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

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

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New post Updated on: 14 Jun 2016, 02:44
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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)

Originally posted by YAObolt on 14 Jun 2016, 02:40.
Last edited by Bunuel on 14 Jun 2016, 02:44, edited 1 time in total.
Edited the question.
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Re: If the curve represented by y = x^2 – 5x + t intersects with the  [#permalink]

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New post 14 Jun 2016, 07:39
Curve represented by y = x^2 – 5x + t intersects with the x-axis at points where y = 0.
One of the points is (-1,0) Plugging in the equation => 0 = 1 + 5 + t => t = -6
Plugging t back to the equation now we have y = x^2 – 5x - 6, where y = 0. This equation gives 2 solution x= -1 and 6. We already know about the point (-1,0) so the other point is (6,0). Hence D.
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If the curve represented by y = x^2 – 5x + t intersects with the  [#permalink]

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New post 14 Jun 2016, 09:12
YAObolt wrote:
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)


Hi,

we can work without finding value of t also..
when the curve cuts x axis, y = 0..

so the quadratic equation = \(x^2-5x+t = 0\)...
in a normal equation \(ax^2+bx+c=0\)..
SUM of the values of x or roots is \(\frac{-b}{a}\)and product =\(\frac{c}{a}\)..

here a=1, b = -5 and c = t..
now one value of x = -1....
so \(-1+ x_2 = -(\frac{-5}{1}) = 5....... x_2 = 5+1 = 6\)....

so point of intersection = (6,0)
D
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If the curve represented by y = x^2 – 5x + t intersects with the   [#permalink] 14 Jun 2016, 09:12
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