SOLUTION: what are the approximate values of x that are solutions to f(x) = 0, where f(x) = -6x^2 + 9x + 5. I am not even sure how to start this. I see that it is in quadratic equation fo

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Question 857318: what are the approximate values of x that are solutions to f(x) = 0, where
f(x) = -6x^2 + 9x + 5. I am not even sure how to start this. I see that it is in quadratic equation format, and my thinking is that f(x)=0 means that 0=-6x^2 + 9x +5, then I think I am supposed to multiply the -6 by the 5 and find the common factor of -30 that equals 9, but nothing does so I'm confused...Am I doing it all wrong?
Found 2 solutions by Fombitz, Theo:
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
You can graph it and get approximate values.


Looks like about and would be approximate values.
For exact values you can complete the square or use the quadratic formula.
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
the equation is quadratic and it is equal to:
-6x^2 + 9x + 5
you do set the equation equal to 0 to get:
-6x^2 + 9x + 5 = 0
at this point in time you would want to factor out the greatest common factor, if there is one.
there doesn't appear to be one, so you go to the next step.
the factoring method is always easier when the x^2 term is positive, so you would want to multiply both sides of the equation by -1 to get:
(-1) * (6x^2 - 9x - 5) = 0
you would now want to factor 6x^2 - 9x - 5, if, in fact, it can be factored using the factoring method.
if it can't, then you would have to resort to the quadratic formula.
we'll try to factor it using the factoring method first.
start with:
6x^2 - 9x - 5 = 0
this is in standard form of ax^2 + bx + c = 0 where:
a = 6
b = -9
c = -5
multiply the a term by the c term to get 6*-5 = -30.
the possible factors of 30 are:
1*30
2*15
3*10
5*6
it doesn't appear that any of these factors will work because none of them will be equal to 9 when you add them together or subtract them from each other.
1+/-30 = 31 or 29
2+/-15 = 17 or 13
3+/-10 = 13 or 7
5+/-6 = 11 or 1
i didn't take into account the signs at this point in time. that would come later if i found a combination that would work.
it does not appear that there is one, so we will need to resort to the quadratic formula.
that formula is x = (-b +/- sqrt(b^2-4ac))/(2a).
we will use the quadratic equation that we generated after we factored out the greatest common factor. that would have been before factoring out the -1, so the equation we will start with is going to be:
-6x^2 + 9x + 5 = 0
this equation is in the standard form of ax^2 + bx + c = 0, so we can determine what a, b, and c are.
a = -6
b = 9
c = 5
the quadratic formula is, once again:
x = (-b +/- sqrt(b^2-4ac))/(2a).
we replace a, b, and c in this formula with their respective values to get:
x = (-9 +/- sqrt(9^2 - 4*-6*5))/(2*-6) which is equal to:
x = (-9 +/- sqrt(201)) / -12
the fact that sqrt(201) is not a perfect square is the reason why you couldn't find the factor using the factoring method.
you have 2 roots to this equation.
they are:
x1 = (-9 + sqrt(201)) / -12 which is roughly equal to -.43.
x2 = (-9 - sqrt(201)) / -12 which is roughly equal to 1.9.
a graph of your equation is shown below:

you can see that the graph crosses the x-axis at x = roughly -.43 and 1.9.
this agrees with the results derived from the formula.

if you can't find the factors using the factoring method, it's usually because there are either no zero crossing points on the graph of the equation or because the roots are not rational.

rational roots are either integers or rational numbers.
rational numbers are composed of numerators and denominators that are both integers.

3/4 is a rational number.
sqrt(2) is not a rational number because there are no integers that can form a fraction that is exactly equal to sqrt(2).

if the graph doesn't cross the x-axis, it's because the roots are not real.
this means they contain an imaginary component.
the imaginary component is the square root of a negative number.

the discriminant of the quadratic formula will tell you whether you have rational roots, real roots, or imaginary roots.

if b^2 - 4ac is positive and a perfect square , then you have 2 real roots that are also rational.
if b^2 - 4ac is positive but not a perfect square, then you have 1 real roots that are not rational.
if b^2 - 4ac is 0 then you have 1 real root that is rational.
if b^2 - 4ac is negative than you have 0 real roots. in this case, the roots still exist, but not in the real numbering system and so you would say you have no roots because you are dealing in the real number system only.

if the equation has real roots, the roots are the x-axis crossing points.
if the equation has imaginary roots, you will not be able to see the roots on the graph because the graph will not cross the x-axis at those points.



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