# Solve the following equation for x

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

Answer

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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.

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.

Last updated date: 20th Sep 2023

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