Consider this system of equations:

Mathematics

2 answers:

kumpel [21]3 years ago

4 0

for which value of k does the system have no real number solutions? = -2

for which value of k does the system have one real number solution? = -0.25

for which value of k does the system have two real number solutions? = 2

natali 33 [55]3 years ago

3 0

We know that

In a quadratic equation of the form
ax²+bx+c=0
the discriminant is equal to
(b²-4ac)
if (b²-4ac) > 0 then the system has two real number solutions
if (b²-4ac) =0 then the system has one real number solution
if (b²-4ac) < 0 then the system has no real number solutions

we have
y = x² and y = x + k

to resolve the system equate both equations
x²=x+k-----------> x²-x-k=0
a=1
b=-1
c=-k
find the disciminant
(b²-4ac)=(-1)²-4*(1)*(-k)-----> 1+4k

Part 1)For which value of k does the system have no real number solutions?
we know that
if (b²-4ac) < 0 then the system has no real number solutions
so
(1+4k) < 0-------> 4k < -1
k< -0.25

the answer part 1) is
for k < -0.25 the system has no real number solutions
example
if k=-2
-2 < -0.25-----> is ok
the system has no real number solutions
see the attached figure N 1

Part 2) For which value of k does the system have one real number solution?
we know that
if (b²-4ac) =0 then the system has one real number solution
so
(1+4k) = 0-------> k=-0.25

the answer part 2) is
for k=-0.25 the system has one real number solution
see the attached figure N 2

Part 3) For which value of k does the system have two real number solutions?
we know that
if (b²-4ac) > 0 then the system has two real number solutions
so
(1+4k) > 0------> 4k > -1
k > -0.25

the answer Part 3) is
for k > -0.25 the system has two real number solutions
example
if k=2
2 > -0.25-----> is ok
the system has two real number solutions
see the attached figure N 3

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Find f(-1) (inverse) if y=4x+2/3x-1 and x is not equal with 1/3

Setler [38]

Before we find $f^{-1}$ , let's think for a minute about what $f$ does, and what it means to find the inverse of a function. A function $f$ essentially takes a value $x$ from the domain and maps it to another value $y$ in that function's range according to a set of rules. Here, those rules are defined by the formula

$y= \frac{4x+2}{3x-1} , x \neq \frac{1}{3}$

All the inverse does is swap the domain and range of the function. Now, instead of trying to map $x$ to $y$ , we're trying to find a set of rules that'll map $y$ back to $x$ . To find those rules, all we have to do is solve the above equation for $x$ .

First, we'll multiply both sides of the equation by $3x-1$ to get it out of the denominator:

$(3x-1)y=( \frac{4x+2}{3x-1})(3x-1)$

Cancelling on the right side and distributing on the left, we get:

$3xy-y=4x+2$

Next, we collect all of our x terms on one side, and all of our non-x terms on the other:

$(3xy-y)+y=(4x+2)+y\\(3xy)-4x=(4x+2+y)-4x\\3xy-4x=2+y$

Let's rearrange the right side and factor out an x on the left:

$x(3y-4)=y+2$

And finally, we divide both sides by $3y-4$ to obtain our answer:

$[x(3y-4)]/(3y-4)=(y+2)/(3y-4)\\\\x= \frac{y+2}{3y-4}$

This equation gives us the "rules" for mapping any given y in the range back to an x in the domain. If we swap the domain and the range, we can define our function

$f^{-1}(x)= \frac{x+2}{3x-4}$