Algebra II help plz??=D?

Question

how do you solve for x?
1) x^2 -4x+20=0
2) x^2 -25 =0
3)2x^2-x-3=0
how do you factor these?
y=x^2 + x -6

y=x^2 -7x

y= x^2 + 20x + 100

y= 3x^2 -x-24

square root of -9 + square root of -16

5i times 6i

Answer 1

1) x^2-4x+20=0
have to use quadratic formula

2) x^2 -25 =0
(x+5)(x-5)=0 difference of squares, x= 5 or -5

3) 2x^2-x-3=0
(2x-3)(x+1)=0 x= 3/2 or -1

y=x^2 + x -6
y=(x-2)(x+3)

y=x^2-7x
y=x(x-7) always take out common factor first

y= x^2 + 20x + 100
y=(x+10)^2

y= 3x^2 -x-24
y=(3x-3)(x+8)

sqrt(-9) +sqrt(-16)
3i + 4i = 7i

5i * 6i = 30i^2 i = sqrt(-1) so i^2 = -1
so 30 * -1 = -30

Answer 2

1. x2 - 4x + 20 = 0

*subtract 20 from both sides

x2 - 4x + 20 - 20 = 0 - 20
x2 - 4x = -20

*find a common factor between x^2 and 4x. Break break each set apart: x^2 is (x)(x) and (4)(x) stays as it is. Common factor for both sets are one "x." Bring "x" to the front and multiply by the remaining parts:

x(x - 4) = - 20
you have two variables to solve for. Set each "x" to equal -20

x = -20 and

x - 4 = - 20
x - 4 + 4 = - 20 + 4
x = -16

2. x^2 - 25 = 0
x^2 - 25 + 25 = 0 + 25
x^2 = 25

*get rid of the exponent by finding the square root of both sides.

square root of x^2 is "x." Find the square root of a positive 25 and negative 25

x = -5, 5

3. 2x^2 - x - 3 = 0
2x^2 - x - 3 + 3 = 0 + 3
2x^2 - x = 3

*factor 2x^2 which is 2(x)(x) and "x" stays the same.
common factor: “x”

x(2x - 1) = 3
x = 3 and

2x -1 = 3
2x -1 + 1 = 3 + 1
2x = 4

2x/2 = 4/2
x = 2

4. y = x^2 – x + 6

Use a big X method: write a big “X” on paper. Take the first coefficient (1) and multiply it by the last number (6) which equals 6. Write 6 at the top of the big “X.”

Take the middle coefficient (-1) and write in on the bottom of the big “X.”

Find two numbers that give you (6) when multiplied together and (–1) when added or subtracted from each other.

You can use: 3, -2 and y = (x +3)(x – 2)

5. y = x^2 – 7x

you break them apart: (x)(x) - 7(x)
common factor: one “x” and brinf “x” to the front and multiply it by the remanining parts.

y = x(x – 7)

6. y = x^2 + 20x + 100
use the big “X” method

the two numbers to give you 100 when multiplied are 10 and 10. They give you 20 when added together.

Y = (x + 10) (x + 10)

7. 3x^2 -x-24
use the big “X” method

top number is 3(24) = 72
bottom number is middle coefficient: -1

two numbers that give me 72 when multiplied: 6(7) and give me –1 when (–7 + 6)

y = (x – 7) (x + 6)

8. square root of –9 + the sqaure root of –16

its impossible to find the square root of a negative number so, the negative sign becomes an imaginary number know as: “i”

Bring out an “i” beside the square root of 9 and bring another “i” out beside the square root of 16. Find the square root of 9 which is 3 and the square root of 16 is 4.

We have 3i + 4i and add them together = 7i

9. (5i)(6i) = 30i^2

and i^2 = -1 then replace i^2 with (-1)

30(-1) = -30

Answer 3

1) x^2 - 4x + 20 = 0 ; (x-5)(x+4) = 0
except it comes out to negative 20 so I don't know if you typed it wrong or maybe my brain isn't working... either way: x= 5 & -4
2) add 25 to both sides -> x^2 = 25 , then square root. So x = 5
3) (2x-3)(x+1) = 0 ; x = -1 & 1.5

y=x^2 + x - 6
y = (x-2)(x+3)

y = x^2 + 20x + 100
y = (x+10) (x+10)

y= 3x^2 - x - 24
y = (3x - 8)(x + 3)

square root of -9 is like 3i or something? So then for -16 it would be 4i. 3i + 4i = 7i

5i x 6i = -30 (because i^2 = -1

Answer 4

to solve the first group of problems, you need to be able to factor or use the quadratic formula. Factoring is by far the best way.
y = ( x + 3)( x - 2)
y = x( x - 7)
y = ( x + 10 )^2
y = (x - 3)( 3x + 8)
3i + 4i = 7i
30i^2 = -30

Answer 5

in case you pass Algebra 2, then you certainly ought to circulate to Precalculus. Math diagnosis is two jumps. Going over a e book is diverse than doing a years nicely worth of homework. I advise you're taking the type in simple terms for adventure on the grounds that Algebra 2 is the commencing up for something of the mathematics instructions you will take. yet once you relatively drilled your self, then choose for Precal, no longer math diagnosis.

Answer 6

Ahh yes. Algebra two. So nice. Look at your own notes and maybe you can figure it out cause I sure can't. XD