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If the curve represented by y = x^2 – 5x + t intersects with the x-axi

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If the curve represented by y = x^2 – 5x + t intersects with the x-axi  [#permalink]

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New post 07 Feb 2020, 03:37
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Question Stats:

80% (01:42) correct 20% (01:30) wrong based on 44 sessions

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If the curve represented by y = x^2 – 5x + t intersects with the x-axis at two points and one of the points is (–1, 0), what is the other point?

(A) (1, 0)
(B) (–2, 0)
(C) (5, 0)
(D) (6, 0)
(E) (3, 0)

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Re: If the curve represented by y = x^2 – 5x + t intersects with the x-axi  [#permalink]

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New post 07 Feb 2020, 03:47
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If the curve represented by y = x^2 – 5x + t intersects with the x-axis at two points and one of the points is (–1, 0), what is the other point?

(A) (1, 0)
(B) (–2, 0)
(C) (5, 0)
(D) (6, 0)
(E) (3, 0)

y = x^2 – 5x + t intersects at (-1,0) -> 0=1+5+t -> t=-6

Equation -> y = x^2 – 5x - 6

For other point to intersect with X-Axis -> y=0 -> x^2 – 5x - 6 = 0 -> (x-6)(x+1) = 0 -> x=6 y=0 is the other point -> Answer - D
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If the curve represented by y = x^2 – 5x + t intersects with the x-axi  [#permalink]

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New post Updated on: 08 Feb 2020, 21:48
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Standard form of quadratic equation is \(ax^2+bx+c=0\)
Given quadratic equation is \(y = x^2 – 5x + t\), comparing \(a=1, b=-5, c=t\)

One root is -1 and let other be \(M\)

As sum of the roots is \(\frac{-b}{a}\)

-1 + \(M\) = \(\frac{-(-5)}{1}\)
-1+\(M\) = 5
\(M\) = 6

IMO D

Originally posted by kapil1 on 07 Feb 2020, 03:48.
Last edited by kapil1 on 08 Feb 2020, 21:48, edited 1 time in total.
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Re: If the curve represented by y = x^2 – 5x + t intersects with the x-axi  [#permalink]

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New post 08 Feb 2020, 21:36
1
Solve for T, by putting values of x and y co-ordinates
y=x^2-5x+t
0=(-1)^2-5*-1+t
t=-6

Replace t in the original equation,
x^2-5x-6=0

Solve for X by splitting the middle term, we get x=-1 and x=6

Thus, D(6,0) is the correct option.
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Re: If the curve represented by y = x^2 – 5x + t intersects with the x-axi   [#permalink] 08 Feb 2020, 21:36
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