Question 197024: thank you in advance for any help with these:
1. 15x^4 - 11x^2 + 2 = 0 use radical as needed
2. x^2 + 6x - 4 = 0 simply, use radical if needed
3. 4x(x - 1)- 5x(x)= 3 solve for x
4. 8x^2 - 6x - 2 = 0 state value of discriminant; one or two real soulution? or two imaginary solution?
thank you for any help you can provide...
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. 15x^4 - 11x^2 + 2 = 0 use radical as needed
15x^4 -6x^2-5x^2+2 = 0
3x^2(5x^2-2) - (5x^2-2) = 0
(5x^2-2)(3x^2-1) = 0
---
5x^2-2 = 0 or 3x^2-1 = 0
x^2 = 2/5 or x^2 = 1/3
x = +/-sqrt(2/5) or x = +/-sqrt(1/3)
=================================================
2. x^2 + 6x - 4 = 0 simply, use radical if needed
x = [-6 +-sqrt(36-4*1*-4)]/2
x = [-6 +-sqrt(52)]/2
x = [-3 +- sqrt(13)]
=================================================
3. 4x(x - 1)- 5x(x)= 3 solve for x
4x^2-4 -5x^2 - 3 = 0
-x^2 - 7 = 0
x^2 = -7
x = +/-isqrt(7)
==================================================
4. 8x^2 - 6x - 2 = 0 state value o
4x^2 - 3x - 1 = 0
4x^2 - 4x + x -1 = 0
4x(x-1) + (x-1) = 0
(x-1)(4x+1) = 0
x = 1 or x = -1/4
==================================================
f discriminant; one or two real soulution? or two imaginary solution?
If discriminant = 0 ; two equal Real solutions
If discriminant > 0 ; two unequal Real solutions
If discriminant < 0 ; two unequal complex solutions
==========================================================
Cheers,
Stan H.
|
|
|
