Welcome to our community

Be a part of something great, join today!

Evaluate (ab + a + 1)/b.

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
If a and b are two real numbers satisfying the relations \(\displaystyle 2a^2-3a-1=0\) and \(\displaystyle b^2+3b-2=0\), and \(\displaystyle ab\ne 1\), evaluate \(\displaystyle \frac{ab+a+1}{b}\).
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Re: Evaluate (ab+a+1)/b.

Here's my solution
Notice that $a + \dfrac{a}{b} + \dfrac{1}{b}$ can be written as $\left(a+1\right)\left(\dfrac{1}{b}+1\right) - 1$. Furthermore we have $a$ satisfying $2a^2-3a-1 = 0$ and $b$ satisfying $b^2+3b - 2=0$ or $\dfrac{2}{b^2} - \dfrac{3}{b} -1$. If we let $x=a+1$ and $y = \dfrac{1}{b}+1$ then we want to find the value of $xy-1$ where both $x$ and $y$ satisfy $2x^2-7x+4 = 0$. Now the condition $ab \ne 1$ excludes the possibility of choosing $x = y$ so we want the two different roots of our quadratic and their product being $\dfrac{4}{2}$. Thus, the answer we seek is $2-1=1$.
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
Re: Evaluate (ab+a+1)/b.

Hi Jester, thanks for participating in this problem and I really like your solution! :)

I'd like to share another method (which isn't my solution) on how to tackle this problem too...here goes:

We are given

\(\displaystyle 2a^2-3a-1=0\) and \(\displaystyle b^2+3b-2=0\)

We see that we can do some algebraic manipulation to the second equation to make it be another quadratic equation that is similar to that of first equation:

\(\displaystyle \frac{b^2}{b^2}+\frac{3b}{b^2}-\frac{2}{b^2}=\frac{0}{b^2}\)

\(\displaystyle 1+\frac{3}{b}-\frac{2}{b^2}=0\)

\(\displaystyle 2\left(\frac{1}{b^2}\right)-3\left(\frac{1}{b}\right)-1=0\)

Therefore, \(\displaystyle a\) and \(\displaystyle \frac{1}{b}\) are the roots for the equation \(\displaystyle 2x^2-3x-1=0\) and the intended expression could be found by finding the sum of roots and product of roots of the equation above as follows:

\(\displaystyle a+\frac{1}{b}=\frac{3}{2}\)

\(\displaystyle a\left(\frac{1}{b}\right)=-\frac{1}{2}\)

\(\displaystyle \therefore a+\frac{1}{b}+a\left(\frac{1}{b}\right)=\frac{ab+b+1}{b}=\frac{3}{2}-\frac{1}{2}=1\)
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Re: Evaluate (ab+a+1)/b.

Definitely cleaner than mine.
 
  • Thread starter
  • Admin
  • #5

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
Re: Evaluate (ab+a+1)/b.

I say both are cool and clever methods!:cool::)
 

Albert

Well-known member
Jan 25, 2013
1,225
Re: Evaluate (ab+a+1)/b.

If a and b are two real numbers satisfying the relations \(\displaystyle 2a^2-3a-1=0\) and \(\displaystyle b^2+3b-2=0\), and \(\displaystyle ab\ne 1\), evaluate \(\displaystyle \frac{ab+a+1}{b}\).
\(\displaystyle 2a^2-3a-1=0---(1)\) and \(\displaystyle b^2+3b-2=0---(2)\), and \(\displaystyle ab\ne 1\),
$(1)\times b+(2)\times a :$ rearrange and we get :
$(ab-1)(b+2a)=0$
$\because ab\neq 1 \therefore b=-2a$
\(\displaystyle \dfrac{ab+a+1}{b}=\dfrac{-2a^2+3a+1-2a}{-2a}=1\)
 
Last edited: