Mixing
How can I use quadratic formula with an inequation like this $(x-3)(x^2-5x+5)le0$
$begingroup$
I know how to use the quadratic formula with an inequation.
How can I make use the quadratic formula?
$frac{-bpmsqrt{b^2-4ac}}{2a}$
With an inequation like this
$(x-3)(x^2-5x+5)le0$
quadratics
asked Jan 17 at 3:48
b.benb.ben
1254
$endgroup$
add a comment |
$begingroup$
I know how to use the quadratic formula with an inequation.
How can I make use the quadratic formula?
$frac{-bpmsqrt{b^2-4ac}}{2a}$
With an inequation like this
$(x-3)(x^2-5x+5)le0$
quadratics
asked Jan 17 at 3:48
b.benb.ben
1254
$endgroup$
4
$begingroup$
Find roots. Determine sign on each interval.
$endgroup$
– Brevan Ellefsen
Jan 17 at 3:51
$begingroup$
@BrevanEllefsen Simplest solution
$endgroup$
– b.ben
Jan 17 at 4:14
add a comment |
$begingroup$
I know how to use the quadratic formula with an inequation.
How can I make use the quadratic formula?
$frac{-bpmsqrt{b^2-4ac}}{2a}$
With an inequation like this
$(x-3)(x^2-5x+5)le0$
quadratics
asked Jan 17 at 3:48
b.benb.ben
1254
$endgroup$
I know how to use the quadratic formula with an inequation.
How can I make use the quadratic formula?
$frac{-bpmsqrt{b^2-4ac}}{2a}$
With an inequation like this
$(x-3)(x^2-5x+5)le0$
quadratics
quadratics
asked Jan 17 at 3:48
b.benb.ben
1254
asked Jan 17 at 3:48
b.benb.ben
1254
asked Jan 17 at 3:48
b.benb.ben
1254
asked Jan 17 at 3:48
b.benb.ben
1254
asked Jan 17 at 3:48
b.benb.ben
1254
1254
4
$begingroup$
Find roots. Determine sign on each interval.
$endgroup$
– Brevan Ellefsen
Jan 17 at 3:51
$begingroup$
@BrevanEllefsen Simplest solution
$endgroup$
– b.ben
Jan 17 at 4:14
add a comment |
4
$begingroup$
Find roots. Determine sign on each interval.
$endgroup$
– Brevan Ellefsen
Jan 17 at 3:51
$begingroup$
@BrevanEllefsen Simplest solution
$endgroup$
– b.ben
Jan 17 at 4:14
4
4
$begingroup$
Find roots. Determine sign on each interval.
$endgroup$
– Brevan Ellefsen
Jan 17 at 3:51
$begingroup$
Find roots. Determine sign on each interval.
$endgroup$
– Brevan Ellefsen
Jan 17 at 3:51
$begingroup$
@BrevanEllefsen Simplest solution
$endgroup$
– b.ben
Jan 17 at 4:14
$begingroup$
@BrevanEllefsen Simplest solution
$endgroup$
– b.ben
Jan 17 at 4:14
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Firstly, inequality. Please try to be accurate with terminology.
For the question: First, you want to find the critical values, that is, the $x$ where $(x-3)(x^2-5x+5)=0$
The first solution $x=3$ is trivial, then use the quadratic formula on $x^2-5x+5$ to find the other two.
I'll call them $x_1$ and $x_2$ (you should work them out) and note that $x_1<3<x_2$.
We now have four different intervals to test. Notice that between two adjacent zeroes of a function, the function is always positive or always negative. Hence if we pick an arbitrary $x$ in each of the intervals:
$$x<x_1$$
$$x_1<x<3$$
$$3<x<x_2$$
$$x>x_2$$
And see if the result is positive or negative when we put it in our function. If its negative, the interval it is in fits the $leq 0$ criterion.
answered Jan 17 at 4:04
Rhys HughesRhys Hughes
6,8601530
$endgroup$
add a comment |
$begingroup$
Using the formula we have $$f(x)=(x-a)(x-3)(x-b)le0$$
where $a=dfrac{5-sqrt5}2<3<b=dfrac{5+sqrt5}2$
For $f(x)=0,x=a,b$ or $3$
For $f(x)<0,$ odd number of multiplicands need to be $<0$
which will occur if $3<x<b$ or if $x<a$
answered Jan 17 at 4:03
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Can we guarantee it was $(x-a)(x-3)(x-b)le0$ not $(a-x)(x-3)(b-x)le0$ ?
$endgroup$
– b.ben
Jan 17 at 4:09
$begingroup$
@b.ben, $$(x-a)(x-b)=(a-x)(b-x)$$
$endgroup$
– lab bhattacharjee
Jan 17 at 4:11
$begingroup$
Hmm, $(a-x)(x-3)(x-b)le0$ answer will not the same as $(x-a)(x-3)(x-b)le0$ in inequation.
