Math Help Binomials?

Question

heres the problem. i have to set this quadratic inequality to zero...here it is: x*squared* -5x-4<-9....which simplifies to x*squared* -5x +5x< 0. then i need to find the binomial that makes x*squared* -5x +5x. HELP!!!! so far people at my school have told me that its impossible...sos.

Answer 1

x^2 - 5x + 5x = 0
this is factorable but not by whole numbers

so use the quadratic formula

a=1
b=-5
c=5

-b + or - sqrt of b^2 - 4ac all over 2a

5+ or - sqrt of 25-20 all over 2
so ure final answer is x= 5+ or - the sqrt of 5 all over 2

Answer 2

I'm going to assume that the problem is:
"Solve x^2-5x-4<-9"
x^2-5x-4+9<0
x^2-5x+5<0
x^2 -5x+....?<-5....+?
x^2-5x+25/4<-5+25/4
When completing the square, the term you need is 1/2the coefficient if the x term, squared. Be aware that the coefficient of the x^2 term must be +1 before you do this.
(x-5/2)^2<-5+25/4,=-20/4+25/4, =5/4
Take the square root of both sides
(x-5/2)< +/- rt(5)/2
x<5/2 +rt(5)/2, or x<5/2-rt(5)/2
x<[5+rt(5)]/2 or x<[5-rt(5)]/2

You can convert this stuff to hard numbers
rt(5)=2.236
Therefore x<[5+2.236]/2,=<[7.236]/2,=<3.618
Or, x<[5-2.236]/2,=<[2.764]/2,=<1.382

We have to check these answers.
For x=<3.618 let's use x=3
x^2-5x+5=(3)^2-5(3)+5,=9-15+5, =-1. It works,-1 is
certainly less than zero.
For x<1.382, let's use 1
x^2-5x+5=(1)^2-5(1)+5,=1-5=5, =1. This doesn't work
because 1 is greater than 0, not less.
In fact, picking ANY number less than x<1.382, you get a positive result. You can see that any negative number plugged into your equation will make
x^2-5x+5 have nothing but positive terms.

Similarly, any number greater than 3.618 gives
a positive result.
Well, what about x=1.382 exactly?
x^2-5x+5 becomes (1.382)^2-5(1.382)+5,
=1.91-6.91+5, =0
We can draw only one conclusion from all this.
x has to be greater than 1.382 and at the same time
smaller than 3.618 to satisfy x^2 -5x+5<0
We write this as 1.382
All the above could be useless if I have mis-interpreted the problem. Anyway, good luck!

Answer 3

x^2 - 5x - 4 < -9 (add '9' to both sides to preserve inequality)

x^2 - 5x - 4 + 9 < -9 + 9
x^2 - 5x + 5 < 0

Your statement x^2 - 5x + 5x <0 is INCORRECT!!!!

Then use 'Completing the Square' rule, which is:-

x = {- b +/- sq rt [b^2 - 4ac]} / 2a
Where a,b,c follow the convention ax^2 + bx + c
In this case a = 1, b = -5 & c = 5
So!!!

x = {--5 +/- sq rt[(-5)^2 - 4(1)(5)]} / 2(1)
x = {5 +/- sq rt[25 - 20]} / 2
x = {5 +/- sqrt(5)}/2
x = {5 +/-2.236} / 2
x = {5+ 2.236} / 2 = 3.618

& x = {5-2.236}/2 = 1.382

Hence the values of 'x' which satisfy the inequality
x^2 - 5x +5<0 are 1.382 < x < 3.618

NB Values of 'x' outside this range will make the inequality greater than zero, hence the inequality will not be satisfied.

Answer 4

hello
lets formulate your case in steps

consider the quad eqn. to be

x^2 -5x+5 = 0 ,

a = 5/2+sqrt(25-20)/2 = 2.5 + 1.12
b = 5/2-sqrt(25-20)/2 = 2.5 - 1.12

so the period between a & b , is the one that satisfies
your inequality

you may mail me , for any further questions

are both roots to our case

then we can re-formulate it as follows
(x-a)(x-b) = 0