SOLUTION: There is a two-digit number whose digits are the same, and has got the following property: When squared, it produces a four-digit number, whose first two digits are the same and eq

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Question 1100502: There is a two-digit number whose digits are the same, and has got the following property: When squared, it produces a four-digit number, whose first two digits are the same and equal to the original’s minus one, and whose last two digits are the same and equal to the half of the original’s. Find that number. Please solve it using Quadratic formula!!!!
(Please solve it using Quadratic formula.)*
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
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%2Bx.

The number SQUARED will be %2810x%2Bx%29%5E2=%2811x%29%5E2=121x%5E2.

Following the description, the resulting squared original number must be 1000%28x-1%29%2B100%28x-1%29%2B10%28x%2F2%29%2Bx%2F2.


This means the two expressions must be equal:
121x%5E2=1000%28x-1%29%2B100%28x-1%29%2B10%28x%2F2%29%2Bx%2F2-------simplify and solve this equation.

The equation should be found 121x%5E2-1105%261%2F2x%2B1100=0
or in whole number-coefficients,
highlight_green%28242x%5E2-2211x%2B2200=0%29
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%28x=8%29.
Original Number: 88.

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Why the two-digit original numbers starts as 10x%2Bx?
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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%2B1%2Ax=10x%2Bx=11x.