$endgroup$
– b.ben
Jan 17 at 4:13
add a comment |
$begingroup$
Let $f(x)=x^2-5x+5$, we see that here the coefficient of $x^2$ term is positive and $f(3)=-1<0$, so 3 lies between the roots of $f(x)$. Let the smaller root be $alpha$. So we have,
$$(x-alpha)(x-3)(x-beta)leq 0$$
Determine the sign on each of the intervals $(-infty,alpha), (alpha,3), (3,beta)$ and $(beta,infty)$. Therefore, the required solution is $$xin (-infty,alpha]cup [3,beta]$$ where the values of $alpha$ and $beta$ are to be found by the quadratic formula.
Hope it helps:)
edited Jan 17 at 4:16
Andrei
12.4k21128
answered Jan 17 at 4:09
MartundMartund
1,645213
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
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$begingroup$
Firstly, inequality. Please try to be accurate with terminology.
For the question: First, you want to find the critical values, that is, the $x$ where $(x-3)(x^2-5x+5)=0$
The first solution $x=3$ is trivial, then use the quadratic formula on $x^2-5x+5$ to find the other two.
I'll call them $x_1$ and $x_2$ (you should work them out) and note that $x_1<3<x_2$.
We now have four different intervals to test. Notice that between two adjacent zeroes of a function, the function is always positive or always negative. Hence if we pick an arbitrary $x$ in each of the intervals:
$$x<x_1$$
$$x_1<x<3$$
$$3<x<x_2$$
$$x>x_2$$
And see if the result is positive or negative when we put it in our function. If its negative, the interval it is in fits the $leq 0$ criterion.
answered Jan 17 at 4:04
Rhys HughesRhys Hughes
6,8601530
$endgroup$
add a comment |
$begingroup$
Firstly, inequality. Please try to be accurate with terminology.
For the question: First, you want to find the critical values, that is, the $x$ where $(x-3)(x^2-5x+5)=0$
The first solution $x=3$ is trivial, then use the quadratic formula on $x^2-5x+5$ to find the other two.
I'll call them $x_1$ and $x_2$ (you should work them out) and note that $x_1<3<x_2$.
We now have four different intervals to test. Notice that between two adjacent zeroes of a function, the function is always positive or always negative. Hence if we pick an arbitrary $x$ in each of the intervals:
$$x<x_1$$
$$x_1<x<3$$
$$3<x<x_2$$
$$x>x_2$$
And see if the result is positive or negative when we put it in our function. If its negative, the interval it is in fits the $leq 0$ criterion.
answered Jan 17 at 4:04
Rhys HughesRhys Hughes
6,8601530
$endgroup$
add a comment |
$begingroup$
Firstly, inequality. Please try to be accurate with terminology.
For the question: First, you want to find the critical values, that is, the $x$ where $(x-3)(x^2-5x+5)=0$
The first solution $x=3$ is trivial, then use the quadratic formula on $x^2-5x+5$ to find the other two.
I'll call them $x_1$ and $x_2$ (you should work them out) and note that $x_1<3<x_2$.
We now have four different intervals to test. Notice that between two adjacent zeroes of a function, the function is always positive or always negative. Hence if we pick an arbitrary $x$ in each of the intervals:
$$x<x_1$$
$$x_1<x<3$$
$$3<x<x_2$$
$$x>x_2$$
And see if the result is positive or negative when we put it in our function. If its negative, the interval it is in fits the $leq 0$ criterion.
answered Jan 17 at 4:04
Rhys HughesRhys Hughes
6,8601530
$endgroup$
Firstly, inequality. Please try to be accurate with terminology.
For the question: First, you want to find the critical values, that is, the $x$ where $(x-3)(x^2-5x+5)=0$
The first solution $x=3$ is trivial, then use the quadratic formula on $x^2-5x+5$ to find the other two.
I'll call them $x_1$ and $x_2$ (you should work them out) and note that $x_1<3<x_2$.
