Question

If `x^2-10ax-11b=0` and `x^2-10cx-11d=0` then what is `a+b +c+d`?

Answer 1

`10a+10c`

Explanation:

The sum of roots of a quadratic equation is easily given as `-b/a` if it is in the form `ax^2+bx+c=0`

For the first equation:
`c+d=-(-10a)/1=10a`

For the second:
`a+b=-(-10c)/1=10c`

Thus `a+b+c+d=10a+10c`

To learn more, look up Vieta's Formulas.

Answer 2

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

Explanation:

If we have a quadratic equation `px^2+qx+r=0`, sum of roots is `-q/p` and product of roots is `r/p`

Hencce as `x²-10ax-11b=0` has roots `c` and `d`, we have

`c+d=-(-10a)/1=10a` and `cd=-11b`

Further as `x²-10cx-11d=0` gas roots `a` and `b`, we have

`a+b=10c` and `ab=-11d`

Hence `a+b+c+d=10a+10c=10(a+c)`

Answer 3

`a+b +c+d= 10a+10c`

Explanation:

There is a relationship between the roots of a quadratic equation and the coefficients:

If the quadratic equation:

`ax^2 + bx + c = 0`

Has roots `alpha` and `beta` then:

`alpha + beta = -b/a`, and `alpha \ beta = c/a`

So for the first equation `x^2-10ax-11b=0` with roots `c` and `d` we have:

`c+d = -(10a)/1 = 10a`

Similarly, for the second equation `x^2-10cx-11d=0` with roots `a` and `b` we have:

`a+b = -(10c)/1 = 10c`

And adding these two results we get:

`a+b +c+d= 10a+10c`