Solve the equation

[Math Tutor]

(x + 1)(2x + 3)(2x + 5)(x + 3) = 945

[MM]

This is [tex](2x+2)(2x+3)(2x+5)(2x+6)=3780[/tex]. Write [tex]t=2x+4[/tex] and we get [tex]t^4-5t^2-3776=0[/tex] which is quadratic equation. The roots are [tex]x=2[/tex] and [tex]x=-6[/tex].

[Guest]

hello everybody
I have a hard PDE to slove. is anybody to help me?
df/dt + udf/dx= sin(kx-wt)dg/du vlasov equation
can i uplode pic?

[Math Tutor]

You can upload an image but you have to register first.

[ranjande007]

sin3[tex]\alpha[/tex]/cos2\alpha < 0 in which value of \alpha it will be possible

[Mathmaven53]

Write equation as
(x+1)(x+3)(2x + 3)(2x + 5) = 945

Multiply first two factors and second two factors

(x^2 + 4 x + 3)(4 x^2 + 16 x +15) = 945

(x^2 + 4 x + 3)(4(x^2 + 4 x) + 15) = 945

Let u = x^2 + 4 x

(u + 3)(4 u + 15) = 945

Expand as a quadratic in u and solve for u
For each u solve
x^2 + 4 x = u
for x

[leesajohnson]

Sorry this time I'm not able to understand the question.

[Guest]

Which question.??

[Guest]

YES, mathmaven53, this solution works........... a simple solution to an awkward problem.....
......................................................
Write equation as
(x+1)(x+3)(2x + 3)(2x + 5) = 945

Multiply first two factors and second two factors

(x^2 + 4 x + 3)(4 x^2 + 16 x +15) = 945

(x^2 + 4 x + 3)(4(x^2 + 4 x) + 15) = 945

Let u = x^2 + 4 x

(u + 3)(4 u + 15) = 945

Expand as a quadratic in u and solve for u
For each u solve
x^2 + 4 x = u
for x
..................................

and simpler than MM's solution, but gets same answer.....so must be right...???

Just looking at it.......

(x+1)(x+3)(2x + 3)(2x + 5) = y
y = 945

the function is Zero for x=-1, x=-3. x= -1.5, x=-2.5
and so zero from x = -1 to x= -3 and positive outside that...U shaped.....???
Solution is where cuts the line y = 945....????