How to show that the equation #ax^2-(a+b)x+b=0# has a solution for all values of a and b?

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Answer 1 · 2018-01-24T10:41:00

See the explanation below

The quadratic equation is

#ax^2-(a+b)x+b=0#

In order for this quadratic equation to have solutions, the discriminant #>=0#

The discriminant is

#Delta=(-(a+b))^2-4*(a)*(b)#

#=(a+b)^2-4ab#

#=a^2+2ab+b*2-4ab#

#=a^2-2ab+b^2#

#=(a-b)^2#

#AA a,b in RR^2#, #=>#, #Delta>=0#

The roots are

#x=((a+b)+-sqrt((a-b)^2))/(2a)#

#x=(a+b)+-(a-b)/(2a)#

#x_1=(a+b+a-b)/(2a)=1#

#x_2=(a+b-a+b)/(2a)=b/a#

The roots are #=1# and #b/a#

Therefore, whatever values of #a!=0 " and "b#, the quadratic equation has a solution.

Answer 2 · 2018-01-24T12:33:58

Please see be,ow.

#ax^2-(a+b)x+b# can be factored to get

#ax^2-(a+b)x+b = (ax-b)(x-1) = 0#

So, #x=1# is always a solution.

Also, if #a != 0#, then #x = b/a# is a solution as well.