SOLUTION: Find the values of a and b that result in the given sum. [(a+1)x^2 + bx + 2] + [(b-2)x^2 - ax + 5] = x^2 - 4x + 7 Please help. Thanks

Algebra.Com
Question 201065: Find the values of a and b that result in the given sum.
[(a+1)x^2 + bx + 2] + [(b-2)x^2 - ax + 5] = x^2 - 4x + 7

Please help. Thanks

Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
[(a+1)x^2 + bx + 2] + [(b-2)x^2 - ax + 5] = x^2 - 4x + 7 ... Start with the given equation.


(a+1)x^2 + bx + 2 + (b-2)x^2 - ax + 5 = x^2 - 4x + 7 ... Remove the brackets


[(a+1)x^2+ (b-2)x^2] + [ bx - ax] + [2 + 5] = x^2 - 4x + 7 ... Group like terms.


(a+1+b-2)x^2 + (b-a)x + 7 = x^2 - 4x + 7 ... Combine like terms.


(a+b-1)x^2 + (b-a)x + 7 = x^2 - 4x + 7 ... Simplify


Now just equate the coefficients to get the two equations:

a+b-1=1
b-a=-4


So simply solve this system (which I'll let you do) using any method to get the answers
a=3, b=-1



If you have any questions, email me at jim_thompson5910@hotmail.com.
Check out my website if you are interested in tutoring.

RELATED QUESTIONS

Given that f(x)=ax^2+bx+5, f(3)=2 and f'(3)=2 find the values of a and... (answered by Fombitz)
If F(x)=ax^2 +bx+c and F(x+5)=x^2+9x-7, find the sum of... (answered by Alan3354,ikleyn)
Find the values of a and b that will make the function continuous every where. Justify. (answered by KMST)
given f(x) = ax^2 + bx + 5. Find a and b such that f(x + 1) - f(x) = 8x +... (answered by jim_thompson5910)
if ax^2+bx+c=0 and x=2 or 6/7 what are the values of a b... (answered by ikleyn)
1)consider the polynomial p(x)=x^3+ax^2+bx-12 given that (x+3)and (x-4)are factors of... (answered by Edwin McCravy)
Find the values of a and b so that the function f(x)=(1/3)x^3 + ax^2 +bx will have a... (answered by greenestamps)
Find the sum of a, b, c, and d if (x^3 - 2x^2 + 3x + 5)/(x + 2) = ax^2 + bx + c +... (answered by josgarithmetic,timofer)
the sum 2/x+1 + 7/x+2 can be expressed in the form ax+b/(x+1)(x+2). what are the values... (answered by Edwin McCravy)