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![If 2 + √3 is one root of the equation, [m]x^2 - ax + b = 0[/m], ......](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If 2 + √3 is one root of the equation, [m]x^2 - ax + b = 0[/m], ......"){kind=icon}
Updated on: 26 Jun 2019, 03:42
Originally posted by EgmatQuantExpert on 19 Jun 2019, 06:44.
Last edited by EgmatQuantExpert on 26 Jun 2019, 03:42, edited 1 time in total.
Last edited by EgmatQuantExpert on 26 Jun 2019, 03:42, edited 1 time in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (01:56) correct
36%
(02:29)
wrong
based on 489
sessions
History
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If 2 + √3 is one root of the equation, \(x^2 - ax + b = 0\), where a, b are positive numbers, then what is the value of a + b?
- (A) 0
(B) 1
(C) 2√3
(D) 4
(E) 5
19 Jun 2019, 11:30
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If one root of the quadratic equation is \(2+\sqrt{3},\)the other root is \(2-\sqrt{3}.\)Now, sum of roots = a = \(2+\sqrt{3}+2-\sqrt{3}\) = 4
product of roots = b = \((2+\sqrt{3})(2-\sqrt{3})\)= \(2^2 -{\sqrt{3}}^2\) = 4-3 = 1
Hence, a + b = 4 + 1 = 5 (Option E)
product of roots = b = \((2+\sqrt{3})(2-\sqrt{3})\)= \(2^2 -{\sqrt{3}}^2\) = 4-3 = 1
Hence, a + b = 4 + 1 = 5 (Option E)
26 Jun 2019, 03:44
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Solution
Given:
- • 2 + √3 is one root of the quadratic equation, \(x^2 - ax + b = 0\)
• a, b are positive numbers
To find:
- • The value of a + b
Approach and Working Out:
- • We are given that 2 + √3 is one root of the quadratic equation, \(x^2 - ax + b = 0\)
• And, we know that, if a +√ b is one root of a quadratic equation, then the other root will be a - √b.
- o Thus, 2 + √3, and 2 - √3 are the two roots
• Now, from the given quadratic equation, \(x^2 - ax + b = 0\)
- o Sum of the roots = -(-a) = a = (2 + √3) + (2 - √3) = 4
o Product of the roots = b = (2 + √3) * (2 - √3) = 4 – 3 = 1
• Therefore, a + b = 4 + 1 = 5
Hence, the correct answer is Option E.
Answer: E
