Mathematical and Physical Journal
for High Schools
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Problem C. 1118. (March 2012)

C. 1118. Solve the equation 4x^2 + \frac34 = 2\sqrt x on the set of real numbers.

(5 pont)

Deadline expired on April 10, 2012.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Elsőként megjegyezzük, hogy \(\displaystyle x\ge 0\). Az egyenlet mindkét oldalához adjunk \(\displaystyle 4x+\frac{1}{4}\)-et: \(\displaystyle 4x^2+4x+1=4x+2\sqrt x + \frac 14\) egyenlet felírható \(\displaystyle (2x+1)^2=\left (2\sqrt x +\frac 12 \right )^2\) alakba, ahonnan \(\displaystyle |2x+1|=\left |2\sqrt x +\frac 12 \right |\). Mivel \(\displaystyle 2x+1>0\) és \(\displaystyle 2\sqrt x + \frac 12 >0\), ezért az abszolút-érték elhagyható: \(\displaystyle 2x+1=2\sqrt x + \frac 12\). Ez az egyenlet \(\displaystyle \sqrt x\)-re nézve másodfokú, egyetlen megoldása a \(\displaystyle \sqrt x =\frac 12\), ahonnan \(\displaystyle x=\frac14\).


Statistics:

175 students sent a solution.
5 points:112 students.
4 points:16 students.
3 points:24 students.
2 points:7 students.
1 point:5 students.
0 point:7 students.
Unfair, not evaluated:4 solutions.

Problems in Mathematics of KöMaL, March 2012