algebra II problem?

Question

4y^2+25=20y I'm just hoping someone can help me so with this one example so that I can finish the rest of my h rest of my homework. Thanks in advance

Answer 1

Convert to standard form

4x² + 25 = 20y

4x² - 20y + 25 = 20y - 20y

4x² - 20y + 25 = 0. . . .This is standard form

The middle term is - 20y

Find the sum of the middle term

Multiply the first term 4 by the last term 25 equals 100 and factor

Factors of 100 =

1 x 100
2 x 50
4 x 25
5 x 20
10 x 10. . .<=. .use these factors

- 10 and - 10 satisfy the sum of the middle term

insert - 10y and - 10y into the euation

4x² - 20y + 25 = 0

Group factor

4x² - 10y - 10y + 25

2y(2y - 5) - 5(2y -5)

(2y - 5)(2y - 5)

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roots

2y - 5 = 0

2y - 5 + 5 = 0 + 5

2y = 5

2y / 2 = 5 / 2

y = 5/2

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Answer 2

first subtract 20y from both sides to get 4y^2 - 20y + 25 = 0
knowing 25 is a perfect square makes life easy, because the only factor pairs are 25-1 and 5-5. And since you have -20y + 25, you know that both factors will subtract the whole #: (a-b)(c-d).

So you have either (2y-5)(2y-5) = 0 OR (y-5)(4y-5) = 0. FOIL them and see what you get. Failing that, use the quadratic formula.

Answer 3

First: set the equation to equal "0" - subtract "20y" from both sides (when you move a term to the opposite side, always use the opposite sign).

4y^2 + 25 - 20y = 20y -20y
4y^2 + 25 - 20y = 0

*Make sure the expression is written in alpha descending order.

4y^2 - 20y + 25 = 0

Sec: factor... multiply the 1st & 3rd term to get 100. find two numbers that give you 100 when multiplied & -20 (2nd term) when added/subtracted. the numbers are (-10 & -10).

*Rewrite the expression with the new middle terms.

4y^2 - 10y - 10y + 25 = 0

*With 4 terms, group "like" terms & factor both sets of parenthesis.

(4y^2 - 10y) - (10y + 25) = 0
2y(2y - 5) - 5(2y - 5) = 0
(2y-5)(2y-5) = 0
(2y - 5)^2 = 0

Third: solve the "y" variable - eliminate the exponent by finding the square root of both sides.

V`(2y - 5)^2 = V`0
2y - 5 = 0

*Add 5 to both sides....

2y - 5 + 5 = 0 + 5
2y = 0 + 5
2y = 5

*ISolate "y" by dividing both sides by 2.

2y/2 = 5/2
y = 5/2

Answer 4

4y^2-20y+25 =0
y=((20 +-sqrt(400-16*25))/8 = 20/8= 5/2 double root
4y^2-20y+25 = (2y-5)^2

Answer 5

To solve this equation isolate everything on one side with zero on the other, so that you have

4y^2 - 20y + 25 = 0.

Now factor, or use the quadratic formula. In this case it factors into:

(2y-5)(2y-5)

or (2y-5)^2. This follows the identity that (ax-b)^2 = ax^2 - 2abx + b^2.

If you want the solution you'd solve 2y-5 = 0, or y = 5/2

Answer 6

4y^2+25=20y

4y^2 - 20y + 25 = 0

( 2y - 5) ( 2y - 5) = 0

y = 5/2

Answer 7

Rewrite the equation in standard form:
4y^2 - 20y + 25 = 0

Now you can plunk it into the quadratic formula and solve.

Alternatively, this one is fairly easy to solve by factoring

Answer 8

You can write it as (2y-5)^2=0
so it must be that 2y-5=0
therefore, y=2 and 1/2

Answer 9

Firstly, you have to factorise it (do the working yourself)

Answer : (2y-5)(2y-5)=0. You MUST show your working.

Then find out the value for y

y=5/2

PS Do not copy me and hand this in as your homework. You are only cheating yourself.

Answer 10

y^2-5y+25/4 = 0
(y-5/2)^2 = 0
y= 5/2