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18 Jan 2022, 02:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
60% (01:18) correct
40%
(01:19)
wrong
based on 42
sessions
History
Date
Time
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Not Attempted Yet
What is the product of the roots of the equation \(x^2 - 4x + k = 3\) ?
(1) One of the roots of the equation \(x^2 - 4x + k = 3\) is 1
(2) \(k = 6\)
(1) One of the roots of the equation \(x^2 - 4x + k = 3\) is 1
(2) \(k = 6\)
Updated on: 18 Jan 2022, 03:51
Originally posted by ElninoEffect on 18 Jan 2022, 03:20.
Last edited by ElninoEffect on 18 Jan 2022, 03:51, edited 1 time in total.
Last edited by ElninoEffect on 18 Jan 2022, 03:51, edited 1 time in total.
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Bunuel
Equation: x^2 - 4x + k = 3
Can be rewritten as x^2 - 4x + k-3 = 0
So product of the equation = c/a = k-3/1. | c= k-3 a =1
So basically we need the value of K.
(1) One of the roots of the equation \(x^2 - 4x + k = 3\) is 1
Let the other root be R.
Sum of the roots = -b/a = 4
One root is 1
So, 1+ R = 4
R = 3
Product = 3.
Sufficient.
(2) \(k = 6\)
We are getting the value of K, we can find out the product of the roots.
Sufficient.
D is the answer.
18 Jan 2022, 03:42
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1) Tells us that one root of \(x^2 - 4x + k = 3\) is 1.
For the form \(ax^2 + bx + c\), the sum of the roots will equal \(-b\). If one of the roots is 1, then we know that after factorising the quadratic we will have a bracket with \((x - 1)\). As the sum of the roots must equal to \(-b\), the other root will have to be \(3\).
With the roots being 1 and 3, their product will be \(3\)
SUFFICIENT
2) Plugging in k = 6 into \(x^2 - 4x + k = 3\) gives:
\(x^2 - 4x + 6 = 3\)
\(x^2 - 4x + 3 = 0\)
\((x - 3)(x - 1)\)
x = 3 or x = 1
Product of roots: \(3*1 = 3\)
SUFFICIENT
Answer D
For the form \(ax^2 + bx + c\), the sum of the roots will equal \(-b\). If one of the roots is 1, then we know that after factorising the quadratic we will have a bracket with \((x - 1)\). As the sum of the roots must equal to \(-b\), the other root will have to be \(3\).
With the roots being 1 and 3, their product will be \(3\)
SUFFICIENT
2) Plugging in k = 6 into \(x^2 - 4x + k = 3\) gives:
\(x^2 - 4x + 6 = 3\)
\(x^2 - 4x + 3 = 0\)
\((x - 3)(x - 1)\)
x = 3 or x = 1
Product of roots: \(3*1 = 3\)
SUFFICIENT
Answer D
