algebra help ??????????

Question

if u forgot....squared = to the 2nd power and cubed is means to the 3rd power...i put the names instead of the #s beacuse i would get confusing......u may answer as many as u want...i have to do corrections on my test, and i got these wrong..im in the 8th grade....please help..........and if possible, explain.....thank you

complete the square:

x(squared) -6x + 7 = 0

(hint: you will get 2 answers)


slove for the variable:

3x(squared) - x = 0

(hint: you will get 2 answers.) x=0 and x=?


simplify:

5p(cubed) q(-squared)
---------------------------------- (over)
15p(to the -4th power) q


simplify this expression:

2 squareroot of 44 + 5 squareroot of 11 =


slove the equation:

squareroot of 4x + 32 + 1 =1

4x+32 is all under one squareroot, the + 1 is not


thank you

Answer 1

x^2-6x +7= 0
x^2-6x +9 +7 =9 [9 = (6/2)^2]
(x-3)^2 =9-7 = 2
x-3 = +/- sqrt(2)
x = 3 +sqrt(2) and x = 3- sqrt(2)

3x^2-x =0
x(3-x)=0
3-x = 0 so x = 3 [ x=0 is other answer]

5p^3q^-2/15p^-4q = p^7/3q^3 [Remember x^-2=1/x^2]

2sqrt(44) +5sqrt(11)
= 2 sqrt(4)sqrt(11) +5sqrt(11)
=4sqrt(11) +5sqrt(11) = 9sqrt(11)

sqrt(4x+32) +1 = 1
sqrt(4x+32) = 0
4x+32 = 0
4x = -32
x = -8

Answer 2

Completing the square is an easy process once you've done it a few times. given ax^2 + bx + c = 0, you "add zero" in the form of +(b/2)^2 - (b/2)^2. Then you have the equation ax^2 + bx + (b/2)^2 + c = 0 which you can then simplify to (ax + b/2)^2 + c=0. In your case, a = 1, b = -6, and c = 7. So "add zero" to your equation for (x^2 - 6x + 9) - 9 + 7 = 0 (do you see where the zero is?) Then you can see that this is the same as (x-3)^2 -2 = 0 or (x-3)^2 = 2. Now take the square root and add three to both sides to get x = 3 +/- sqrt(2)... plug the two answers in and you will find that they both work.

Now for 3x^2-x = 0... obviously 0 is an answer (this is usually called the trivial solution) so divide out the 'extra' x to get the equation 3x - 1 = 0... so then 3x = 1 or x = 1/3. If you want to be sure that this is true, plug it back into the original equation.

Now for the next one... remember that dividing by an exponentiated variable of one sign is the same as multiplying by the variable with the same exponent, but opposite sign... that is, dividing by x^2 is the same as multiplying by x^(-2)... likewise dividing by x^(-2) is the same as multiplying by x^2. Now, convert all the dividing into multiplying (except the 5/15 = 1/3)... (1/3) * p^3 * p^4 * q^(-2) * q^(-1)... now remember that multiplying two exponentials with like bases yields you a single exponential with the same base and the exponents added together. So p^3 * p^4 becomes p^7... Now simply the equation... (1/3) * p^7 * q^(-3). If you want, you could then turn this back into a fraction, but there's really no reason to.

continuing on... 2 * sqrt( 44 ) + 5 sqrt( 11 ) is the same as 2*sqrt( 4* 11) + 5*sqrt(11) = 2*sqrt(4)*sqrt(11) + 5 * sqrt(11) = (4+5)*sqrt(11) = 9 * sqrt(11)

And lastly sqrt( 4x + 32 ) + 1 = 1 ---> sqrt( 4x + 32 ) = 0 right? Well now, square both sides (keep in mind if you do this that you will always need to check all your answers to make sure they are valid)... then you have 4x+32 = 0 so x = -8. Plugging that back in to the original equation... yup... works great...

You should make sure that you really understand these concepts... these are some of the most fundamental strategies you'll be exposed to in algebra classes... especially completing the square.

Answer 3

1) complete the square

x^2 -6x +7 = 0

first, move constant to right side

x^2 -6x = -7

next add 1/2 of term before x squared to both sides..
that is add (1/2 of -6 )^2 = 9 to both sides

x^2 - 6x + 9 = -7 + 9 = 2

then factor left side...

(x -3)^2 = 2

x - 3 = +/- (2)^.5
so x = sqrt 2 + 3 and x = 3 - sqrt 2

2) solve....

3 x^2 - x = 0

factor....
x (3x -1) = 0

so x = 0
and 3x -1 = 0 ---> 3x = 1, x = 1/3


3) simplify

(5 p^3 * q ^-2) / (15 p^-4 * q)

divide numbers. subtract powers...

5 / 15 and p ^(3 - (-4)) and q ^ (-2 -1)

= 1/3 x p^7 x q ^-3 = p^7 / (3 q^3)

4) simplify

2 x 44^.5 + 5 x 11^.5
2 x (4 x 11)^.5 + 5 x 11^.5
2 x 4^.5 x 11^.5 + 5 x 11^.5
2 x 2 x 11^.5 + 5 x 11^.5
4 x 11^.4 + 5 x 11^.5
9 x 11^.5

5) solve....

(4x + 32) ^ .5 + 1 = 1
(4x + 32)^.5 = 0

4x + 32 = 0
4x = -32
x = -8

Answer 4

x^2- 6x + 7 = 0 move 7 to right side

x^2 - 6x +9 = -7 + 9 add (b/2)^2 = 9 to both sides

(x-3)^2 = 2 now take square root (2 +/- answers)

x-3 = +/- SQRT(2)
x = 3 + SQRT (2) or x = 3 - SQRT (2)

3x^2 - x = 0 common factor is x
x(3x - 2) = 0
x = 0 or 3x - 2 = 0 now solve
Ans: x = 0 or x = 2/3


5p^3q^-2/(15p^-4q) Find reciprocals for neg. exp.

= (5p^3p^4)/(15 q q^2) add exponents for letters

= 1p^7/(3q^3) reduce as a fraction

2SQRT44 + 5SQRT11 break down 44 = 4*11
2*SQRT4*SQRT11 +5SQRT11
2*2*SQRT11 + 5SQRT11 now combine like terms
= 9SQRT 11

SQRT(4X + 32) + 1 = 1 Subtract one
SQRT(4X + 32) = 0 Now square both sides
4X + 32 = 0 Finally, solve for X
x= -8