Mixing
For which values of $a$ we will get two different roots?
$begingroup$
In given the following system of equations:
$$ |x-1| > 2x+2 $$
$$ x^2 + ax + a -1 = 0 $$
For which values of $a$ we will get two different roots?
systems-of-equations roots quadratics absolute-value
edited Jan 26 at 16:49
Michael Rozenberg
108k1895200
asked Jan 26 at 16:29
StackUserStackUser
61
$endgroup$
add a comment |
$begingroup$
In given the following system of equations:
$$ |x-1| > 2x+2 $$
$$ x^2 + ax + a -1 = 0 $$
For which values of $a$ we will get two different roots?
systems-of-equations roots quadratics absolute-value
edited Jan 26 at 16:49
Michael Rozenberg
108k1895200
asked Jan 26 at 16:29
StackUserStackUser
61
$endgroup$
$begingroup$
Use the fact that we get two different real roots when $b^2-4ac gt 0.$
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:31
$begingroup$
What do you need the first inequality for?
$endgroup$
– Dr. Mathva
Jan 26 at 16:43
add a comment |
$begingroup$
In given the following system of equations:
$$ |x-1| > 2x+2 $$
$$ x^2 + ax + a -1 = 0 $$
For which values of $a$ we will get two different roots?
systems-of-equations roots quadratics absolute-value
edited Jan 26 at 16:49
Michael Rozenberg
108k1895200
asked Jan 26 at 16:29
StackUserStackUser
61
$endgroup$
In given the following system of equations:
$$ |x-1| > 2x+2 $$
$$ x^2 + ax + a -1 = 0 $$
For which values of $a$ we will get two different roots?
systems-of-equations roots quadratics absolute-value
systems-of-equations roots quadratics absolute-value
edited Jan 26 at 16:49
Michael Rozenberg
108k1895200
asked Jan 26 at 16:29
StackUserStackUser
61
edited Jan 26 at 16:49
Michael Rozenberg
108k1895200
asked Jan 26 at 16:29
StackUserStackUser
61
edited Jan 26 at 16:49
Michael Rozenberg
108k1895200
edited Jan 26 at 16:49
Michael Rozenberg
108k1895200
edited Jan 26 at 16:49
Michael Rozenberg
108k1895200
108k1895200
asked Jan 26 at 16:29
StackUserStackUser
61
asked Jan 26 at 16:29
StackUserStackUser
61
asked Jan 26 at 16:29
StackUserStackUser
61
61
$begingroup$
Use the fact that we get two different real roots when $b^2-4ac gt 0.$
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:31
$begingroup$
What do you need the first inequality for?
$endgroup$
– Dr. Mathva
Jan 26 at 16:43
add a comment |
$begingroup$
Use the fact that we get two different real roots when $b^2-4ac gt 0.$
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:31
$begingroup$
What do you need the first inequality for?
$endgroup$
– Dr. Mathva
Jan 26 at 16:43
$begingroup$
Use the fact that we get two different real roots when $b^2-4ac gt 0.$
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:31
$begingroup$
Use the fact that we get two different real roots when $b^2-4ac gt 0.$
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:31
$begingroup$
What do you need the first inequality for?
$endgroup$
– Dr. Mathva
Jan 26 at 16:43
$begingroup$
What do you need the first inequality for?
$endgroup$
– Dr. Mathva
Jan 26 at 16:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The first gives
$$x-1>2x+2$$ or $$x-1<-2x-2,$$ which gives $$x<-frac{1}{3}.$$
Now, let $f(x)=x^2+ax+a-1$ and solve the following system:
$$fleft(-frac{1}{3}right)>0$$
$$-frac{a}{2}<-frac{1}{3}$$ and
$$a^2-4(a-1)>0.$$
answered Jan 26 at 16:45
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Why you want that $f(-frac{1}{3}) > 0 $ ?
$endgroup$
– StackUser
Jan 26 at 21:31
$begingroup$
@StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
$endgroup$
– Michael Rozenberg
Jan 26 at 21:38
add a comment |
$begingroup$
The equation has two roots $x_1=-1$ and $x_2=1-a$. It is easy to see $x_1neq x_2$ if $aneq 2$.
Clearly if $x=x_1$, then the inequality $|x-1|>2x+2$ holds. For $x=x_2$, then the inequality $|x-1|>2x+2$ becomes
$$ |a|>4-2a. tag{1}$$
If $age2$, (1) holds. If $0le a<2$, then the solution of (1) is $frac43<a<2$. If $a<0$, then (1) does not hold.
So if $frac43<a<2$, the equation has two different roots satisfying the inequality.
answered Jan 26 at 16:52
xpaulxpaul
23.3k24655
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first gives
$$x-1>2x+2$$ or $$x-1<-2x-2,$$ which gives $$x<-frac{1}{3}.$$
Now, let $f(x)=x^2+ax+a-1$ and solve the following system:
$$fleft(-frac{1}{3}right)>0$$
$$-frac{a}{2}<-frac{1}{3}$$ and
$$a^2-4(a-1)>0.$$
answered Jan 26 at 16:45
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Why you want that $f(-frac{1}{3}) > 0 $ ?
$endgroup$
– StackUser
Jan 26 at 21:31
$begingroup$
@StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
$endgroup$
– Michael Rozenberg
Jan 26 at 21:38
add a comment |
$begingroup$
The first gives
$$x-1>2x+2$$ or $$x-1<-2x-2,$$ which gives $$x<-frac{1}{3}.$$
Now, let $f(x)=x^2+ax+a-1$ and solve the following system:
$$fleft(-frac{1}{3}right)>0$$
$$-frac{a}{2}<-frac{1}{3}$$ and
$$a^2-4(a-1)>0.$$
answered Jan 26 at 16:45
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Why you want that $f(-frac{1}{3}) > 0 $ ?
