Question

Solve the following equation:
$\sqrt{2{{x}^{2}}-9x+4}+3\sqrt{2x-1}=\sqrt{2{{x}^{2}}+21x-11}$

Answer 1

Here we have been given one equation in x and we have to solve for x. For this, we will first observe this equation carefully and we will notice that all the terms in this equation are under roots. To solve this, we will first have to remove the under roots. To do that, we will square the equation on both sides and see if there are any more terms left under root. If there are, we will collect them on one side and then square both sides again. We will repeat this process until and unless all the under roots are removed from this equation. Once that is done, it will be a simple polynomial in x which can be solved by factorization. Hence, we will get the required answer.

Complete step-by-step solution:
We here have been given the following equation:
$\sqrt{2{{x}^{2}}-9x+4}+3\sqrt{2x-1}=\sqrt{2{{x}^{2}}+21x-11}$
Now, since the equation has under roots on both sides, we will try to remove them so as to solve this equation.
For this, we will square this equation on both sides.
Squaring the given equation on both sides we get:

& \sqrt{2{{x}^{2}}-9x+4}+3\sqrt{2x-1}=\sqrt{2{{x}^{2}}+21x-11} \\\ & \Rightarrow {{\left( \sqrt{2{{x}^{2}}-9x+4}+3\sqrt{2x-1} \right)}^{2}}={{\left( \sqrt{2{{x}^{2}}+21x-11} \right)}^{2}} \\\ \end{aligned}$$ Now, we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ Thus, using this formula in LHS and solving this equation we get: $$\begin{aligned} & {{\left( \sqrt{2{{x}^{2}}-9x+4}+3\sqrt{2x-1} \right)}^{2}}={{\left( \sqrt{2{{x}^{2}}+21x-11} \right)}^{2}} \\\ & \Rightarrow {{\left( \sqrt{2{{x}^{2}}-9x+4} \right)}^{2}}+{{\left( 3\sqrt{2x-1} \right)}^{2}}+2.\sqrt{2{{x}^{2}}-9x+4}.3.\sqrt{2x-1}=2{{x}^{2}}+21x-11 \\\ & \Rightarrow 2{{x}^{2}}-9x+4+9\left( 2x-1 \right)+6\sqrt{2{{x}^{2}}-9x+4}\sqrt{2x-1}=2{{x}^{2}}+21x-11 \\\ & \Rightarrow 2{{x}^{2}}-9x+4+18x-9+6\sqrt{2{{x}^{2}}-9x+4}\sqrt{2x-1}=2{{x}^{2}}+21x-11 \\\ & \Rightarrow 6\sqrt{2{{x}^{2}}-9x+4}\sqrt{2x-1}=12x-6 \\\ \end{aligned}$$ Now, taking 6 common on both the sides of the equation, we get: $\begin{aligned} & 6\sqrt{2{{x}^{2}}-9x+4}\sqrt{2x-1}=12x-6 \\\ & \Rightarrow 6\sqrt{2{{x}^{2}}-9x+4}\sqrt{2x-1}=6\left( 2x-1 \right) \\\ & \Rightarrow \sqrt{2{{x}^{2}}-9x+4}\sqrt{2x-1}=2x-1 \\\ \end{aligned}$ Now, we can see that the LHS of this equation is still in the form of roots. Thus, to remove it, we will have to square the equation again on both the sides. Thus, squaring both sides of the equation, we get: $\begin{aligned} & \sqrt{2{{x}^{2}}-9x+4}\sqrt{2x-1}=2x-1 \\\ & \Rightarrow {{\left( \sqrt{2{{x}^{2}}-9x+4}\sqrt{2x-1} \right)}^{2}}={{\left( 2x-1 \right)}^{2}} \\\ & \Rightarrow \left( 2{{x}^{2}}-9x+4 \right)\left( 2x-1 \right)={{\left( 2x-1 \right)}^{2}} \\\ \end{aligned}$ Now, solving this equation, we get: $$\begin{aligned} & \left( 2{{x}^{2}}-9x+4 \right)\left( 2x-1 \right)={{\left( 2x-1 \right)}^{2}} \\\ & \Rightarrow \left( 2{{x}^{2}}-9x+4 \right)\left( 2x-1 \right)-{{\left( 2x-1 \right)}^{2}}=0 \\\ & \Rightarrow \left( 2x-1 \right)\left[ \left( 2{{x}^{2}}-9x+4 \right)-\left( 2x-1 \right) \right]=0 \\\ & \Rightarrow \left( 2x-1 \right)\left( 2{{x}^{2}}-11x+5 \right)=0 \\\ & \Rightarrow \left( 2x-1 \right)\left( 2{{x}^{2}}-10x-x+5 \right)=0 \\\ & \Rightarrow \left( 2x-1 \right)\left[ 2x\left( x-5 \right)-\left( x-5 \right) \right]=0 \\\ & \Rightarrow \left( 2x-1 \right)\left( 2x-1 \right)\left( x-5 \right)=0 \\\ & \therefore x=\dfrac{1}{2},\dfrac{1}{2},5 \\\ \end{aligned}$$ **Thus, the required values of x are $\dfrac{1}{2}$ and 5.** **Note:** Here, when we open the LHS of the equation when we square both sides for the first time, we do not multiply the under roots together. This is because, at last, we had to factorize and they were already in that form. Multiplying them and then factorizing them again just would have been another way of making this question complicated and we need to avoid that as much as we can.