Example 1: Solve by completing the square:
Do we have the correct coefficient of y2 for using the completing-the-square method?
Yes, we must have a coefficient of 1, as here.
Add 3 to both sides to prepare for completing the square.
What do we add to both sides to get a perfect square on the left side?
Half of the "7", squared.
Do it.
Rewrite with the left side as a perfect square.
What do we do next?
Take the square root on both sides.
Do it.
, or
Solve for y.
Use a calculator to get decimal values.
y » 0.405, -7.405
Check these in the original equation.
y » 0.405: (0.405)2 + 7*(0.405) - 3 = -9.75*10-4 » 0.000
y » -7.405: (-7.405)2 + 7*(-7.405) - 3 » 0.000
Example 2: Solve
by completing the square.
In order to use "completing the square", we the coefficient of x2 to be 1. Can we achieve this here?
Yes, we can divide by 2 everywhere.
Do it.
After dividing by 2, we have
We need to have just x2 and x terms on the left side. How can we achieve this?
Add 15 to both sides, as usual.
Do it.
What do we need to add to both sides in order to get a perfect square on the left side?
We need to add "half of the 6, squared".
Do it.
, or
Rewrite the left side as a perfect square.
What is the next step?
Take the square root of both sides.
Do it.
, or
Solve for x.
Subtracting 3 from both sides, we get
Use a calculator to get decimal values.
x » -7.90, 1.90
Check each in the original equation.
x »
-7.90:
2(-7.90)2 + 12(-7.90) - 30 »
0.00
x »
1.90:
2(1.90)2 + 12(1.90) - 30 = 0.02 »
0.00
OK