$endgroup$
– StackUser
Jan 26 at 21:31
$begingroup$
@StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
$endgroup$
– Michael Rozenberg
Jan 26 at 21:38
add a comment |
$begingroup$
The first gives
$$x-1>2x+2$$ or $$x-1<-2x-2,$$ which gives $$x<-frac{1}{3}.$$
Now, let $f(x)=x^2+ax+a-1$ and solve the following system:
$$fleft(-frac{1}{3}right)>0$$
$$-frac{a}{2}<-frac{1}{3}$$ and
$$a^2-4(a-1)>0.$$
answered Jan 26 at 16:45
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
The first gives
$$x-1>2x+2$$ or $$x-1<-2x-2,$$ which gives $$x<-frac{1}{3}.$$
Now, let $f(x)=x^2+ax+a-1$ and solve the following system:
$$fleft(-frac{1}{3}right)>0$$
$$-frac{a}{2}<-frac{1}{3}$$ and
$$a^2-4(a-1)>0.$$
answered Jan 26 at 16:45
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 16:45
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 16:45
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 16:45
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
$begingroup$
Why you want that $f(-frac{1}{3}) > 0 $ ?
$endgroup$
– StackUser
Jan 26 at 21:31
$begingroup$
@StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
$endgroup$
– Michael Rozenberg
Jan 26 at 21:38
add a comment |
$begingroup$
Why you want that $f(-frac{1}{3}) > 0 $ ?
$endgroup$
– StackUser
Jan 26 at 21:31
$begingroup$
@StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
$endgroup$
– Michael Rozenberg
Jan 26 at 21:38
$begingroup$
Why you want that $f(-frac{1}{3}) > 0 $ ?
$endgroup$
– StackUser
Jan 26 at 21:31
$begingroup$
Why you want that $f(-frac{1}{3}) > 0 $ ?
$endgroup$
– StackUser
Jan 26 at 21:31
$begingroup$
@StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
$endgroup$
– Michael Rozenberg
Jan 26 at 21:38
$begingroup$
@StackUser Because we need that roots of the quadratic equation would be less that $-frac{1}{3}.$ Draw the graph of $f$.
$endgroup$
– Michael Rozenberg
Jan 26 at 21:38
add a comment |
$begingroup$
The equation has two roots $x_1=-1$ and $x_2=1-a$. It is easy to see $x_1neq x_2$ if $aneq 2$.
Clearly if $x=x_1$, then the inequality $|x-1|>2x+2$ holds. For $x=x_2$, then the inequality $|x-1|>2x+2$ becomes
$$ |a|>4-2a. tag{1}$$
If $age2$, (1) holds. If $0le a<2$, then the solution of (1) is $frac43<a<2$. If $a<0$, then (1) does not hold.
So if $frac43<a<2$, the equation has two different roots satisfying the inequality.
answered Jan 26 at 16:52
xpaulxpaul
23.3k24655
$endgroup$
add a comment |
$begingroup$
The equation has two roots $x_1=-1$ and $x_2=1-a$. It is easy to see $x_1neq x_2$ if $aneq 2$.
Clearly if $x=x_1$, then the inequality $|x-1|>2x+2$ holds. For $x=x_2$, then the inequality $|x-1|>2x+2$ becomes
$$ |a|>4-2a. tag{1}$$
If $age2$, (1) holds. If $0le a<2$, then the solution of (1) is $frac43<a<2$. If $a<0$, then (1) does not hold.
So if $frac43<a<2$, the equation has two different roots satisfying the inequality.
answered Jan 26 at 16:52
xpaulxpaul
23.3k24655
$endgroup$
add a comment |
$begingroup$
The equation has two roots $x_1=-1$ and $x_2=1-a$. It is easy to see $x_1neq x_2$ if $aneq 2$.
Clearly if $x=x_1$, then the inequality $|x-1|>2x+2$ holds. For $x=x_2$, then the inequality $|x-1|>2x+2$ becomes
$$ |a|>4-2a. tag{1}$$
If $age2$, (1) holds. If $0le a<2$, then the solution of (1) is $frac43<a<2$. If $a<0$, then (1) does not hold.
So if $frac43<a<2$, the equation has two different roots satisfying the inequality.
answered Jan 26 at 16:52
xpaulxpaul
23.3k24655
$endgroup$
The equation has two roots $x_1=-1$ and $x_2=1-a$. It is easy to see $x_1neq x_2$ if $aneq 2$.
Clearly if $x=x_1$, then the inequality $|x-1|>2x+2$ holds. For $x=x_2$, then the inequality $|x-1|>2x+2$ becomes
$$ |a|>4-2a. tag{1}$$
If $age2$, (1) holds. If $0le a<2$, then the solution of (1) is $frac43<a<2$. If $a<0$, then (1) does not hold.
So if $frac43<a<2$, the equation has two different roots satisfying the inequality.
answered Jan 26 at 16:52
xpaulxpaul
23.3k24655
answered Jan 26 at 16:52
xpaulxpaul
23.3k24655
answered Jan 26 at 16:52
xpaulxpaul
23.3k24655
answered Jan 26 at 16:52
xpaulxpaul
23.3k24655
23.3k24655
add a comment |
add a comment |
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$begingroup$
Use the fact that we get two different real roots when $b^2-4ac gt 0.$
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 16:31
$begingroup$
What do you need the first inequality for?
$endgroup$
– Dr. Mathva
Jan 26 at 16:43