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Intern  Joined: 29 Nov 2015
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If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a  [#permalink]

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Question Stats: 60% (02:05) correct 40% (02:07) wrong based on 124 sessions

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If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a positive number t?

A. 2√15
B. 4√15
C. 3√13
D. 4√13
E. 6√15

Originally posted by fattty on 18 Jan 2016, 23:32.
Last edited by Bunuel on 18 Jan 2016, 23:37, edited 1 time in total.
Edited the question.
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Re: If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a  [#permalink]

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fattty wrote:
If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a positive number t?

A. 2√15
B. 4√15
C. 3√13
D. 4√13
E. 6√15

What is the meaning of this:
f(x), a quadratic in x, is tangent to x axis? It means that the graph of the quadratic touches x axis at only one point. So when y = 0, both roots are the same. So expression in x must be a perfect square when y = 0.

$$3x^2 - tx + 5 = 0$$

$$(\sqrt{3}x)^2 - tx + (\sqrt{5})^2 = 0$$

To get a perfect square which looks like $$a^2 - 2ab + b^2 = (a - b)^2$$, we must put $$tx = 2ab = 2*(\sqrt{3}x)*(\sqrt{5})$$

So t must be $$2*\sqrt{15}$$

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If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a  [#permalink]

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fattty wrote:
If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a positive number t?

A. 2√15
B. 4√15
C. 3√13
D. 4√13
E. 6√15

f(x) = $$3x^2 - tx + 5$$, is a quadratic equation.
Given , X-axis is a tangent to this Quadratic equation , implies there is only one root for this equation .

As a result,the Discriminant (D) for this quadratic equation must be equal to 0.
We know that $$D=b^2-4*a*c$$
Ergo, since D=0

$$t^2-4*3*5=0$$
$$t^2=60$$
$$t=2\sqrt{15}$$

KUDOS if you liked the solution.
Kanishk

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Math Expert V
Joined: 02 Aug 2009
Posts: 7958
Re: If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a  [#permalink]

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fattty wrote:
If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a positive number t?

A. 2√15
B. 4√15
C. 3√13
D. 4√13
E. 6√15

Hi,
the equation is of a parabola..
and since its a tangent to x-axis, it means the parabola has min value as 0..
and where does this occur at x=-b/2a, so here, at x= t/6, or t=6x..
substitute the value of eq as 0..
f(x) = 3x^2 - tx + 5 ..
3x^2 - tx + 5 = 0..
3x^2 - 6x*x + 5 =0..
or x=$$\sqrt{\frac{5}{3}}$$...
so t=6x=6$$\sqrt{\frac{5}{3}}$$=2*$$\sqrt{15}$$
A
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GRE 1: Q169 V154 Re: If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a  [#permalink]

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fattty wrote:
If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a positive number t?

A. 2√15
B. 4√15
C. 3√13
D. 4√13
E. 6√15

I am still confused about this Tangent to the axis thing...
How do you know that both the roots are the same..
chetan2u i did not understand your solution either..
can you Elaborate or provide the theory on the same..
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Math Expert V
Joined: 02 Aug 2009
Posts: 7958
Re: If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a  [#permalink]

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Chiragjordan wrote:
fattty wrote:
If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a positive number t?

A. 2√15
B. 4√15
C. 3√13
D. 4√13
E. 6√15

I am still confused about this Tangent to the axis thing...
How do you know that both the roots are the same..
chetan2u i did not understand your solution either..
can you Elaborate or provide the theory on the same..

Hi,

let me divide the Q in different sections..

A Linear Equation- a straight line
QUADRATIC equation:- the equation is of a parabola..
so if you have an equation like ax^2+bx+c=y.. and you plot the corresponding value of (x,y) on a graph
you will get a parabola- a U shaped or an inverted U shaped..
this has minimum value at the center of the Curve if it is U shaped and MAX value at the center of Curve if it is inverted U..

Now the equation or the curve is tangent to X-axis, meaning the curve just touches the X-axis at one point and thus this is the min value..
At X- axis y=0..
And the Curve is Symmetrical about a line determined by x=-b/2a..
so this will ofcourse meet the curve at the max/min value
----- we can derive this formula but not required here---

so the min value=0 at x=-b/2a
in equation f(x) = 3x^2 - tx + 5 .. b=-t and a=3
so x=-(-t)/2*3=t/6..
so t=6x..
and thereafter its all calculations..

f(x) = 3x^2 - tx + 5 ..
at x=t/6, y=0.. SO
3x^2 - tx + 5 = 0..
3x^2 - 6x*x + 5 =0..
or x=$$\sqrt{\frac{5}{3}}$$...
so t=6x=6$$\sqrt{\frac{5}{3}}$$=2*$$\sqrt{15}$$
A

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If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a  [#permalink]

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1
fattty wrote:
If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a positive number t?

A. 2√15
B. 4√15
C. 3√13
D. 4√13
E. 6√15

A more straightforward way is to realize that for a quadratic equation to be a perfect square ---> Discriminant = D = 0 ---> $$\sqrt{b^2-4*a*c} = 0$$

---> $$\sqrt{(-t)^2 -4*5*3} = 0$$ ---> $$t^2 -4*5*3 = 0$$ ---> $$t^2 = 60$$ ---> $$t = 2 \sqrt{15}$$

P.S.: For any quadratic equation $$ax^2+bx+c = 0$$--->

1. 2 real unequal roots ---> D > 0
2. 2 equal roots (in the case of perfect squares) ---> D= 0
3. No real roots ---> D < 0
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Re: If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a  [#permalink]

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_________________ Re: If f(x) = 3x^2 - tx + 5 is tangents to x-axis, what is the value of a   [#permalink] 19 Mar 2019, 23:53
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