MHB What values of b give a common root for these equations?

[anemone]

For what values of b do the equation $$1988x^2+bx+8891=0$$ and $$8891x^2+bx+1988=0$$ have a common root?

 

[mathmaniac1]

For what values of b do the equation $$1988x^2+bx+8891=0$$ and $$8891x^2+bx+1988=0$$ have a common root?

$$1988x^2+bx+8891=0=8891x^2+bx+1988$$

$$(8891-1988)x^2-(8891-1988)=0$$
$$x^2=1$$

So x=1

 

[anemone]

$$1988x^2+bx+8891=0=8891x^2+bx+1988$$

$$(8891-1988)x^2-(8891-1988)=0$$
$$x^2=1$$

So x=1

Thanks for participating, mathmaniac but hey, we're asked to solve for b instead...(Wondering)

 

[mathmaniac1]

oooH!
I am fast,didn't even read the question carefully!
Ok...
b=-(8891+1988)=-10879Honestly I felt no challenge in this one while most of your other questions seemed challenging...
Hoping to get more challenges from you...Edit:I got the numbers wrong in my mind...The same carelessness I show in exams

 

[MarkFL]

Let's let $r$ be the common root, and we may express the first quadratic as:

$$1988(x-r)\left(x-\frac{8891}{1988r} \right)=1988x^2-(8891+1988r)x+8891$$

and the second quadratic may be expressed:

$$8891(x-r)\left(x-\frac{1988}{8891r} \right)=8891x^2-(1988+8891r)x+1988$$

Thus, we must have:

$$-b=8891+1988r=1988+8891r\implies r=1\implies b=-10879$$

Thus, we find that when $b=-10879$ the two quadratics share the root $r=1$.

 

[MarkFL]

...
b=-(8891+1988)=-9089

Check your arithmetic...you have the correct expression, you just added incorrectly. (Nod)

 

[mathmaniac1]

I got the numbers wrong in my mind!

 

[Jameson]

I got the numbers wrong in my mind!

One of the implied conditions of saying a problem is "easy" is to get it correct. ;) You seem to be in a hurry often when you are posting. We really encourage our users to take a few minutes for every post so we have a higher "post:content" ratio than most forums. It also helps cut down on mistakes and is more efficient for communicating. The same thing goes for formatting equations in Latex - it looks more professional and is easier for others to read. In general, making quality posts is a sign of respect to those who take time to read them. Everything you post is read by multiple people believe it or not :)

I remember being similar when I was 15 - doing problems in my head, rushing through things because answers were trivial, etc. but try to fight your instincts while posting here and I think you'll get more of out MHB. :)

 

[MarkFL]

In my haste, I also made an error. I should have written:

Let's let $r$ be the common root, and we may express the first quadratic as:

$$1988(x-r)\left(x-\frac{8891}{1988r} \right)=1988x^2-\left(\frac{8891}{r}+1988r \right)x+8891$$

and the second quadratic may be expressed:

$$8891(x-r)\left(x-\frac{1988}{8891r} \right)=8891x^2-\left(\frac{1988}{r}+8891r \right)x+1988$$

Thus, we must have:

$$-b=\frac{8891}{r}+1988r=\frac{1988}{r}+8891r \implies r^2=1 \implies b=\pm10879$$

Thus, we find that when $b=\pm10879$ the two quadratics share the root $r=\pm1$.

 

[mathmaniac1]

We both made the same mistake!(me once again)

 

[kaliprasad]

There is another solution

We can eliminate x^2 from both to get
8891(1988x2+bx+8891) – 1988(8891x2+bx+1988) = 0

Or bx(8891-1988) = 1988^2- 9981 => bx = - (8891 + 1988) = - 10879 ..3

Subtracting 2nd one from 1st we get (1988-8891) x^2 + (8891 – 1988) = 0 or x^2 = 1 +> x = 1 or – 1

From (3) if x = 1 then b = - 10879 and if x = - 1 then b = 10879

 

[MarkFL]

Do you perhaps mean:

There is another solution method:

We are given:

(1) $$1988x^2+bx+8891=0$$

(2) $$8891x^2+bx+1988=0$$

We can eliminate $x^2$ from both to get:

$$8891(1988x^2+bx+8891)–1988(8891x^2+bx+1988)=0$$

Or:

$$bx(8891-1988)=1988^2-8891^2$$

(3) $$bx=-(8891 + 1988)=-10879$$

Subtracting (2) one from (1) we get:

$$(1988-8891)x^2+(8891–1988)=0\,\therefore\,x^2=1\, \therefore\,x=\pm1$$

From (3) if $x=1$ then $b=-10879$ and if $x=-1$ then $b=10879$