Question 1100502
The way you want to start this, if quadratic equation should be used,
Let your original number's same two digits be x.

Your original number is then  {{{10x+x}}}.


The number SQUARED will be  {{{(10x+x)^2=(11x)^2=121x^2}}}.


Following the description, the resulting squared original number must be {{{1000(x-1)+100(x-1)+10(x/2)+x/2}}}.



This means the two expressions must be equal:
{{{121x^2=1000(x-1)+100(x-1)+10(x/2)+x/2}}}-------simplify and solve this equation.


The equation should be found {{{121x^2-1105&1/2x+1100=0}}}
or in whole number-coefficients,
{{{highlight_green(242x^2-2211x+2200=0)}}}
which you should find the solutions for x will be either 1.136363, or 8.
(You can substitute into quadratic solution formula yourself and evaluate each expression).
The solution which works as the digit is {{{highlight_green(x=8)}}}.
Original Number:  88.


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Why the two-digit original numbers starts as {{{10x+x}}}?
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GIVEN:   There is a two-digit number whose digits are the same, ...
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Let the ten's digit and the unit's digit each be x.  This means that the two-digit number can be expressed as  {{{10x+1*x=10x+x=11x}}}.