solve each equation by solving the square.?

Question

x^2-16x+23=0

i need you to explein it please.

Answer 1

x^2 - 16x + 23 = 0
x^2 - 16x = -23
find half of 16, square it, and add to both sides
x^2 - 16x + 64 = 41
factor the left side into a perfect square
(x - 8)^2 = 41
sqrt both sides
x - 8 = sqrt(41)
add 8 to both sides
x = 8 ± sqrt(41)

Answer 2

x^2-16x+23=0

first, you start by isolating the x.

using like terms (numbers of the same type... like 3 and 4, or 14x and 42x. :-p), we can subtract the 23 to the other side.

x^2-16x+23=0
-23 -23
_____________
x^2-16x=-23

then, we can divide both sides by 16.

thus making x^2= -23/16

then... square both sides... which makes:
x=sqrt of -23 divided by -23/16

but this looks like a quadratic problem...
I sorta remember it from Algebra II (which was a few semesters ago for me... around 6 months ago.)

The formula for Quadratic expressions is this:

-b ± (plus-minus) SQRT OF (b^2 - 4ac)
x = _________________________________

...........................2a


think of it this way: x^2 is the same as 1(ONE)x^2... right?
well, let this '1' be the A in this case...

let the '16' from 16x be the B, and at the same time... let C be 23.
remember this order. it applies to all quadratics. the first number in the quadratic is the A, the second is the B, and the third is the C. The zero is excluded from this... so if you had 14x^3-3x^2-44x=0... this would stilll work in the ABC, thing... as long as they're descending values. Anyway... I'm rambling.

Plug in the A, B, and C values into the formula I've provided above and you shall have your correct Algebraic Quadratic answer. :-)

Email me if you have any more math trouble. I'll try to help you... that is... if I'm not braindead or forgot anything else from math class, lol. I remember this, though... I had fun.

Answer 3

I'm assuming you mean completing the square?

It is just like this one. I answered this question for somebody else. All you have to do is change the numbers to match your problem. There are a lot of steps, but they aren't hard steps.

Completing the square isn't too bad. The first thing I have my students do is identify a, b, and c.
a=1 (this is good, it makes the problem even easier)
b=-4 (this is VERY good, whenever b is even, it divides by 2, which makes our problem much easier)
c=-5

Now, the only steps for completing the square are to:

1) Find b
2) Take half of b
3) Square that
4) Add that number to (x^2 - 4x + ) - 5

1) b= -4
2) half b is -2
3) square that is +4
4) Add that number to (x^2 - 4x + 4) - 5

The trick now is that you changed the problem by adding +4, so you minus it back out to keep things fair. We now have:

(x^2 - 4x + 4) - 5 - 4 simplify to
(x^2 - 4x + 4) - 9

Now, we factor (x^2 - 4x + 4), which is easy.

It will always be (x )(x ), and the blank will always be the half b.

(x - 2)(x - 2) - 9
(x - 2)^2 - 9 = 0 to solve for x...add 9, then square root both sides.

(x - 2) = sqrt (9)

(x - 2) = +3 OR (x - 2) = -3, add 2 to both sides:

x = +3 + 2 OR x = -3 + 2

x = 5, or x = -1

Once you get the hang of completing the square, it is really easy. Plus, it is really nice because doing so puts your function in whats called vertex form, making it really easy to find the vertex, and then the shape of the graph.

Answer 4

i don't know how to explain it , i think the teacher did a good job on that i can tell you i solve it OK:
X*2 - 16x +23 = 0
A= 1
B= -4 and C= 1/2 B (squared) = 2

(x*2 -4x +4 ) -5 - 4=0
(x*2 - 4x + 4 ) -9 = 0
Square root the equation and you get this

( x-2 )( x-2 ) = + or - 3
x-2 = 3 OR x-2 = -3
x=3 +2 x= -3 +2
x= 5 X= -1

F (x) = ]-1 ; 5 [

Answer 5

x^2 - 16x + 23 = 0

x^2 - 16x + 64 = 41

(x-8)(x-8) = 41 or (x-8)^2= 41

x-8= sqrt of 41

x= 8+ sqrt of 41
note: square roots have 2 values positive and negative

Answer 6

eqn of the form ax^2+bx+c=0
solve by discriminant method.
let D=(b^2-4ac)

x=(-b+sqrt(D))/2a and (-b-sqrt(D))/2a are the solutions for finding x..substitute and find x values..

Answer 7

no possible solution