Question

Solve $\sqrt {3{x^2} - 7x - 30} - \sqrt {2{x^2} - 7x - 5} = x - 5$

Answer 1

Hint:
Firstly, write the equation $\sqrt {3{x^2} - 7x - 30} - \sqrt {2{x^2} - 7x - 5} = x - 5$ as $\sqrt {3{x^2} - 7x - 30} = x - 5 + \sqrt {2{x^2} - 7x - 5} $ .
Then, square both sides of the equation and solve it further.
On getting a linear equation in terms of x, factorize it and get the two roots of the equation.
Thus, verify whether the roots satisfy the given equation by substituting their values in the given equation.
Finally, write the root of the equation which satisfies the given equation.

Complete step by step solution:
It is given that, $\sqrt {3{x^2} - 7x - 30} - \sqrt {2{x^2} - 7x - 5} = x - 5$
 $\Rightarrow \sqrt {3{x^2} - 7x - 30} = x - 5 + \sqrt {2{x^2} - 7x - 5} $
Now, squaring both the sides and solving further will give
 $
  \Rightarrow {\left( {\sqrt {3{x^2} - 7x - 30} } \right)^2} = {\left( {x - 5 + \sqrt {2{x^2} - 7x - 5} } \right)^2} \\
  \Rightarrow 3{x^2} - 7x - 30 = {\left( {x - 5} \right)^2} + {\left( {\sqrt {2{x^2} - 7x - 5} } \right)^2} + 2\left( {x - 5} \right)\left( {\sqrt {2{x^2} - 7x - 5} } \right) \\
  \Rightarrow 3{x^2} - 7x - 30 = {\left( {x - 5} \right)^2} + 2{x^2} - 7x - 5 + 2\left( {x - 5} \right)\left( {\sqrt {2{x^2} - 7x - 5} } \right) \\
  \Rightarrow 3{x^2} - 7x + 7x - 30 = {x^2} - 10x + 25 + 2{x^2} - 5 + 2\left( {x - 5} \right)\left( {\sqrt {2{x^2} - 7x - 5} } \right) \\
  \Rightarrow 3{x^2} - 7x + 7x - 30 = 3{x^2} - 10x - 20 + 2\left( {x - 5} \right)\left( {\sqrt {2{x^2} - 7x - 5} } \right) \\
  \Rightarrow 3{x^2} - 3{x^2} + 10x - 30 - 20 = 2\left( {x - 5} \right)\left( {\sqrt {2{x^2} - 7x - 5} } \right) \\
  \Rightarrow 10x - 50 = 2\left( {x - 5} \right)\left( {\sqrt {2{x^2} - 7x - 5} } \right) \\
  \Rightarrow 10\left( {x - 5} \right) = 2\left( {x - 5} \right)\left( {\sqrt {2{x^2} - 7x - 5} } \right) \\
  \Rightarrow \dfrac{{10\left( {x - 5} \right)}}{{2\left( {x - 5} \right)}} = \sqrt {2{x^2} - 7x - 5} \\
  \Rightarrow 5 = \sqrt {2{x^2} - 7x - 5} \\
 $
Again, squaring both sides and solving further will give
 \[
  \Rightarrow {5^2} = {\left( {\sqrt {2{x^2} - 7x - 5} } \right)^2} \\
  \Rightarrow 25 = 2{x^2} - 7x - 5 \\
  \Rightarrow 2{x^2} - 7x - 5 - 25 = 0 \\
  \Rightarrow 2{x^2} - 7x - 30 = 0 \\
 \]
Now, we will factorize the equation \[2{x^2} - 7x - 30 = 0\] using the splitting of the middle term method.
 $
  \Rightarrow 2{x^2} - 12x + 5x - 30 = 0 \\
  \Rightarrow 2x\left( {x - 6} \right) + 5\left( {x - 6} \right) = 0 \\
  \Rightarrow \left( {x - 6} \right)\left( {2x + 5} \right) = 0 \\
  \Rightarrow x - 6 = 0{\text{ or }}2x + 5 = 0 \\
  \Rightarrow x = 6{\text{ or }}2x = - 5 \\
