Need help with Polynomials?

Question

1. 4x2 - 16 = 0
2. 2x2 = 16
3. x2 + 5 = 0
4. x2 + 5x - 6 = 0
5. x2 - 25 = 0
6. 3x2 = -9x
7. 2x2 - 4x + 1 = 0
8. 3x2 + 2x = 7
9. x2 + 8x + 9 = -7


1 is x=2
2 is x=2sqrt2
3 is x=sqrt5
... or am I doing this all wrong? Help is very much appreciated.

Answer 1

Answers:
1. x= 2,-2
2. x= 2sqrt2, -2sqrt2
3.x= isqrt5, -isqrt5 i is sqrt-1, an imaginary number
4.( x+6)(x-1)=0 => x= -6, 1
5. x = +5, -5
6. x2+3x=0 =>x(x+3)=0 => x=0, -3
7. 2 (x2-2x+1)=1 => (x-1)^2=1/2 => x= 1+ 1/sqrt2, 1-1/sqrt2
8. 3x2+2x-7=0 => x= (-2 + sqrt88)/6, (-2 - sqrt88)/6
9. x2+8x+16=0 => x=-4

For 1,2,3: keep in mind to take the negative of answer too.
For 7 and 8 use quadratic formula
For others, simply factorize.

Hope this helps!

Answer 2

1 is partially correct
4x^2-16=0
4x^2=16
x^2=4
x=+-2

Because you are square rooting a lot of people forget that the negative answer.

2 is partially correct
2x^2=16
x^2=8
x=+-sqrt(8)
x=+-2*sqrt(2)

3 is x=+-sqrt(-5) seeing you change the positive 5 to a negative side when taking is over to the other side
This can also be writen as x=sqrt(5)i where i =sqrt(-1) and is a complex number. If you have no covered complex numbers then leave answers in the form x=sqrt(-5)

Looks like you know how to rearrange the polynomials in order to find the x values, just need to watch out for the signs and the negative answer when square rooting

Answer 3

reported unquestionably, polynomial applications are applications with x as an enter variable, made up of countless words, each and each term is made up of two factors, the 1st being a actual quantity coefficient, and the 2d being x raised to a pair non-destructive integer power eg:f(x) = 4x3 + 8x2 + 2x + 3 g(x) = 2.5x5 + 5.2x2 + 7 h(x) = 3x2 the cost of n could be an nonnegative integer. that's, that's going to be total quantity; that's comparable to 0 or a favorable integer. The coefficients, as they're referred to as, are an, an-a million, ..., a1, a0. those are actual numbers. The degree of the polynomial function is the optimum fee for n the place an isn't equivalent to 0.

Answer 4

4x² - 16

(2x - 4)(2x + 4)
- - - - - - - - - - - -

2x² = 16

2x² / 2 = 16 / 2

x² = 8

√x² = ± √8

√x² = ± √4 √2

x = ± 2 √2

- - - - - - - - - - -

x² + 5 = 0

x² + 5 - 5 = 0 - 5

x² = - 5

√x² = ± √- 5

x = ± √- 5 no real number solution

- - - - - -

x² + 5x - 6

(x + 6)(x - 1)

- - - - - - - - - -

x² - 25

(x - 5)(x + 5)

- - - - - - - - - -

3x² = - 9

3x² / 3 = - 9 / 3

x² = - 9 / 3

x² = - 3

√x² = ± √- 3

x = ± √- 3. .no real number solution

- - - - - - - - - - -

x² + 8x + 9 = - 7

x² + 8x + 9 + 7 = - 7 + 7

x² + 8x + 16 = 0

(x + 4)(x + 4)

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

Answer 5

You have often forgotten the negative root! I give two examples

1 ) x=+2 and x=-2
2) x = 2 sqrt2 and its opposite -2 sqrt 2
3) x= sqrt 5 and its opposite
4) x=1 and x=-6
5) x=5 and X=-5
6) x= sqrt3 and -sqrt3
7)x=1+sqrt2 and 1-sqrt2
8)x =(-2+sqrt88)/6 and x =-(2-sqrt88)/6
9) x^2+8x+2=0 x= -4 +sqrt2 and -4-sqrt2

Answer 6

1) 4x^2-16 = 0

4x^2 = 16

x^2 = 4

x = ±2 ...don't forget ±

2) 2x^2 = 16

x^2 = 8

x = √8

x = √[(4)(2)]

x = ±2√(2) ...don't forget ±

3) x^2 +5 = 0

x^2 = -5

x = √(-5)

x = ±i√(5)

4) x^2 + 5x - 6 = 0

(x+6)(x-1) = 0

x + 6 = 0

x = -6

x - 1 = 0

x = 1

5) x^2 - 25 = 0

x^2 = 25

x = ±5

6) 3x^2 = -9x

x = -3

7) 2x^2 - 4x + 1 = 0

a = 2, b = -4, c = 1

x = {-b±√(b^2-4ac)}/(2a)

x = {4±√(-4^2)-4(2)(1)}/(2)(2)

x = {4±√(8)}/4

x = {4±2√(2)}/4

x = 1±(√2)/2

8) 3x^2 + 2x = 7

3x^2 + 2x - 7 = 0

a = 3, b = 2, c = -7

x = {-b±√(b^2-4ac)}/(2a)

x = {2±√(2^2)-4(3)(-7)}/(2)(3)

x = {2±√(88)}/6

x = {2±2√(22)}/6

x = {1±√22}/3

9) x^2 + 8x + 9 = -7

x^2 + 8x +16 = 0

(x+4)^2 = 0

x+4=0

x = -4

Answer 7

in the first question
x^2=16/4
x^2=4
here x can be +2 or -2
in the second question
x^2=8
x=2*2^2 (correct)
in the third,
x=-5^2 that is imaginary
so your is wrong

Answer 8

yes, what do you mean 4x2-16=0 ?? (four times x times 2 is what you are doing??)

Answer 9

i dont know if this is a polynomial i can't see some power of....or raise to..........but if it is not a polynomial u can use PMDAS to calculate it just P for parentheses M for multiplication D for division A for Addition and S for subtraction!!! i wish i can help you w/ that!!

Answer 10

Correct your question pls because it makes no sense as it is written now!