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08 Jan 2019, 05:00
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aeros232
\(? = n - m\)
\(n > m\,\,\,{\rm{are}}\,\,{\rm{the}}\,\,{\rm{roots}}\,\,{\rm{of}}\,\,{\rm{the}}\,\,{\rm{equation}}\,\,\,{x^2} + ax + b = 0\,\,\,\,\left( * \right)\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{sum/product}}} \,\,\,\,\,\,\,\,\left\{ \matrix{\\
\,m + n = - a \hfill \cr \\
\,mn = b \hfill \cr} \right.\)
\(\left( 1 \right)\,\,4b = {a^2} - 4\,\,\,\,\, \Rightarrow \,\,\,\,\,\Delta = {a^2} - 4b = 4\)
\(\left( * \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{\\
\,n = {{ - a + \sqrt \Delta } \over 2} = - {a \over 2} + 1 \hfill \cr \\
\,m = {{ - a - \sqrt \Delta } \over 2} = - {a \over 2} - 1 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = n - m = 2\)
\(\left( 2 \right)\,\,b = 0\,\,\,\, \Rightarrow \,\,\,\,mn = 0\,\,\,\,\,\)
\(\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {n,m} \right) = \left( {1,0} \right)\,\,\,\, \Rightarrow \,\,\,\,a = - 1\,\,\,\,\, \Rightarrow \,\,\,\,{x^2} - x = 0\,\,\,\,\left( {{\rm{viable}}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 1 \hfill \cr \\
\,{\rm{Take}}\,\,\left( {n,m} \right) = \left( {2,0} \right)\,\,\,\, \Rightarrow \,\,\,\,a = - 2\,\,\,\,\, \Rightarrow \,\,\,\,{x^2} - 2x = 0\,\,\,\,\left( {{\rm{viable}}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 2\,\, \hfill \cr} \right.\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Updated on: 15 Jan 2019, 06:57
Originally posted by BrentGMATPrepNow on 14 Jan 2019, 09:16.
Last edited by BrentGMATPrepNow on 15 Jan 2019, 06:57, edited 1 time in total.
Last edited by BrentGMATPrepNow on 15 Jan 2019, 06:57, edited 1 time in total.
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aeros232
Given: The graph of y = x² + ax + b passes through the points (m, 0) and (n, 0)
Target question: What is the value of n – m ?
This is a good candidate for rephrasing the target question.
The quadratic formula tells us that, if ax² + bx + c = 0, then \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
We're told that (m, 0) and (n, 0) are on the graph of y = x² + ax + b.
In other words, m and n are the two x-values that result in y = 0
In other words, m and n are the two solutions to the equation x² + ax + b = 0
When we apply the quadratic formula to the equation x² + ax + b = 0, we get two solutions:
SOLUTION 1
\(x=\frac{-a+\sqrt{a^2-4(1)(b)}}{2(1)}\)
\(=\frac{-a+\sqrt{a^2-4b}}{2}\)
SOLUTION 2
\(x=\frac{-a-\sqrt{a^2-4(1)(b)}}{2(1)}\)
\(=\frac{-a-\sqrt{a^2-4b}}{2}\)
So, one of the two solutions (above) is equal to m and the other is equal to n
We're told that m < n
So, it must be the case that n = solution 1, and m = solution 2 (since one solution has + \(\sqrt{{a^2 - 4(1)(b)}}\) and the other solution has - \(\sqrt{{a^2 - 4(1)(b)}}\)
That is, \(m=\frac{-a-\sqrt{a^2-4b}}{2}\)
and \(n=\frac{-a+\sqrt{a^2-4b}}{2}\)
So, \(n - m = \frac{-a+\sqrt{a^2-4b}}{2} - \frac{-a-\sqrt{a^2-4b}}{2}\)
\(= \frac{2\sqrt{a^2-4b}}{2}\)
\(= \sqrt{a^2-4b}\)
REPHRASED target question: What is the value of √(a² - 4b)?
Now that we've REPHRASED the target question, it will be easy to analyze the statements
Aside: I've posted a video (below) with tips on rephrasing the target question
Statement 1: 4b = a² – 4
Take √(a² - 4b) and replace 4b with a² – 4
We get: √[a² - (a² – 4 )]
Simplify to get: √(4)
Evaluate to get: 2
So, the answer to the REPHRASED target question is √(a² - 4b) = 2
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT
Statement 2: b = 0
Take √(a² - 4b) and replace b with 0
We get: √[a² - 4(0)]
Simplify to get: √(a²)
Since we don't know the value of a, there's no way to determine the value of √(a²)
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT
Answer: A
Cheers,
Brent
RELATED VIDEO FROM OUR COURSE
14 Jan 2019, 23:11
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GMATPrepNow
Dear GMATPrepNow Brent,
Although you reached solution, you swapped m & n.
The question states that n>m.....So n must have the term with positive (solution 1) and m with negative (solution 2) to match.
Please see the highlight, it must be n-m = 2 to match the question stem.
