Maths Help ~ (1 Viewer)

[xXmuffin0manXx]

Solve:

i) x^4 + 5x^2 - 36 = 0

ii) 2x + 9rootx = 5

iii) 1/t^2 + 8/t + 15 = 0



thanks..
working would be appreciated !

 

[tommykins]

Solve:

i) x^4 + 5x^2 - 36 = 0

ii) 2x + 9rootx = 5

iii) 1/t^2 + 8/t + 15 = 0



thanks..
working would be appreciated !

i) let m = x²
m² +5m - 36 = 0
(m+9)(m-4) = 0
m = -9 or 4
x² = -9 or 4, .'. 4 is only soln.
x² = 4
x = ±2

ii) 2x + 9rootx = 5
2x + 9sqrtx - 5 = 0
Let m = sqrtx
2m² + 9m - 5 = 0
(2m-1)(m+5) = 0
m = 1/2 or -5
sqrtx = 1/2 or -5, 1/2 only soln.
sqrtx = 1/2
x = 1/4

iii) 1/t^2 + 8/t + 15 = 0

1 + 8t + 15t² = 0
(5t+1)(3t+1) = 0

thus t = -1/5 and -1/3

 

[ItsColinoOo0o]

haha is that like ur maths hw?

 

[Dr. Zoidberg]

Q1.

P(2)=0

Therefore, (x-2) is a factor

P(-2)=0

Therefore (x+2) is a factor

Therefore, (x^2-4) is a factor

Use long division

............x^2+9
..........._________________
x^2-4 |x^4 + 5x^2 - 36
...........x^4 - 4x^2
......................9x^2 - 36
......................9x^2 - 36
...................................0

Therefore, (x^2-4)(x^2+9)=0 <= original polynomial in factored form

Therefore, x= 2 or -2

Lol, tommykins way is better

 

[xXmuffin0manXx]

maybe

 

[tommykins]

Q1.

P(2)=0

Therefore, (x-2) is a factor

P(-2)=0

Therefore (x+2) is a factor

Therefore, (x^2-4) is a factor

Use long division

x^2+9
_________________
x^2-4 |x^4 + 5x^2 - 36
...........x^4 - 4x^2
......................9x^2 - 36
......................9x^2 - 36
...................................0

Therefore, (x^2-4)(x^2+9)=0 <= original polynomial in factored form

Therefore, x= 2 or -2

Look at my solution for a quicker way.