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If f(x) = 3x^2  tx + 5 is tangents to xaxis, what is the value of a
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Updated on: 18 Jan 2016, 23:37
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If f(x) = 3x^2  tx + 5 is tangents to xaxis, what is the value of a positive number t? A. 2√15 B. 4√15 C. 3√13 D. 4√13 E. 6√15
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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.
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Re: If f(x) = 3x^2  tx + 5 is tangents to xaxis, what is the value of a
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18 Jan 2016, 23:58
fattty wrote: If f(x) = 3x^2  tx + 5 is tangents to xaxis, 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 \(2*\sqrt{15}\) Answer (A)
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If f(x) = 3x^2  tx + 5 is tangents to xaxis, what is the value of a
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13 Mar 2016, 22:13
fattty wrote: If f(x) = 3x^2  tx + 5 is tangents to xaxis, 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 , Xaxis 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 \(D=b^24*a*c\) Ergo, since D=0 \(t^24*3*5=0\) \(t^2=60\) \(t=2\sqrt{15}\) KUDOS if you liked the solution. Kanishk
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Re: If f(x) = 3x^2  tx + 5 is tangents to xaxis, what is the value of a
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19 Jan 2016, 00:50
fattty wrote: If f(x) = 3x^2  tx + 5 is tangents to xaxis, 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 xaxis, 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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Re: If f(x) = 3x^2  tx + 5 is tangents to xaxis, what is the value of a
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15 Mar 2016, 01:23
fattty wrote: If f(x) = 3x^2  tx + 5 is tangents to xaxis, 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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Re: If f(x) = 3x^2  tx + 5 is tangents to xaxis, what is the value of a
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15 Mar 2016, 06:16
Chiragjordan wrote: fattty wrote: If f(x) = 3x^2  tx + 5 is tangents to xaxis, 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 Xaxis, meaning the curve just touches the Xaxis 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 xaxis, what is the value of a
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23 Mar 2016, 12:16
fattty wrote: If f(x) = 3x^2  tx + 5 is tangents to xaxis, 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 > \(\sqrt{b^24*a*c} = 0\) > \(\sqrt{(t)^2 4*5*3} = 0\) > \(t^2 4*5*3 = 0\) > \(t^2 = 60\) > \(t = 2 \sqrt{15}\) 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 xaxis, what is the value of a
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Re: If f(x) = 3x^2  tx + 5 is tangents to xaxis, what is the value of a
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