Answer
Verified
423.3k+ views
Hint: Square both sides to get and substitute $\dfrac{7x-3}{2x+1}=t$. Take care of extraneous roots which will be due to the quantity inside the square root is $\ge 0$ and denominator of each fraction $\ne 0$. Remove those values at which the above conditions are not satisfied.
Complete step-by-step solution -
Let $t=\dfrac{7x-3}{2x+1}$
So, we have
$\begin{align}
& \dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{250\sqrt{2x+1}}{3\sqrt{7x-3}} \\
& \Rightarrow \dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{250}{3}\sqrt{\dfrac{2x+1}{7x-3}} \\
& \Rightarrow \dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{250}{3\sqrt{\dfrac{7x-3}{2x+1}}} \\
& \Rightarrow 18t=\dfrac{250}{3\sqrt{t}} \\
\end{align}$
Cross multiplying, we get
$54{{t}^{\dfrac{3}{2}}}=\dfrac{250}{3}$
Dividing both sides by 54 we get
$\begin{align}
& \dfrac{54{{t}^{\dfrac{3}{2}}}}{54}=\dfrac{250}{3\times 54} \\
& \Rightarrow {{t}^{\dfrac{3}{2}}}=\dfrac{125}{81} \\
& \Rightarrow {{t}^{\dfrac{3}{2}}}={{\left( \dfrac{5}{3} \right)}^{3}} \\
& \Rightarrow t={{\left( \dfrac{5}{3} \right)}^{3\times \dfrac{2}{3}}} \\
& \Rightarrow t={{\left( \dfrac{5}{3} \right)}^{2}} \\
& \Rightarrow t=\dfrac{25}{9} \\
& \Rightarrow \dfrac{7x-3}{2x+1}=\dfrac{25}{9} \\
\end{align}$
Cross multiplying, we get
$9\left( 7x-3 \right)=25\left( 2x+1 \right)$
Simplifying we get
$63x-27=50x+25$
Transposing 27 to RHS and 50x to LHS we get
$63x-50x=25+27$
Simplifying we get
$13x=52$
Dividing both sides by 13, we get
$\dfrac{13x}{13}=\dfrac{52}{13}$
x = 4
Also, we need to check if the terms under square root are non-negative
We have $7x-3=28-3=25\ge 0$ which satisfies the condition
$2x+1=8+1=9\ge 0$ which also satisfies the condition
Also, we need to check that the terms in the denominator are not equal to 0
Since 2x+1 = 9 and 7x – 3 = 25 are both not equal to 0 we have all the conditions satisfied.
Hence x = 4 is the root of the equation $\dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{250\sqrt{2x+1}}{3\sqrt{7x-3}}$
Note: [1] Verification: $\dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{18\times 25}{9}=50$
$\dfrac{250\sqrt{2x+1}}{3\sqrt{7x-3}}=\dfrac{250\sqrt{9}}{3\sqrt{25}}=\dfrac{250\times 3}{3\times 5}=50$
Hence $\dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{250\sqrt{2x+1}}{3\sqrt{7x-3}}$.
Hence it is verified that x = 4 is the root of the given equation.
[2] Substitution of $t=\dfrac{7x-3}{2x+1}$ was the primary step in solving the question. It reduced the effort in solving the question. Substitutions usually help in solving equations faster. e.g. we can solve the question $\left( x+3 \right)\left( x+4 \right)\left( x+5 \right)\left( x+6 \right)=100$ by expanding and forming a biquadratic and solving that biquadratic which would be very difficult or by substituting ${{x}^{2}}+9x+18=t$ in which case we have
$\left( x+3 \right)\left( x+4 \right)\left( x+5 \right)\left( x+6 \right)=\left[ \left( x+3 \right)\left( x+6 \right) \right]\left[ \left( x+5 \right)\left( x+4 \right) \right]$
= $\left( {{x}^{2}}+9x+18 \right)\left( {{x}^{2}}+9x+20 \right)=t\left( t+2 \right)=\text{ }100$
In this case, it is equivalent to solving two quadratic equations and is much easier than solving the biquadratic equation.