  \Rightarrow x = 6{\text{ or }}x = - \dfrac{5}{2} \\
 $
Thus, we get $x = 6$ or $x = - \dfrac{5}{2}$ .
Now, we will substitute $x = 6$ and $x = - \dfrac{5}{2}$ in the equation $\sqrt {3{x^2} - 7x - 30} - \sqrt {2{x^2} - 7x - 5} = x - 5$ and verify whether the values satisfy the equation or not.
Substituting $x = 6$ in $\sqrt {3{x^2} - 7x - 30} - \sqrt {2{x^2} - 7x - 5} = x - 5$ .
 $
  \sqrt {3{{\left( 6 \right)}^2} - 7\left( 6 \right) - 30} - \sqrt {2{{\left( 6 \right)}^2} - 7\left( 6 \right) - 5} = 6 - 5 \\
  \Rightarrow \sqrt {3\left( {36} \right) - 42 - 30} - \sqrt {2\left( {36} \right) - 42 - 5} = 1 \\
  \Rightarrow \sqrt {108 - 72} - \sqrt {72 - 47} = 1 \\
  \Rightarrow \sqrt {36} - \sqrt {25} = 1 \\
  \Rightarrow 6 - 5 = 1 \\
  \Rightarrow 1 = 1 \\
 $
Thus, $x = 6$ satisfies the equation $\sqrt {3{x^2} - 7x - 30} - \sqrt {2{x^2} - 7x - 5} = x - 5$ .
Now, substituting $x = - \dfrac{5}{2}$ in $\sqrt {3{x^2} - 7x - 30} - \sqrt {2{x^2} - 7x - 5} = x - 5$
 $
  \sqrt {3{{\left( { - \dfrac{5}{2}} \right)}^2} - 7\left( { - \dfrac{5}{2}} \right) - 30} - \sqrt {2{{\left( { - \dfrac{5}{2}} \right)}^2} - 7\left( { - \dfrac{5}{2}} \right) - 5} = - \dfrac{5}{2} - 5 \\
  \Rightarrow \sqrt {3\left( {\dfrac{{25}}{4}} \right) + \dfrac{{35}}{2} - 30} - \sqrt {2\left( {\dfrac{{25}}{4}} \right) + \dfrac{{35}}{2} - 5} = - \dfrac{{5 + 10}}{2} \\
  \Rightarrow \sqrt {\dfrac{{75}}{4} + \dfrac{{35}}{2} - 30} - \sqrt {\dfrac{{25}}{2} + \dfrac{{35}}{2} - 5} = - \dfrac{{15}}{2} \\
  \Rightarrow \sqrt {\dfrac{{75 + 70 - 120}}{4}} - \sqrt {\dfrac{{25 + 35}}{2} - 5} = - \dfrac{{15}}{2} \\
  \Rightarrow \sqrt {\dfrac{{25}}{4}} - \sqrt {\dfrac{{60}}{2} - 5} = - \dfrac{{15}}{2} \\
  \Rightarrow \dfrac{5}{2} - \sqrt {30 - 5} = - \dfrac{{15}}{2} \\
  \Rightarrow - \sqrt {25} = - \dfrac{{15}}{2} - \dfrac{5}{2} \\
  \Rightarrow - 5 = - \dfrac{{15 + 5}}{2} \\
  \Rightarrow - 5 = - \dfrac{{20}}{2} \\
 $
 $\Rightarrow - 5 = - 10$ (which is not possible)
So, $x = - \dfrac{5}{2}$ does not satisfy the equation $\sqrt {3{x^2} - 7x - 30} - \sqrt {2{x^2} - 7x - 5} = x - 5$.

Thus, $x = 6$ is the only root of the equation.

Note:
The steps to factorize the equation of form $a{x^2} + bx + c = 0$ by the method of splitting of the middle term are as follows:
1) Find the product of first and last term $a \times c$
2) Find the factors of b such that the sum of the factors is the product of first and last term. Let, p and q be the factors of the equation $a{x^2} + bx + c = 0$ . So, $p + q = a \times c$ .
3) Write the middle term as the sum of the factors as $a{x^2} + px + qx + c = 0$ .
Thus, solve the equation above to get the factors of the equation.