Question 728593
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You can get good mental exercise by solving these equations using logical reasoning and simple mental arithmetic.<br>
Start by assuming (hoping) that the solutions are integers.  Then use logical reasoning to determine the solution(s).<br>
(1) {{{sqrt(2x-3)=x-1}}}<br>
In this equation, we want (2x-3) to be a perfect square. And if x is an integer, the (2x-3) is an odd perfect square.  So look at values of x for which (2x-3) is an odd perfect square.<br>
(a) 2x-3 = 1; 2x = 4; x = 2.  sqrt(2x-3) = sqrt(1) = 1; and 2-1 = 1.  So x = 2 is a solution.<br>
(b) 2x-3 = 9; 2x = 12; x = 6.  sqrt(2x-3) = 3 but x-1 = 6.  x = 6 is not a solution.<br>
For larger values of x, "x-1" will increase faster than "sqrt(2x-3)", so there are no more solutions.<br>
ANSWER: x = 2<br>
(2) {{{sqrt(3x-3)-sqrt(x)=1}}}<br>
In this equation, we want x and (3x-3) to both be perfect squares.<br>
(a) x = 1; 3x-3 = 0; both are perfect squares.  However, trying x = 1 in the equation gives us "0-1 = 1", which is not true.  Note that the two sides of the equation evaluate to numbers that are opposites.  This is a common occurrence in equations like this, where the formal solution process including squaring both sides of an equation leads to extraneous solutions.<br>
So x = 1 is NOT a solution.<br>
(b) x = 4; 3x-3 = 9 is also a perfect square; and sqrt(3x-3) = 3 and x-1 = 3, so x = 4 is a solution.<br>
Similarly to the first problem, for larger values of x "sqrt(3x-3)" will increase faster than "sqrt(x)", so again there are no more solutions.<br>
ANSWER: x = 4<br>