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fattty
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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fattty
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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fattty
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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IMO A

f(X) is a parabola facing up.

So if f(X) is tangent to X axis - meaning intersects X axis at (a,0) ................(1)

The minimum value of f(x) is given by (-b/2a, c- (b^2/4a))

From (1) we can infer that c - b^2 /4a must be zero.

5 - t^2/12 = 0

>> t = 2 * Sqrt(5)
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If the quadratic formula is tangent to x-axis. That being said, there is only one answer.
Which means b^2-4ac=0
t^2-4(3)(5)=0
t^2=60
t=2√15
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is this actually in the gmat seems like its reaching geometry.
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Yep, this question can be asked under focus edition as well

onlymalapink
is this actually in the gmat seems like its reaching geometry.
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