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16 Jun 2021, 09:02
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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:
75%
(hard)
Question Stats:
50% (01:53) correct
50%
(01:49)
wrong
based on 213
sessions
History
Date
Time
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At how many points does the graph \(y = ax^2 + bx + c\) intersect X-axis?
(1) a + b = 0
(2) a*b = 0 and c ≠ 0
(1) a + b = 0
(2) a*b = 0 and c ≠ 0
Source: https://www.GMATinsight.com
16 Jun 2021, 09:44
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Asked: At how many points does the graph \(y = ax^2 + bx + c\) intersect X-axis?
1) a+b = 0
b = - a
\(y = ax^2 - ax + c = a(x^2 - x + 1/4) + (c-a/4) = a(x-1/2)^2 + (c-a/4) = 0\)
\(x = 1/2 + - \sqrt{(a/4 - c)/a} \)
(a-4c)/4a >= 0
If a = 4c ; the graph intersects x-axis at 1 point.
If (a-4c)/4a > 0 ; the graph intersects x-axis at 2 points
If (a-4c)/4a < 0 ; the graph does not intersects x-axis at any point.
NOT SUFFICIENT
2) a*b = 0 and c ≠ 0
If a=0; y = bx + c = 0; x = - c/b; the graph intersects x-axis at 1 point
If b=0; y = ax^2 + c = 0; x^2 = -c/a; If c/a <0 the graph intersects x-axis at 2 points; But if c/a>0 the graph does not intersect x-axis at any point.
If a=b=0; y = -c; the graph does not intersect x-axis at any point
NOT SUFFICIENT
(1) + (2)
1) a+b = 0
2) a*b = 0 and c ≠ 0
Since a+b = 0 and a*b=0; a = b = 0
y = c; the graph does not intersect x-axis since c ≠ 0
SUFFICIENT
IMO C
1) a+b = 0
b = - a
\(y = ax^2 - ax + c = a(x^2 - x + 1/4) + (c-a/4) = a(x-1/2)^2 + (c-a/4) = 0\)
\(x = 1/2 + - \sqrt{(a/4 - c)/a} \)
(a-4c)/4a >= 0
If a = 4c ; the graph intersects x-axis at 1 point.
If (a-4c)/4a > 0 ; the graph intersects x-axis at 2 points
If (a-4c)/4a < 0 ; the graph does not intersects x-axis at any point.
NOT SUFFICIENT
2) a*b = 0 and c ≠ 0
If a=0; y = bx + c = 0; x = - c/b; the graph intersects x-axis at 1 point
If b=0; y = ax^2 + c = 0; x^2 = -c/a; If c/a <0 the graph intersects x-axis at 2 points; But if c/a>0 the graph does not intersect x-axis at any point.
If a=b=0; y = -c; the graph does not intersect x-axis at any point
NOT SUFFICIENT
(1) + (2)
1) a+b = 0
2) a*b = 0 and c ≠ 0
Since a+b = 0 and a*b=0; a = b = 0
y = c; the graph does not intersect x-axis since c ≠ 0
SUFFICIENT
IMO C
Archit3110
16 Jun 2021, 10:50
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given is eqn of parabola \(y = ax^2 + bx + c\)
where IF value + a upward curve and -ve a is downward curve and
curvature depends on value of B ; C is Y intercept
also we have three conditions
if
b^2-4ac>0 intersect at 2 points x axis
b^2-4ac=0 touches at x axis
b^2-4ac<0 does not not touch x axis
target of question At how many points does the graph \(y = ax^2 + bx + c\) intersect X-axis?
#1
a+b = 0
or say b=-a for b^2-4ac
we get a^2+4ac ; a *(a+4c)
value of c is not known
we can get values of b^2-4ac as
b^2-4ac>0 intersect at 2 points x axis
b^2-4ac=0 touches at x axis
b^2-4ac<0 does not not touch x axis
insufficient
#2
a*b=0 and c≠ 0
so either of a & b is 0 or both are =0
again not unique graph value can be determined
insufficient
from 1 &2
we get a=b=0 and c≠ 0
so b^2-4ac=0
one vertex at x axis
option C
sufficient
where IF value + a upward curve and -ve a is downward curve and
curvature depends on value of B ; C is Y intercept
also we have three conditions
if
b^2-4ac>0 intersect at 2 points x axis
b^2-4ac=0 touches at x axis
b^2-4ac<0 does not not touch x axis
target of question At how many points does the graph \(y = ax^2 + bx + c\) intersect X-axis?
#1
a+b = 0
or say b=-a for b^2-4ac
we get a^2+4ac ; a *(a+4c)
value of c is not known
we can get values of b^2-4ac as
b^2-4ac>0 intersect at 2 points x axis
b^2-4ac=0 touches at x axis
b^2-4ac<0 does not not touch x axis
insufficient
#2
a*b=0 and c≠ 0
so either of a & b is 0 or both are =0
again not unique graph value can be determined
insufficient
from 1 &2
we get a=b=0 and c≠ 0
so b^2-4ac=0
one vertex at x axis
option C
sufficient
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