MHB Sum and product of roots of quadratic equation

[Mathsonfire]

#85

 

[MarkFL]

You've posted 4 questions, and I have given all 4 threads more useful titles, now I would ask that you show what you've tried in each thread. This way we can see where you're stuck.

 

[Mathsonfire]

Well i am stuck at this point
I found sum of zeroes in both cases and then got stuck.what should i do

 

[MarkFL]

Please post what you have so far. :)

 

[MarkFL]

We know:

$$c+d=10a$$

$$a+b=10c$$

Hence:

$$a+b+c+d=10(a+c)$$

We also know:

$$c^2-11b=a^2-11d$$

$$11d-11b=a^2-c^2$$

$$11(d-b)=(a+c)(a-c)$$

And we obtain by subtraction of the first 2 equations:

$$(d-b)-(a-c)=10(a-c)$$

Or:

$$d-b=11(a-c)$$

Hence:

$$121(a-c)=(a+c)(a-c)$$

Assuming \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:

$$a+c=121$$

Hence:

$$a+b+c+d=10\cdot121=1210$$