We now have four different intervals to test. Notice that between two adjacent zeroes of a function, the function is always positive or always negative. Hence if we pick an arbitrary $x$ in each of the intervals:
$$x<x_1$$
$$x_1<x<3$$
$$3<x<x_2$$
$$x>x_2$$
And see if the result is positive or negative when we put it in our function. If its negative, the interval it is in fits the $leq 0$ criterion.
answered Jan 17 at 4:04
Rhys HughesRhys Hughes
6,8601530
answered Jan 17 at 4:04
Rhys HughesRhys Hughes
6,8601530
answered Jan 17 at 4:04
Rhys HughesRhys Hughes
6,8601530
answered Jan 17 at 4:04
Rhys HughesRhys Hughes
6,8601530
6,8601530
add a comment |
add a comment |
$begingroup$
Using the formula we have $$f(x)=(x-a)(x-3)(x-b)le0$$
where $a=dfrac{5-sqrt5}2<3<b=dfrac{5+sqrt5}2$
For $f(x)=0,x=a,b$ or $3$
For $f(x)<0,$ odd number of multiplicands need to be $<0$
which will occur if $3<x<b$ or if $x<a$
answered Jan 17 at 4:03
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Can we guarantee it was $(x-a)(x-3)(x-b)le0$ not $(a-x)(x-3)(b-x)le0$ ?
$endgroup$
– b.ben
Jan 17 at 4:09
$begingroup$
@b.ben, $$(x-a)(x-b)=(a-x)(b-x)$$
$endgroup$
– lab bhattacharjee
Jan 17 at 4:11
$begingroup$
Hmm, $(a-x)(x-3)(x-b)le0$ answer will not the same as $(x-a)(x-3)(x-b)le0$ in inequation.
$endgroup$
– b.ben
Jan 17 at 4:13
add a comment |
$begingroup$
Using the formula we have $$f(x)=(x-a)(x-3)(x-b)le0$$
where $a=dfrac{5-sqrt5}2<3<b=dfrac{5+sqrt5}2$
For $f(x)=0,x=a,b$ or $3$
For $f(x)<0,$ odd number of multiplicands need to be $<0$
which will occur if $3<x<b$ or if $x<a$
answered Jan 17 at 4:03
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Can we guarantee it was $(x-a)(x-3)(x-b)le0$ not $(a-x)(x-3)(b-x)le0$ ?
$endgroup$
– b.ben
Jan 17 at 4:09
$begingroup$
@b.ben, $$(x-a)(x-b)=(a-x)(b-x)$$
$endgroup$
– lab bhattacharjee
Jan 17 at 4:11
$begingroup$
Hmm, $(a-x)(x-3)(x-b)le0$ answer will not the same as $(x-a)(x-3)(x-b)le0$ in inequation.
$endgroup$
– b.ben
Jan 17 at 4:13
add a comment |
$begingroup$
Using the formula we have $$f(x)=(x-a)(x-3)(x-b)le0$$
where $a=dfrac{5-sqrt5}2<3<b=dfrac{5+sqrt5}2$
For $f(x)=0,x=a,b$ or $3$
For $f(x)<0,$ odd number of multiplicands need to be $<0$
which will occur if $3<x<b$ or if $x<a$
answered Jan 17 at 4:03
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Using the formula we have $$f(x)=(x-a)(x-3)(x-b)le0$$
where $a=dfrac{5-sqrt5}2<3<b=dfrac{5+sqrt5}2$
For $f(x)=0,x=a,b$ or $3$
For $f(x)<0,$ odd number of multiplicands need to be $<0$
which will occur if $3<x<b$ or if $x<a$
answered Jan 17 at 4:03
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 17 at 4:03
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 17 at 4:03
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 17 at 4:03
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
Can we guarantee it was $(x-a)(x-3)(x-b)le0$ not $(a-x)(x-3)(b-x)le0$ ?
$endgroup$
– b.ben
Jan 17 at 4:09
$begingroup$
@b.ben, $$(x-a)(x-b)=(a-x)(b-x)$$
$endgroup$
– lab bhattacharjee
Jan 17 at 4:11
$begingroup$
Hmm, $(a-x)(x-3)(x-b)le0$ answer will not the same as $(x-a)(x-3)(x-b)le0$ in inequation.
$endgroup$
– b.ben
Jan 17 at 4:13
add a comment |
$begingroup$
Can we guarantee it was $(x-a)(x-3)(x-b)le0$ not $(a-x)(x-3)(b-x)le0$ ?
$endgroup$
– b.ben
Jan 17 at 4:09
$begingroup$
@b.ben, $$(x-a)(x-b)=(a-x)(b-x)$$
$endgroup$
– lab bhattacharjee
Jan 17 at 4:11
$begingroup$
Hmm, $(a-x)(x-3)(x-b)le0$ answer will not the same as $(x-a)(x-3)(x-b)le0$ in inequation.
$endgroup$
– b.ben
Jan 17 at 4:13
$begingroup$
Can we guarantee it was $(x-a)(x-3)(x-b)le0$ not $(a-x)(x-3)(b-x)le0$ ?
$endgroup$
– b.ben
Jan 17 at 4:09
$begingroup$
Can we guarantee it was $(x-a)(x-3)(x-b)le0$ not $(a-x)(x-3)(b-x)le0$ ?
