Use complete sentences to explain how the quadratic formula is related to the process of completing the square.

SOS

Completing the square is a method that will solve all quadratic equations. By completing the square on the general quadratic equation, we obtain a formula that is worth learning and that will solve all quadratic equations.

The quadratic formula does the same thing as completing the square.  The formula reads:

The a, b and c terms are the coefficients of a polynomial where a is the coefficient of the squared term, b the linear term and c the constant when the equation is set equal to zero [ax squared plus bx plus c = 0].

Solving quadratics with this formula is then just as simple as substituting the numbers from an equation into these placeholder variables. Then, you need to simplify the expression to get the two solutions, one where you use the plus sign and the other the minus sign.

Hope this helps!!

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

b is the final answer! : )