SOLUTION: Could someone help me? Solve the following equations. Leave radicals in radical form(do not calculate there approximate values). a. 4x^2+3x-10=0 b. 2x^2=3x c. x^3-4x^2-

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Could someone help me? Solve the following equations. Leave radicals in radical form(do not calculate there approximate values). a. 4x^2+3x-10=0 b. 2x^2=3x c. x^3-4x^2-      Log On


   

Question 60886: Could someone help me?
Solve the following equations. Leave radicals in radical form(do not calculate there approximate values).
a. 4x^2+3x-10=0
b. 2x^2=3x
c. x^3-4x^2-x+4=0
d. x^4-10x^2+9=0
e. x^3-3x^2+x-3=0
f. 3x^2-5x+1=0
I also need to know how to put radicals signs on here.
Thank you for your help!!
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!

Solve the following equations. Leave radicals in radical form(do not calculate there approximate values).
a. 4x%5E2%2B3x-10=0
Replace the middle term with two numbers that multiply to give you 4x^2*-10=-40x^2, but add to give you 3x. 8x*-5x=-40x^2, 8x+-5x=3x.
:4x%5E2%2B8x-5x-10=0 Factor by grouping, group the first two terms and the last two.
%284x%5E2%2B8x%29%2B%28-5x-10%29=0 Factor the GCF out of each parentheses
4x%28x%2B2%29-5%28x%2B2%29=0 Factor out x+2
%284x-5%29%28x%2B2%29=0 Set each parentheses equal to 0 and solve for x.
4x-5=0 and x+2=0
4x=5 and x=-2
4x/4=5/4 and x=-2
x=5/4 and x=-2
:
b. 2x^2=3x Set equal to 0
2x%5E2-3x=0 Factor out the GCF.
x(2x-3)=0
x=0 and 2x-3=0
x=0 and 2x=3
x=0 and 2x/2=3/2
x=0 and x=3/2
:
c. x^3-4x^2-x+4=0 Factor by grouping
%28x%5E3-4x%5E2%29%2B%28-x%2B4%29=0
x%5E2%28x-4%29-1%28x-4%29=0
%28x%5E2-1%29%28x-4%29=0 x^2-1 is the difference of squares
(x+1)(x-1)(x-4)=0
x+1=0 and x-1=0 and x-4=0
x=-1 and x=1 and x=4
:
d. x^4-10x^2+9=0
%28x%5E2-9%29%28x%5E2-1%29=0
(x+3)(x-3)(x+1)(x-1)=0
x+3=0 and x-3=0 and x+1=0 and x-1=0
x=-3 and x=3 and x=-1 and x=1
:
e. x^3-3x^2+x-3=0
%28x%5E3-3x%5E2%29%2B%28x-3%29=0
x%5E2%28x-3%29%2B1%28x-3%29=0
%28x%5E2%2B1%29%28x-3%29=0
(x^2+1 is prime, so if you need only real numbers you can ignore it, otherwise it's solution is +/-i)
x-3=0
x=3
:
f. 3x^2-5x+1=0
a=3 b=-5 and c=1
The quadratic formula is highlight%28x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29%29
x=%28-%28-5%29%2B-sqrt%28%28-5%29%5E2-4%283%29%281%29%29%29%2F%282%283%29%29
x=%285%2B-sqrt%2825-12%29%29%2F6
x=%285%2B-sqrt%2813%29%29%2F6
:
An abreviation for radical is "sqrt(whatever you want)" like the square root of x+5 is sqrt(x+5)
:
Happy Calculating!!!