$endgroup$
– b.ben
Jan 17 at 4:09
$begingroup$
@b.ben, $$(x-a)(x-b)=(a-x)(b-x)$$
$endgroup$
– lab bhattacharjee
Jan 17 at 4:11
$begingroup$
@b.ben, $$(x-a)(x-b)=(a-x)(b-x)$$
$endgroup$
– lab bhattacharjee
Jan 17 at 4:11
$begingroup$
Hmm, $(a-x)(x-3)(x-b)le0$ answer will not the same as $(x-a)(x-3)(x-b)le0$ in inequation.
$endgroup$
– b.ben
Jan 17 at 4:13
$begingroup$
Hmm, $(a-x)(x-3)(x-b)le0$ answer will not the same as $(x-a)(x-3)(x-b)le0$ in inequation.
$endgroup$
– b.ben
Jan 17 at 4:13
add a comment |
$begingroup$
Let $f(x)=x^2-5x+5$, we see that here the coefficient of $x^2$ term is positive and $f(3)=-1<0$, so 3 lies between the roots of $f(x)$. Let the smaller root be $alpha$. So we have,
$$(x-alpha)(x-3)(x-beta)leq 0$$
Determine the sign on each of the intervals $(-infty,alpha), (alpha,3), (3,beta)$ and $(beta,infty)$. Therefore, the required solution is $$xin (-infty,alpha]cup [3,beta]$$ where the values of $alpha$ and $beta$ are to be found by the quadratic formula.
Hope it helps:)
edited Jan 17 at 4:16
Andrei
12.4k21128
answered Jan 17 at 4:09
MartundMartund
1,645213
$endgroup$
add a comment |
$begingroup$
Let $f(x)=x^2-5x+5$, we see that here the coefficient of $x^2$ term is positive and $f(3)=-1<0$, so 3 lies between the roots of $f(x)$. Let the smaller root be $alpha$. So we have,
$$(x-alpha)(x-3)(x-beta)leq 0$$
Determine the sign on each of the intervals $(-infty,alpha), (alpha,3), (3,beta)$ and $(beta,infty)$. Therefore, the required solution is $$xin (-infty,alpha]cup [3,beta]$$ where the values of $alpha$ and $beta$ are to be found by the quadratic formula.
Hope it helps:)
edited Jan 17 at 4:16
Andrei
12.4k21128
answered Jan 17 at 4:09
MartundMartund
1,645213
$endgroup$
add a comment |
$begingroup$
Let $f(x)=x^2-5x+5$, we see that here the coefficient of $x^2$ term is positive and $f(3)=-1<0$, so 3 lies between the roots of $f(x)$. Let the smaller root be $alpha$. So we have,
$$(x-alpha)(x-3)(x-beta)leq 0$$
Determine the sign on each of the intervals $(-infty,alpha), (alpha,3), (3,beta)$ and $(beta,infty)$. Therefore, the required solution is $$xin (-infty,alpha]cup [3,beta]$$ where the values of $alpha$ and $beta$ are to be found by the quadratic formula.
Hope it helps:)
edited Jan 17 at 4:16
Andrei
12.4k21128
answered Jan 17 at 4:09
MartundMartund
1,645213
$endgroup$
Let $f(x)=x^2-5x+5$, we see that here the coefficient of $x^2$ term is positive and $f(3)=-1<0$, so 3 lies between the roots of $f(x)$. Let the smaller root be $alpha$. So we have,
$$(x-alpha)(x-3)(x-beta)leq 0$$
Determine the sign on each of the intervals $(-infty,alpha), (alpha,3), (3,beta)$ and $(beta,infty)$. Therefore, the required solution is $$xin (-infty,alpha]cup [3,beta]$$ where the values of $alpha$ and $beta$ are to be found by the quadratic formula.
Hope it helps:)
edited Jan 17 at 4:16
Andrei
12.4k21128
answered Jan 17 at 4:09
MartundMartund
1,645213
edited Jan 17 at 4:16
Andrei
12.4k21128
edited Jan 17 at 4:16
Andrei
12.4k21128
edited Jan 17 at 4:16
Andrei
12.4k21128
12.4k21128
answered Jan 17 at 4:09
MartundMartund
1,645213
answered Jan 17 at 4:09
MartundMartund
1,645213
answered Jan 17 at 4:09
MartundMartund
1,645213
1,645213
add a comment |
add a comment |
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4
$begingroup$
Find roots. Determine sign on each interval.
$endgroup$
– Brevan Ellefsen
Jan 17 at 3:51
$begingroup$
@BrevanEllefsen Simplest solution
$endgroup$
– b.ben
Jan 17 at 4:14