Complete step-by-step solution -
Let $t=\dfrac{7x-3}{2x+1}$
So, we have
$\begin{align}
& \dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{250\sqrt{2x+1}}{3\sqrt{7x-3}} \\
& \Rightarrow \dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{250}{3}\sqrt{\dfrac{2x+1}{7x-3}} \\
& \Rightarrow \dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{250}{3\sqrt{\dfrac{7x-3}{2x+1}}} \\
& \Rightarrow 18t=\dfrac{250}{3\sqrt{t}} \\
\end{align}$
Cross multiplying, we get
$54{{t}^{\dfrac{3}{2}}}=\dfrac{250}{3}$
Dividing both sides by 54 we get
$\begin{align}
& \dfrac{54{{t}^{\dfrac{3}{2}}}}{54}=\dfrac{250}{3\times 54} \\
& \Rightarrow {{t}^{\dfrac{3}{2}}}=\dfrac{125}{81} \\
& \Rightarrow {{t}^{\dfrac{3}{2}}}={{\left( \dfrac{5}{3} \right)}^{3}} \\
& \Rightarrow t={{\left( \dfrac{5}{3} \right)}^{3\times \dfrac{2}{3}}} \\
& \Rightarrow t={{\left( \dfrac{5}{3} \right)}^{2}} \\
& \Rightarrow t=\dfrac{25}{9} \\
& \Rightarrow \dfrac{7x-3}{2x+1}=\dfrac{25}{9} \\
\end{align}$
Cross multiplying, we get
$9\left( 7x-3 \right)=25\left( 2x+1 \right)$
Simplifying we get
$63x-27=50x+25$
Transposing 27 to RHS and 50x to LHS we get
$63x-50x=25+27$
Simplifying we get
$13x=52$
Dividing both sides by 13, we get
$\dfrac{13x}{13}=\dfrac{52}{13}$
x = 4
Also, we need to check if the terms under square root are non-negative
We have $7x-3=28-3=25\ge 0$ which satisfies the condition
$2x+1=8+1=9\ge 0$ which also satisfies the condition
Also, we need to check that the terms in the denominator are not equal to 0
Since 2x+1 = 9 and 7x – 3 = 25 are both not equal to 0 we have all the conditions satisfied.
Hence x = 4 is the root of the equation $\dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{250\sqrt{2x+1}}{3\sqrt{7x-3}}$
Note: [1] Verification: $\dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{18\times 25}{9}=50$
$\dfrac{250\sqrt{2x+1}}{3\sqrt{7x-3}}=\dfrac{250\sqrt{9}}{3\sqrt{25}}=\dfrac{250\times 3}{3\times 5}=50$
Hence $\dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{250\sqrt{2x+1}}{3\sqrt{7x-3}}$.
Hence it is verified that x = 4 is the root of the given equation.
[2] Substitution of $t=\dfrac{7x-3}{2x+1}$ was the primary step in solving the question. It reduced the effort in solving the question. Substitutions usually help in solving equations faster. e.g. we can solve the question $\left( x+3 \right)\left( x+4 \right)\left( x+5 \right)\left( x+6 \right)=100$ by expanding and forming a biquadratic and solving that biquadratic which would be very difficult or by substituting ${{x}^{2}}+9x+18=t$ in which case we have
$\left( x+3 \right)\left( x+4 \right)\left( x+5 \right)\left( x+6 \right)=\left[ \left( x+3 \right)\left( x+6 \right) \right]\left[ \left( x+5 \right)\left( x+4 \right) \right]$
= $\left( {{x}^{2}}+9x+18 \right)\left( {{x}^{2}}+9x+20 \right)=t\left( t+2 \right)=\text{ }100$
In this case, it is equivalent to solving two quadratic equations and is much easier than solving the biquadratic equation.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE