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Solve the following equations:
$\sqrt {2{x^2} + 5x - 7} + \sqrt {3\left( {{x^2} - 7x + 6} \right)} - \sqrt {7{x^2} - 6x - 1} = 0$

Answer
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Hint: - Here, we just factorize the equation and take out common, to solve the equation.
$\sqrt {2{x^2} + 5x - 7} + \sqrt {3\left( {{x^2} - 7x + 6} \right)} - \sqrt {7{x^2} - 6x - 1} = 0$
We just factorize the equation which are under the root and write in its factor form as
$\sqrt {\left( {2x + 7} \right)\left( {x - 1} \right)} + \sqrt {3\left( {x - 1} \right)\left( {x - 6} \right)} + \sqrt {\left( {7x + 1} \right)\left( {x - 1} \right)} = 0$
And from here we take $\sqrt {\left( {x - 1} \right)} $ as common and write as
$\left( {\sqrt {\left( {x - 1} \right)} } \right)\left( {\sqrt {{\text{2x + 7}}} {\text{ + }}\sqrt {{\text{3}}\left( {{\text{x - 6}}} \right)} {\text{ - }}\sqrt {\left( {{\text{7x + 1}}} \right)} } \right) = 0$
By the above equation either $\sqrt {\left( {x - 1} \right)} = 0{\text{ }}$or $\left( {\sqrt {2x + 7} + \sqrt {3\left( {x - 6} \right)} - \sqrt {\left( {7x + 1} \right)} } \right) = 0$
If $\sqrt {\left( {x - 1} \right)} = 0{\text{ }}$then $x = 1$and
If $\left( {\sqrt {2x + 7} + \sqrt {3\left( {x - 6} \right)} - \sqrt {\left( {7x + 1} \right)} } \right) = 0$
We write it as
$\sqrt {2x + 7} + \sqrt {3\left( {x - 6} \right)} = \sqrt {\left( {7x + 1} \right)} $
On squaring we get
$2x + 7 + 3x - 18 + 2\sqrt {3\left( {2x + 7} \right)\left( {x - 6} \right)} = 7x + 1$
And by solving
$
  2\sqrt {3\left( {2x + 7} \right)\left( {x - 6} \right)} = 2x + 12 \\
 On squaring we get
 3\left( {2x + 7} \right)\left( {x - 6} \right) = {\left( {x + 6} \right)^2} \\
  6{x^2} - 15x - 126 = {x^2} + 36 + 12x \\
  5{x^2} - 27x - 162 = 0 \\
  5{x^2} - 45x + 18x - 162 = 0 \\
  5x\left( {x - 9} \right) + 18\left( {x - 9} \right) = 0 \\
  \left( {x - 9} \right)\left( {5x + 18} \right) = 0 \\
 $
We get $x = 9$ and $x = \dfrac{{ - 18}}{5}$
$\therefore $ The values for $x$ are $1, 9$ and $\dfrac{{ - 18}}{5}$

Note:-Whenever such types of questions are given just try to factorize the equation to take out common from the roots to make the question easy to solve. And then solve the equation to find out the values.