The top left and bottom left
A) 5n- 1
the patterns going up in 5s and to get back to 4 you need to - 1
To complete the square, the second degree term must have a coefficient of 1.
Since the second degree term here has a coefficient of 4, we start by dividing each term on both sides by 4.



Now we can complete the square.
First, we need to find what number completes the square.
We take the coefficient of the first degree term, -7 in this case.
Divide it by 2 and square it. -7 divided by 2 is the fraction -7/2.
Now we square -7/2 to get 49/4.
We add 49/4 to both sides.



Wow ! There's so much extra mush here that the likelihood of being
distracted and led astray is almost unavoidable.
The circle ' O ' is roughly 98.17% (π/3.2) useless to us. The only reason
we need it at all is in order to recall that the tangent to a circle is
perpendicular to the radius drawn to the tangent point. And now
we can discard Circle - ' O ' .
Just keep the point at its center, and call it point - O .
-- The segments LP, LQ, and LO, along with the radii OP and OQ, form
two right triangles, reposing romantically hypotenuse-to-hypotenuse.
The length of segment LO ... their common hypotenuse ... is the answer
to the question.
-- Angle PLQ is 60 degrees. The common hypotenuse is its bisector.
So the acute angle of each triangle at point ' L ' is 30 degrees, and the
acute angle of each triangle at point ' O ' is 60 degrees.
-- The leg of each triangle opposite the 30-degree angle is a radius
of the discarded circle, and measures 6 .
-- In every 30-60 right triangle, the length of the side opposite the hypotenuse
is one-half the length of the hypotenuse.
-- So the length of the hypotenuse (segment LO) is <em>12 </em>.