ax^2-a^2=-bx-b^2

[khavar]

ax^2-a^2=-bx-b^2
Solve the equation for x in terms of a and b.

This is problem number 43, page 196, from A-Plus Notes for Algebra (Algebra 2 and Pre-Calculus) by Rong Yang.

The answer in the back of the book is: X= a-b, a shall not equal -b

I would like to know how to solve this problem in terms of x. Thank you for your help.

 

[pappus]

ax^2-a^2=-bx-b^2
Solve the equation for x in terms of a and b.

This is problem number 43, page 196, from A-Plus Notes for Algebra (Algebra 2 and Pre-Calculus) by Rong Yang.

The answer in the back of the book is: X= a-b, a shall not equal -b

I would like to know how to solve this problem in terms of x. Thank you for your help.

Could it be that the equation reads:

\(\displaystyle ax-a^2=-bx-b^2\)

Otherwise the given result seems to be wrong.

 

[khavar]

Could it be that the equation reads:

\(\displaystyle ax-a^2=-bx-b^2\)

Otherwise the given result seems to be wrong.

Thank you, Pappus. I have found several typographical errors so far in this book, so it appears to be highly likely that this is yet another.
What a relief though. Thank you, thank you, thank you.

 

[JeffM]

ax^2-a^2=-bx-b^2
Solve the equation for x in terms of a and b.

This is problem number 43, page 196, from A-Plus Notes for Algebra (Algebra 2 and Pre-Calculus) by Rong Yang.

The answer in the back of the book is: X= a-b, a shall not equal -b

I would like to know how to solve this problem in terms of x. Thank you for your help.

Quadratic formula (assumes a not 0)

\(\displaystyle ax^2 - a^2 = - bx - b^2 \implies ax^2 + bx + (b^2 - a^2) = 0 \implies x = \dfrac{- b \pm \sqrt{b^2 - 4a(b^2 - a^2)}}{2a} =
\dfrac{- b \pm \sqrt{b^2 - 4ab^2 + 4a^3}}{2a}.\)

Let's check:

\(\displaystyle ax^2 - a^2 = a\left(\dfrac{-b \pm \sqrt{b^2 - 4ab^2 + 4a^3}}{2a}\right)^2 - a^2\) = \(\displaystyle a\left(\dfrac{b^2 \mp 2b\sqrt{b^2 - 4ab^2 + 4a^3} + b^2 - 4ab^2 + 4a^3}{4a^2}\right) - a^2 =\)

\(\displaystyle \dfrac{2b^2 \mp 2b\sqrt{b^2 - 4ab^2 + 4a^3} - 4ab^2 + 4a^3 - 4a^3}{4a} = \dfrac{b^2 \mp b\sqrt{b^2 - 4ab^2 + 4a^3} - 2ab^2}{2a} = (-b)\left(\dfrac{b \pm \sqrt{b^2 - 4ab^2 + 4a^3}}{2a}\right) - b^2 = - bx - b^2.\)

Answer in the book is wrong if you showed the problem correctly.

 

[lookagain]

ax^2-a^2=-bx-b^2
Solve the equation for x in terms of a and b.

This is problem number 43, page 196, from A-Plus Notes for Algebra (Algebra 2 and Pre-Calculus) by Rong Yang.

The answer in the back of the book is: X= a-b, a shall not equal -b

Suppose the equation were supposed to be ax - a^2 = -bx - b^2 instead.

And suppose x = a - b. I don't see why a cannot equal -b.



\(\displaystyle x(a + b) = a^2 - b^2 \ \ is \ \ not \ \ an \ \ equivalent \ \ equation \ \ to \)

\(\displaystyle x \ = \ \ \dfrac{a^2 - b^2}{a + b}\)



In the former equation, there is no restriction that a + b cannot equal 0, as it is
not a denominator.

 

[mmm4444bot]

Suppose the equation were supposed to be ax - a^2 = -bx - b^2 instead.

And suppose x = a - b. I don't see why a cannot equal -b.

If a were to equal -b, we would have

-bx - b^2 = -bx - b^2

which is true for all x.