
Solve the following equations:
$ \sqrt {2{x^2} - 7x + 1} - \sqrt {2{x^2} - 9x + 4} = 1. \\
$
Answer
534.3k+ views
Hint: Simplify the equation using the normal method by bringing common terms together on one side and then solve to find the answer.
The given equation is as
\[ \sqrt {2{x^2} - 7x + 1} - \sqrt {2{x^2} - 9x + 4} = 1 \\
\Rightarrow \sqrt {2{x^2} - 7x + 1} = \sqrt {2{x^2} - 9x + 4} + 1 \\ \]
After squaring both sides, we get,
$2{x^2} - 7x + 1 = 1 + 2{x^2} - 9x + 4 + 2\sqrt {2{x^2} - 9x + 4} \\
2x - 4 = 2\sqrt {2{x^2} - 9x + 4} \\
x - 2 = \sqrt {2{x^2} - 9x + 4} \\$
Again after squaring both sides, we get,
${(x - 2)^2} = 2{x^2} - 9x + 4 \\
{x^2} + 4 - 4x = 2{x^2} - 9x + 4 \\
{x^2} - 5x = 0 \\
x(x - 5) = 0 \\
\Rightarrow x = 0,5 \\$
So, this is the required solution.
NOTE: On squaring the values, we must write the terms carefully, without missing any terms in between.
The given equation is as
\[ \sqrt {2{x^2} - 7x + 1} - \sqrt {2{x^2} - 9x + 4} = 1 \\
\Rightarrow \sqrt {2{x^2} - 7x + 1} = \sqrt {2{x^2} - 9x + 4} + 1 \\ \]
After squaring both sides, we get,
$2{x^2} - 7x + 1 = 1 + 2{x^2} - 9x + 4 + 2\sqrt {2{x^2} - 9x + 4} \\
2x - 4 = 2\sqrt {2{x^2} - 9x + 4} \\
x - 2 = \sqrt {2{x^2} - 9x + 4} \\$
Again after squaring both sides, we get,
${(x - 2)^2} = 2{x^2} - 9x + 4 \\
{x^2} + 4 - 4x = 2{x^2} - 9x + 4 \\
{x^2} - 5x = 0 \\
x(x - 5) = 0 \\
\Rightarrow x = 0,5 \\$
So, this is the required solution.
NOTE: On squaring the values, we must write the terms carefully, without missing any terms in between.
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