complete the square?

Question

what formula do i use and how do i do it?

y=x^2-8x+7

Answer 1

Hi,

y = x² - 8x + 7
Since a = 1, multiply ½ times (-8) from the x term and square your answer. ½(-8) = -4 Now square this number and add it to complete the square. Subtract the same number out on the end to balance out what you added to complete the square.

y = x² - 8x + 16 + 7 - 16
Now factor the first 3 terms as a perfect square trinomial.

y = (x - 4)² - 9

This is the vertex form of a quadratic equation. y = a(x - h)² + k, where (h,k) is the vertex. The vertex here is (4,-9). The parabola would open up because "a" is positive. That means the vertex is the lowest point on the graph.

I hope that helps!! :-)

Answer 2

The formula is complicated. But what you gotta do is find a way to make y(x) into a squared function.
y = x^2 - 8x + 7
Look at x^2 - 8x, if you add and subtract 16, then
x^2 - 8x + 16 - 16 = (x-4)^2 - 16
Now just put that back into y:
y = (x-4)^2 - 16 + 7 = (x-4)^2 - 9

Answer 3

there is no formula, simply its a technique, watc;
1) first step to find (b/2), which is half the co-eff of x = -8/2 = -4
2) add this number ^2 and subtract it to your equation,(that means you did not change your equation),
y = x^2 -8x + (-4)^2 - ( -4)^2 +7
3) the first three terms are perfect square, so you can write it as,
y = ( x -4 )^2 -16 + 7
y = ( x-4 )^2 -9

Answer 4

(x ± b/2)² = x² ± bx + (b/2)²

x² - 8x + 7
x² - 8x + 4² - 4² + 7
(x - 4)² - 9

Answer 5

Hey there!

For a quadratic, x^2+bx+c=0, the formula is:

x^2+bx+c=0 -->
x^2+bx=-c -->
x^2+bx+(b/2)^2=-c+(b/2)^2.

So, here's the answer.

y=x^2-8x+7 --> Write the problem.
y-7=x^2-8x --> Subtract 7 on both sides of the equation.
y-7+(8/2)^2=x^2-8x+(8/2)^2 --> Complete the square.
y-7+16=x^2-8x+16 --> Substitute 16 for (8/2)^2.
y+9=x^2-8x+16 --> Add -7 and 16.
y+9=(x-4)^2 --> Completely factor the right side of the equation.
y=(x-4)^2-9 Subtract 9 on both sides of the equation.

The equation y=(x-4)^2-9, is the vertex form for a parabola, with vertex at (4,-9).

If you're really interested, in order to do completing the square for ax^2+bx+c=0, here's how.

ax^2+bx+c=0 -->
ax^2+bx=-c -->
x^2+(b/a)x=-(c/a) -->
x^2+(b/a)x+(b/2a)^2=-(c/a)+(b/2a)^2 -->
x^2+(b/a)x+(b/2a)^2=-(c/a)+b^2/4a^2 -->
(x+b/2a)^2=-(c/a)+b^2/4a^2 -->
(x+b/2a)^2=-(4ac/4a^2)+b^2/4a^2 -->
(x+b/2a)^2=(b^2-4ac)/4a^2 -->
x+b/2a=±sqrt((b^2-4ac)/4a^2) -->
x+b/2a=±sqrt(b^2-4ac)/sqrt(4a^2) -->
x+b/2a=±sqrt(b^2-4ac)/2a -->
x=(-b/2a)±sqrt(b^2-4ac)/2a -->
x=(-b±sqrt(b^2-4ac))/2a

As you can see, Complete the Square technique proved the Quadratic Formula.

Hope it helps!

Answer 6

y = (x² - 8x + 16) - 16 + 7
y = (x - 4)² - 9

Answer 7

if your talking about factoring. Simply use fearless factoring! This is what I learned in highschool and it works for every factoring question!
Fearless factoring=
Two numbers that would multiply into 7 and two numbers that would add to -8.
(-7) (-1) = 7
(-7)+(-1)= -8
=(x-7)(x-1)

Answer 8

y = x^2 - 8x + 7
y = x^2 - 2*x*4 + 4^2 - 4^2 + 7
y = x^2 - 2*x*4 + 4^2 - (4^2 - 7)
y = (x - 4)^2 - (16 - 7)
y = (x - 4)^2 - 9
y = (x - 4)^2 - 3^2

Answer 9

Completing the square works like this:

y = x^2 + b*x+c
y-(b/2)^2=x^2+b*x+(b/2)^2+c
y-(b/2)^2=(x+(b/2))^2 + c

Answer 10

You could just factor it.
0=(x-7)(x-1)
x-7=0 x-1=0
x=7,x=1