Solve the following equation for x
$\dfrac{18\left( 7x-3 \right)}{2x+1}=\dfrac{250\sqrt{2x+1}}{3\sqrt{7x-3}}$

Answer Verified Verified
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
  & \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}} \\
Cross multiplying, we get
Dividing both sides by 54 we get
  & \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} \\
Cross multiplying, we get
$9\left( 7x-3 \right)=25\left( 2x+1 \right)$
Simplifying we get
Transposing 27 to RHS and 50x to LHS we get
Simplifying we get
Dividing both sides by 13, we get
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.
Bookmark added to your notes.
View Notes