Given Eq:
Solving this:
Squaring both sides,
➝ x + 3 = (x - 3)²
➝ x + 3 = x² - 6x + 9
Now shifting RHS to LHS,
➝ x + 3 - x² + 6x - 9 = 0
➝ -x² + 7x - 6 = 0
➝ x² - 7x + 6 = 0
Now finding the values of x by middle term factorisation,
➝ x² - 6x - x + 6 = 0
➝ x(x - 6) - 1(x - 6) = 0
➝ (x - 1)(x - 6) = 0
➝ x = 1 or 6.
But here, you can see, x = 1 is not a solution to the original equation because when we say about square roots, value can't be negative.
Hence,
- As a result, the only valid solution to the original equation is x = 6.
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