Given the equation of:ax^2+bx+c Deduce that the root is -1 and a+b/ait's ax(squared) +bx+c Deduce that the roots are -1 and a+b(divided by) a

You should use quadratic formula to find the roots such that:

`x_(1,2) = (-b+-sqrt(b^2 - 4ac))/2a`

If one root needs to be equal to -1 , then you should set the equations `(-b+sqrt(b^2 - 4ac))/2a`  and -1 equal such that:

`(-b+sqrt(b^2 - 4ac))/2a= -1`

`(-b+-sqrt(b^2 - 4ac)) = -2a`

`sqrt(b^2 - 4ac) = b - 2a`

You need to raise to square to remove the square root such that:

`b^2 - 4ac = (b-2a)^2`

You need to expand the binomial such that:

`b^2 - 4ac = b^2 - 4ab + 4a^2`

Removing like terms yields:

`- 4ac = - 4ab + 4a^2`

You need to divide by `-4a ` such that:

`c = b - a =gt a+c = b`

Hence, if you need that one root to be equal to -1, then `a +c = b.`

If the second root needs to be equal to `(a+b)/a`  , then you should set the equations `(-b-sqrt(b^2 - 4ac))/2a ` and`(a+b)/a`  equal such that:

`(-b-sqrt(b^2 - 4ac))/2a= (a+b)/a`

`(-b-sqrt(b^2 - 4ac)) = 2(a+b)`

-`sqrt(b^2 - 4ac) = b+2a+2b`

-`sqrt(b^2 - 4ac) = 2a + 3b`

You need to raise to square to remove the square root such that:

`b^2 - 4ac = (2a + 3b)^2`

You need to expand the binomial such that:

`b^2 - 4ac = 4a^2 + 12ab + 9b^` 2

`4a^2 + 12ab + 9b^2 - b^2 + 4ac = 0`

`4a^2 + 12ab + 8b^2 + 4ac = 0`

You need to divide by 4 such that:

`a^2 + 3ab + 2b^2 + ac = 0`

You need to factor out a in the sum `a^2 + ac`  such that:

`a(a+c) + 3ab + 2b^2= 0`

You need to remember that `a+c=b`  (condition imposed by the existence of the root -1) such that:

`ab + 3ab + 2b^2 = ` 0

`4ab + 2b^2 = 0`

`2ab + b^2 = 0`

`b(2a+b) = 0`

`b = 0 or 2a = -b =gt a = -b/2`

Hence, evaluating the conditions for the roots of the quadratic to be equal to -1 and `(a+b)/a`  yields `a+c=b, b=0 or a=-b/2.`