Algebra Help?

Question

I didn't know that there were so many people that loved this stuff! If anyone can help with this with the "how-to's" on how you did it, I would love ya forever!! :

Solve using the quatradic formula: x^2-5x-3= -7
and this one:
Solve by completing the square: x^2+4x+2=0

Thanks!!!

Answer 1

Quadratic Formula

x =- b ± √b² - 4ac / 2a

x² - 5x + 4 = 0

let

a = 1

b = - 5

c = 4

x = - (- 5) ± √(- 5)² - 4(1)(4) / 2(1)

x = 5 ± √25 - 16 / 2

x = 5 ± √9 / 2

x = 5 ± 3 / 2

- - - - - - - - -

Solving for +

x = 5 + 3 / 2

x = 8 / 2

x = 4

- - - - - -

Solving fkor -

x = 5 - 3 / 2

x = 2 / 2

x = 1

- - - - - - - - - - - -

Problem 2

x² + 4x + 2 = 0

x² + 4x + 2 - 2 = 0 - 2

x² + 4x +_______= - 2 + _____

x² + 4x + 4 = - 2 + 4

x² + 4x +4 = 2

(x + 2)(x + 2) = 2

(x + 2)² = 2

(√x + 2) = ± √2

x + 2 = ± √2

x + 2 - 2 = - 2 ± √2

x = - 2 ± √2

- - - - - - - - - - -s-

Answer 2

Hey it's so easy. Answer like this.

First question
x^2-5x-3 = -7
then
x^2-5x-3+7=0
x^2-5x-+4=0
Now we could use the quadratic formula
(-b +/- square root of b^-4ac)/ 2a

Here a = 1 , b = -5 , c = 4
so b^2 - 4ac = (-5 * -5) -(4*1*4)
= 25 - 16
= 9
Then
x = (-b +/- square root of b^-4ac)/ 2a
= (-(-5)+/- square root of 9) / 2*1
= (5 +/- 3) / 2
x = (5 + 3) / 2 or x = (5 - 3) / 2
= 8/2 or = 2/2
= 4 or = 1
so x = 4 or 1
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Second question
x^2+4x+2=0
You want to make it's answer by completing the square.

Then here we want to make the formula (a+b)^2 = x.
for that the equation will be in a^2 + 2ab + b^2 form.

So x^2+4x+2 = 0
x^2+4x+(4/2)^2 = 2 ( x^2 +4x+2+2 = 2)
x^2+4x+2^2 = 2
(x+2)^2 = 2
doing square root in two sides
x+2 = +/-square root of 2
then x = square root of 2 - 2 or x = -square root of 2-2
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Answer 3

the 2nd one is pretty easy. all you need to do is add 2 to each side so you get x^2 +4x+4 = 2. You can then factor it to be (x + 2)^2 = 2. Now what do you need to square to get 2, the answer is the sqr root of 2. So what plus 2 equals the square root of 2. The answer: Sqr root of 2 minus 2. So the answer is (root 2 minus 2).

For the first one, add 7 to both sides so you get (x^2 -5x +4) . Then use the quadratic formula which is X = (-b +- sqroot (b^2 - 4ac)) all over 2a. Where "a" is the coefficient infront of the 2nd degree term. 'b" is infront of the first degree term, and "c" is the constant. So using the formula plug in a = 1 b = -5 c = 4.and you will get your answer.

Answer 4

x^2 - 5x - 3 = -7
Add 7 to both sides:
x^2 - 5x + 4 = 0
You don't really need the quadratic formula
as this factorises quite nicely into:
(x - 1)(x - 4) = 0, so x = 1 or 4.

x^2 + 4x + 2 = 0
Subtract 2 from both sides:
x^2 + 4x = -2
Take half of the 4 and square it. That is, (4/2)^2 = 4.
Then add it to both sides:
x^2 + 4x + 4 = -2 + 4
So, (x + 2)^2 = 2
Take the square root of both sides:
x + 2 = ± sqrt(2)
Subtract 2 from both sides:
x = -2 ± sqrt(2)

Answer 5

x^2-5x-3= -7
=>x^2-5x+4=0
x^2-x-4x+4=0
x(x-1)-4(x-1)=0
=>(x-1)(x-4)=0
Hence x=1 or 4

x^2+4x=-2
=>x^2+2*x*2+(2)^2=-2+4 [adding 2^2 or 4 to both sides
=(x+2)^2=2
=>x+2=+-sqrt 2
=>x= -2+-sqrt 2

Answer 6

x^2-5x-3=-7
x^2-5x+4=0
x^2-4x-x+4=0
X(x-4)-1(x-4)=0
x=4,1 or rather using the quadratic formula
x={-(-5)+/-square root[ (-5)^2-(4*1*4)]}/2*1
=[5+/-square root(25-16)]/2
=(5+/-3)/2=(5+3)/2 or (5-3)/2=4 or 1

Quadratic formula is:
x={-b+/-sqaure root of(b^2-4*a*c)}/(2*a)
where x is the root of the quadratic equation ax^2+bx+c=0

x^2+4x+2=0

x^2+2*2*x+2^2 -2=0

(x+2)^2=2
x+2=(+/-)sqaure root(2)
x= -2(+/-)square root (2)

Answer 7

x^2-5x-3= -7,
x^2-5x+4=0,
x^2-4x-1x+4=0
x^2-1x-4x+4=0
x(x-1)-4(x-1)=0
(x-4)(x-1)=0
Therefore value of x will be 4 or1.


x^2+4x+2=0
(x+2)^2+(2-4)=0
(x+2)^2-2=0
(x+2)^2=2
Therefore value of x will be (2^.5 -2)=2^.5(1-2^.5)
Thank You.

Answer 8

hey..ur first qn.
wd be...x = -b +-sqrt. b^2-4ac /2..so u get..
x= -(-5)+- sqrt. (-5)^2 - 4(1)(4)/2
therefore...after solving u get x=7, and x = -2
second one...
x = (x+b/2)^2 - (b/2)^2 +c
soo..
(x+4/2)^2 - (4/2)^2 +2
= (x+2)^2 -4+2
= x^2+4x+2..final ans.