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14 May 2020, 02:08
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
53% (02:18) correct
47%
(02:21)
wrong
based on 245
sessions
History
Date
Time
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Not Attempted Yet
[GMAT math practice question]
If \(a\) and \(b\) are roots of the quadratic equation \(x^2 - 4x + 1 = 0\), what is \(\sqrt{a}+\sqrt{b}\)?
A. \(2\sqrt{2} \)
B. \(±2\sqrt{2} \)
C. \(\sqrt{6}\)
D. \(±\sqrt{6} \)
E. \(±8\)
If \(a\) and \(b\) are roots of the quadratic equation \(x^2 - 4x + 1 = 0\), what is \(\sqrt{a}+\sqrt{b}\)?
A. \(2\sqrt{2} \)
B. \(±2\sqrt{2} \)
C. \(\sqrt{6}\)
D. \(±\sqrt{6} \)
E. \(±8\)
14 May 2020, 02:43
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Bookmarks
a+b = 4; a*b=1
\((\sqrt{a}+\sqrt{b})^2\)
\(= a+b+2 \sqrt{a*b}\)
\(= 4+2*\sqrt{1}\)
\(=6\)
\(\sqrt{a}+\sqrt{b}\) = \(\sqrt{6}\)
\((\sqrt{a}+\sqrt{b})^2\)
\(= a+b+2 \sqrt{a*b}\)
\(= 4+2*\sqrt{1}\)
\(=6\)
\(\sqrt{a}+\sqrt{b}\) = \(\sqrt{6}\)
MathRevolution
[GMAT math practice question]
If \(a\) and \(b\) are roots of the quadratic equation \(x^2 - 4x + 1 = 0\), what is \(\sqrt{a}+\sqrt{b}\)?
A. \(2\sqrt{2} \)
B. \(±2\sqrt{2} \)
C. \(\sqrt{6}\)
D. \(±\sqrt{6} \)
E. \(±8\)
If \(a\) and \(b\) are roots of the quadratic equation \(x^2 - 4x + 1 = 0\), what is \(\sqrt{a}+\sqrt{b}\)?
A. \(2\sqrt{2} \)
B. \(±2\sqrt{2} \)
C. \(\sqrt{6}\)
D. \(±\sqrt{6} \)
E. \(±8\)
17 May 2020, 19:54
Kudos
Bookmarks
=>
We have \((x - a)(x - b) = x^2 – (a + b)x + ab = x^2 - 4x + 1.\) Thus \(a + b = 4\) and \(ab = 1.\)
\((√a + √b)^2 = (√a + √b)(√a + √b) = a + 2√ab + b = a + b + 2√ab = 4 + 2√1 = 6.\)
Thus, we have \(√a + √b = √6\) since \(√a\) and \(√b\) are greater than or equal to \(0\).
Therefore, C is the answer.
Answer: C
We have \((x - a)(x - b) = x^2 – (a + b)x + ab = x^2 - 4x + 1.\) Thus \(a + b = 4\) and \(ab = 1.\)
\((√a + √b)^2 = (√a + √b)(√a + √b) = a + 2√ab + b = a + b + 2√ab = 4 + 2√1 = 6.\)
Thus, we have \(√a + √b = √6\) since \(√a\) and \(√b\) are greater than or equal to \(0\).
Therefore, C is the answer.
Answer: C
