \\ x & = 6 \end{matrix}$
$\begin{matrix} \sqrt{18 x - 8} - \sqrt{2 x - 4} - 2 \sqrt{2 x + 1} & = 0 \\ {(\sqrt{18 x - 8} - \sqrt{2 x - 4})}^{2} & = {(2 \sqrt{2 x + 1})}^{2} \\ 18 x - 8 - 2 \sqrt{4 (9 x - 4) (x - 2)} + 2 x - 4 & = 4 (2 x + 1) \\ 20 x - 12 - 4 \sqrt{9 x^{2} - 18 x - 4 x + 8} & = 8 x + 4 \\ - 4 \sqrt{9 x^{2} - 22 x + 8} & = - 20 x + 12 + 8 x + 4 \\ - 4 \sqrt{9 x^{2} - 22 x + 8} & = - 12 x + 16 \\ D i v i d i e n d o l a e c u a c i ó n p a r a - 4 \\ {(\sqrt{9 x^{2} - 22 x + 8})}^{2} & = {(3 x - 4)}^{2} \\ 9 x^{2} - 22 x + 8 & = 9 x^{2} - 24 x + 16 \\ - 22 x + 24 x & = 16 - 8 \\ 2 x & = 8 \\ x & = \frac{}{2} \\ x & = 4 \end{matrix}$
$\begin{matrix} \sqrt{8 x + 9} - \sqrt{18 x + 34} + \sqrt{2 x + 7} & = 0 \\ {(\sqrt{8 x + 9} + \sqrt{2 x + 7})}^{2} & = {(\sqrt{18 x + 34})}^{2} \\ 8 x + 9 + 2 \sqrt{(8 x + 9) (2 x + 7)} + 2 x + 7 & = 18 x + 34 \\ 10 x + 16 + 2 \sqrt{16 x^{2} + 56 x + 18 x + 63} & = 18 x + 34 \\ 2 \sqrt{16 x^{2} + 74 x + 63} & = 18 x - 10 x - 16 + 34 \\ 2 \sqrt{16 x^{2} + 74 x + 63} & = 8 x + 18 \\ D i v i d i e n d o l a e c u a c i ó n p a r a 2 \\ {(\sqrt{16 x^{2} + 74 x + 63})}^{2} & = {(4 x + 9)}^{2} \\ 16 x^{2} + 74 x + 63 & = 16 x^{2} + 72 x + 81 \\ 74 x - 72 x & = 81 - 63 \\ 2 x & = 18 \\ x & = \frac{}{2} \\ x & = 9 \end{matrix}$
$\begin{matrix} \sqrt{x - 2} - \sqrt{x - 5} & = \sqrt{4 x - 23} \\ {(\sqrt{x - 2} - \sqrt{x - 5})}^{2} & = {(\sqrt{4 x - 23})}^{2} \\ x - 2 - 2 \sqrt{(x - 2) (x - 5)} + x - 5 & = 4 x - 23 \\ - 2 \sqrt{x^{2} - 7 x + 10} & = 4 x - 2 x - 23 + 7 \\ - 2 \sqrt{x^{2} - 7 x + 10} & = 2 x - 16 \\ D i v i d i e n d o l a e c u a c i ó n p a r a 2 \\ {(- \sqrt{x^{2} - 7 x + 10})}^{2} & = {(x - 8)}^{2} \\ x^{2} - 7 x + 10 & = x^{2} - 16 x + 64 \\ 16 x - 7 x & = 64 - 10 \\ 9 x & = 54 \\ x & = \frac{}{9} \\ x & = 6 \end{matrix}$
$\begin{matrix} \sqrt{x + 6} - \sqrt{9 x + 70} & = - 2 \sqrt{x + 9} \\ {(\sqrt{x + 6} - \sqrt{9 x + 70})}^{2} & = {(- 2 \sqrt{x + 9})}^{2} \\ x + 6 - 2 \sqrt{(x + 6) (9 x + 70)} + 9 x + 70 & = 4 (x + 9) \\ - 2 \sqrt{9 x^{2} + 70 x + 54 x + 420} & = 4 x + 36 - 10 x - 76 \\ - 2 \sqrt{9 x^{2} + 124 x + 420} & = - 6 x - 20 \\ D i v i d i e n d o l a e c u a c i ó n p a r a - 2 \\ {(\sqrt{9 x^{2} + 124 x + 420})}^{2} & = {(3 x + 10)}^{2} \\ 9 x^{2} + 124 x + 420 & = 9 x^{2} + 60 x + 100 \\ 124 x - 60 x & = - 420 + 100 \\ 64 x & = - 320 \\ x & = - \frac{}{64} \\ x & = - 5 \end{matrix}$
$\begin{matrix} \sqrt{x - a} + \sqrt{x + a} & = \sqrt{4 x - 2 a} \\ {(\sqrt{x - a} + \sqrt{x + a})}^{2} & = {(\sqrt{4 x - 2 a})}^{2} \\ x - a + 2 \sqrt{x^{2} - a^{2}} + x + a & = 4 x - 2 a \\ 2 \sqrt{x^{2} - a^{2}} & = 4 x - 2 x - 2 a \\ 2 \sqrt{x^{2} - a^{2}} & = 2 x - 2 a \\ D i v i d i e n d o l a e c u a c i ó n p a r a 2 \\ {(\sqrt{x^{2} - a^{2}})}^{2} & = {(x - a)}^{2} \\ x^{2} - a^{2} & = x^{2} - 2 a x + a^{2} \\ 2 a x & = 2 a^{2} \\ x & = a \end{matrix}$
$\begin{matrix} \sqrt{x - 4 a b} & = - 2 b + \sqrt{x} \\ {(\sqrt{x - 4 a b})}^{2} & = {(- 2 b + \sqrt{x})}^{2} \\ x - 4 a b & = 4 b^{2} - 4 b \sqrt{x} + x \\ - 4 a b - 4 b^{2} & = - 4 b \sqrt{x} \\ D i v i d i e n d o l a e c u a c i ó n p a r a - 4 b \\ {(a - b)}^{2} & = {(\sqrt{x})}^{2} \\ x & = {(a - b)}^{2} \end{matrix}$
$\begin{matrix} \sqrt{x + 4 a} - \sqrt{x + 2 a - 1} & = 1 \\ {(\sqrt{x + 4 a} - 1)}^{2} & = {(\sqrt{x + 2 a - 1})}^{2} \\ x + 4 a - 2 \sqrt{x + 4 a} + 1 & = x + 2 a - 1 \\ - 2 \sqrt{x + 4 a} & = 2 a - 4 a - 1 - 1 \\ - 2 \sqrt{x + 4 a} & = - 2 a - 2 \\ D i v i d i e n d o l a e c u a c i ó n p a r a - 2 \\ {(\sqrt{x + 4 a})}^{2} & = {(a + 1)}^{2} \\ x + 4 a & = a^{2} + 2 a + 1 \\ x & = a^{2} + 2 a - 4 a + 1 \\ x & = a^{2} - 2 a + 1 \\ x & = {(a - 1)}^{2} \end{matrix}$