Problem Solving

[Math Revolution GMAT math practice question]

Which of the following are roots of an equation 10x^{-2}+x^{-1}-21=0

A. -2/3 or 5/7
B. -2/3 or 7/5
C. 2/3 or -5/7
D. 2/3 or -7/5
E. 3/2 or 5/7

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Which of the following are the roots of the equation 10x^{-2}+x^{-1}-21=0 ?

A. -2/3 or 5/7
B. -2/3 or 7/5
C. 2/3 or -5/7
D. 2/3 or -7/5
E. 3/2 or 5/7

$$?\,\,\,:\,\,\,\,{{10} \over {{x^2}}} + {1 \over x} - 21 = 0\,\,\,{\rm{roots}}$$

We can just substitute x with the values presented in the alternative choices. "Luckily" (A) will already be the correct answer.

Another approach, probably even faster (at least when the correct alternative choice is not the first!), is the following:

$${{10} \over {{x^2}}} + {1 \over x} - 21 = 0\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{{\rm{when}}\,\,x\,\, \ne \,0} \,\,\,10 + x - 21{x^2} = 0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,21{x^2} - x - 10 = 0$$
$$21{x^2} - x - 10 = 0\,\,\,{\rm{has}}\,\,{\rm{product}}\,\,{\rm{of}}\,\,{\rm{roots}}\,\,\,{\rm{ = }}\,\,\, - {{10} \over {21}}\,\,\,\,\,\left( {{c \over a}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,\left( A \right)\,\,{\rm{survives}} \hfill \cr
\,\left( B \right)\,\,{\rm{out}} \hfill \cr
\,\left( C \right)\,\,{\rm{survives}} \hfill \cr
\,\left( D \right)\,\,{\rm{out}} \hfill \cr
\,\left( E \right)\,\,{\rm{out}} \hfill \cr} \right.$$
$$21{x^2} - x - 10 = 0\,\,\,{\rm{has}}\,\,{\rm{sum}}\,\,{\rm{of}}\,\,{\rm{roots}}\,\,\,{\rm{ = }}\,\,\, - {{ - 1} \over {21}} = {1 \over {21}}\,\,\,\,\left( { - {b \over a}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,\left( A \right)\,\, - {2 \over 3}{\rm{ + }}{5 \over 7}{\rm{ = }} - {{14} \over {21}} + {{15} \over {21}} = {1 \over {21}}\,\,\,\, \Rightarrow \,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,\left( A \right) \hfill \cr
\,\,\left( C \right)\,\,{\rm{don´t }}\,\,{\rm{need}}\,\,{\rm{checking}}\,\,\,\,\,\left( { - {5 \over 7} + {2 \over 3} = - {{15} \over {21}} + {{14} \over {21}} < 0 \,\,{\rm{out}}} \right) \hfill \cr} \right.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

$$10x^{\left(-2\right)}+x^{\left(-1\right)}-21=0$$
$$\frac{10}{x^2}+\frac{1}{x}-21=0$$
$$10+x-21x^2=0$$
$$-21x^2+x+10=0$$
$$-21x^2+15x-14x+10=0$$
$$-3x\left(7x-5\right)-2\left(7x-5\right)=0$$
$$\left(7x-5\right)\left(-3x-2\right)=0$$
$$7x-5=0$$ or $$-3x-2=0$$
$$\frac{7x}{7}=\frac{5}{7}$$ or $$\frac{-3x}{3}=\frac{2}{3}$$
$$x=\frac{5}{7}\ or\ x=-\frac{2}{3}$$
$$answer\ is\ option\ A$$

=>

10x^{-2}+x^{-1}-21=0
=> 10+x-21x^2=0 by multiplying x^2
=> 21x^2-x-10=0 by multiplying (-1)
=> (3x+2)(7x-5)=0
=> x = -2/3 or x = 5/7

Therefore, the answer is A.
Answer: A