How to find a constant in this quadratic equation?

[Helly123]

Homework Statement

Homework Equations

for equation which has 2 different solutions, D >0

The Attempt at a Solution

(1)[/B] D > 0
b^2 - 4ac > 0
3 - 4root2.k > 0
k < 3 / ( 4root 2 )
k < ( 3 root 2 ) /8

has solution of sin tetha and cos tetha
sin 0 = 0, cos 0 = 1.
when x = 0, and x = 1 --> satisfy the rule?

when x = 0,
k = 0...
when x = 1,
k = 1/(root 3 - root 2)

can anyone help me to finish it?

(2) (x^3 + a/x^2)^5 = -270
something to the power 5 would be -270, means that something is negative
the factor of 270 = 3 x 3 x 3 x 5 x 2
can anyone help me to finish it?

(3) f( g(x) ) = 3gx + 1 / (2gx + 1)
x = 3gx + 1 / (2gx + 1)

and finally I get
2x^2 - 3x + 3 = -2px^2 + 3px + 3
(x1 + x2) on left equation = (x1 + x2) on right ?
-b/a = -b/a
3/2 = -3p/-2p

and x1.x2 on left equation = x1.x2 on right ?
c/a = c/a
3/2 = 3/-2p
p = -1
Is it right way?

 

[Alex Cros]

Hello! please use latex next time thanks!

1) Let x₁=sin(θ) and x₂=cos(θ), then by substitution into the original equation you can find the value of k since you are given the interval of θ.

2) I am not sure what is meant by the "constant term" but I will asume is (x^3+ \frac{a}{x^2})^5=-270.
Then by -(x^3+ \frac{a}{x^2})=(270)^5 and therefore by multipling by x^2 we have (x^5+ a)=-x^2(270)^5 rearranging factors x^2(x^3+(270)^5)=-a shows that a depends on a so there is not a definite answer but an infinite of them... give x a value and you get a value of a and viceversa.

3) By substitution let x in f(x) be x=\frac{px+1}{2x-3} and then impose f(x)=x. As a hint, in the interval x≠-1/2,3/2 you would probably end up having an expresion of the type 1/0...

Hope to help.

 

[Helly123]

I still not find number 2.. can anyone help?

 

[FactChecker]

Expand (x³ +a/x²)⁵ using the binomial theorem and look at the constant term.

 

[Helly123]

using binomial...
$$X^{15} - 5x^{12} \frac {a} {x^2} + 10x^{9} \frac {a^{2}} {x^{4}} - 10x^{6} \frac {a^{3}} {x^{6}} + 5x^{3} \frac {a^{4}} {x^{8}} - \frac {a^{5}} {x^{10}} = -270 $$

now I don't know..
can anyone help?

 

[FactChecker]

Double check your binomial expansion. It is wrong. Then simplify the terms. Combine the x factors in each term by combining the exponents.

 

[SammyS]

using binomial... $$x^{15} - 5x^{12} \frac {a} {x^2} + 10x^{9} \frac {a^{2}} {x^{4}} - 10x^{6} \frac {a^{3}} {x^{6}} + 5x^{3} \frac {a^{4}} {x^{8}} - \frac {a^{5}} {x^{10}} $$
now I don't know..
can anyone help?

Why are the signs alternating?

Which term is the constant term ?

 

[haruspex]

what is meant by the "constant term"

That term in the binomial expansion which is independent of x.

I will asume is (x^3+ \frac{a}{x^2})^5=-270

No, only the constant term equals -270, not the whole function.

 

[Helly123]

That term in the binomial expansion which is independent of x.

No, only the constant term equals -270, not the whole function.

I think it doesn't have constant term. The last term is $$\frac { a^5} {x^{10}}$$

 

[haruspex]

I think it doesn't have constant term. The last term is $$\frac { a^5} {x^{10}}$$

The constant term need not be the last one.

 

[FactChecker]

I think it doesn't have constant term. The last term is $$\frac { a^5} {x^{10}}$$

Simplify each term by combining the exponents of x like xⁿ/x^m = x^n-m. One term will end up with x⁰ = 1. That is the constant term. I shouldn't help more than that on a homework problem.