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# Dealing with Tangents on the GMAT

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Math Expert
Joined: 02 Sep 2009
Posts: 53831
Dealing with Tangents on the GMAT  [#permalink]

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23 Mar 2016, 12:10
2
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Dealing with Tangents on the GMAT

BY Karishma, VERITAS PREP

Considering a two dimensional figure, a tangent is a line that touches a curve at a single point. Here are some examples of tangents:

In each of these cases, the line touches the curve at a single point. In the case of a circle, when you draw the radius of the circle from the center to the point of contact with the tangent, the radius is perpendicular to the tangent (as demonstrated in the figure on the right, above). A question discussing this concept is given in our post here.

Today, we will look at a question involving a tangent to a parabola:

If $$f(x) = 3x^2 – tx + 5$$ is tangent to the x-axis, what is the value of the positive number t?
(A) 2√15
(B) 4√15
(C) 3√13
(D) 4√13
(E) 6√15

Let’s first try to understand what the question is saying.

f(x) is a tangent to the x-axis. We know that the x-axis is a straight line, so f(x) must be a curve. A quadratic equation, such as our given equation of $$f(x) = 3x^2 -tx +5$$, gives a parabola. Since the x^2 term in the equation is positive, the parabola would be facing upwards and touching the x-axis at a single point, such as:

Since the parabola touches the x-axis in only one point, it means the quadratic has only one root, or in other words, the quadratic must be a perfect square.

Therefore, $$f(x) = 3x^2 – tx + 5 = (\sqrt {3}x)^2 – tx + (\sqrt 5)^2$$

To get f(x) in the form $$a^2 – 2ab + b^2 = (a – b)^2$$,

$$tx = 2ab = 2\sqrt 3*x * \sqrt 5$$

$$t = 2\sqrt 15$$

Note that if t takes this value, the quadratic will have only one root.

Plugging this value of t back into our equation, we will get: $$f(x) = (\sqrt {3}x)^2 – 2(\sqrt{15})(x) + (\sqrt 5)^2$$

$$f(x) = (\sqrt 3*x – \sqrt 5)^2$$

We know that the root of f(x) is the point where the value of the y coordinate is 0. Therefore:

$$(\sqrt 3*x – \sqrt 5)^2 = 0$$

$$x = \frac{(\sqrt 5)}{(\sqrt 3)}$$

At this x co-ordinate, the parabola will touch the x axis.

[This calculation was shown only to help you completely understand the question. We could have easily stopped at t = 2(√15).]

Therefore, our answer is A. This question is discussed HERE.

The question can be solved in various other ways – think of how, and write your thoughts in the comments below!

Attachment:

QWQW-1-e1458577171719.jpg [ 15.42 KiB | Viewed 2819 times ]

Attachment:

QWQW-2-e1458577805347.jpg [ 7.64 KiB | Viewed 2804 times ]

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Joined: 02 Sep 2009
Posts: 53831
Re: Dealing with Tangents on the GMAT  [#permalink]

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23 Mar 2016, 12:14
THEORY AND PRACTICE QUESTIONS:

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Joined: 26 Jun 2014
Posts: 78
Re: Dealing with Tangents on the GMAT  [#permalink]

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24 Mar 2016, 22:48
1
This seems to be more of a geometry problem and less like a GMAT question. Have these type of questions ever been asked in the Official GMAT exam ?
Intern
Joined: 21 Sep 2014
Posts: 16
Location: Ukraine
GMAT 1: 700 Q47 V38
GPA: 3.57
Re: Dealing with Tangents on the GMAT  [#permalink]

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21 Jun 2017, 03:39
3
If $$f(x) = 3x^2 – tx + 5$$ is tangent to the x-axis, what is the value of the positive number t?
(A) 2√15
(B) 4√15
(C) 3√13
(D) 4√13
(E) 6√15

Since the parabola touches the x-axis in only one point, it means the quadratic has only one root.
The QE ax^2+bx+c=0 has only one root when its discriminant equals 0
D=b^2-4ac = 0
a=3
b=t
c=5

t^2-4*3*5=0; t^2=60; t=√60=2√15
answer A
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Re: Dealing with Tangents on the GMAT  [#permalink]

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23 Feb 2019, 05:00
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Re: Dealing with Tangents on the GMAT   [#permalink] 23 Feb 2019, 05:00
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# Dealing with Tangents on the GMAT

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