# Math Help - quadratic equation

three points : A=3,2 B=1,2 C=0,k with k being a constant

AC=5BC

FIND THE VALUES OF K

Originally Posted by kitobeirens
three points : A=3,2 B=1,2 C=0,k with k being a constant
AC=5BC
FIND THE VALUES OF K
Solve $(0-3)^2+(k-2)^2=25[(0-1)^2+(k-2)^2]$

Originally Posted by Plato
...... $=25[(0-1)^2+(k-2)^2]$
The 25 should be 5.
You'll get 2 valid solutions.

Originally Posted by Wilmer
The 25 should be 5.
You'll get 2 valid solutions.
Think again: ${\left[{5\sqrt {{{\left( {0 - 1} \right)}^2} + {{\left( {k - 2} \right)}^2}} } \right]^2} = 25\left[ {{{\left( {0 - 1} \right)}^2} + {{\left( {k - 2} \right)}^2}} \right]$

My bad. Jumped on "5" and got nice integer solutions: k=1 or k=3;
both "looked" good on my quickly sketched diagram!

BUT: if the correct "25" is used, then : k = [6 +- sqrt(-6)] / 3;
so no real solution.
Perhaps the OP made a typo....

The 5 does give a very nice solution. So it should have been $AC=\sqrt5 BC$.