If the roots of the quadratic equation x^2-4x-log(base3)a = 0 are real, then least value of \"a\"...

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10 Apr 2013 Quadratic equation

If the roots of the quadratic equation x^2-4x-log(base3)a = 0 are real, then least value of "a".....

OPtion

1) 1/9
2) 1/16
3) 1/25
4) 1/36
5) 1/49
6) 1/64
7) 1/81
8) 1/2
9) 1/18
10) None of these

Solution

Given equation are real
Therefore b^2-4ac is >= 0
Solve 16+4log(base3) a >= 0
Ans 1/81
Option 7

Correct Option: 7 Best Solution (7)

Dr.Dipin Singh   11 years AGO

x^2-4x-log(base3)a = 0

when a=1/81 = 3^(-4)
Then
x^2-4x-log(base3)a = 0 becomes
x^2-4x-log(base3)(3^(-4)) = 0
or
x^2-4x+4 = 0
(x-2)^2=0
x=2

For any value of a < 1/81
log(base3)a will result in equation
x^2-4x+p = 0 where p >4
and then equation will not have any real roots as
16-4p will be negative and
sqrt (16-4p) will be imaginary number

Like? Yes (6) | No (2)  

dan   11 years AGO

roots are real.
whose square root part must be >=0;
so,16+4log(base3)a>=0
or,a>=1/81.

Like? Yes (1) | No   

Mukesh Kr Singh   11 years AGO

for real root
4^2-4*1*(-log(base3)a)=0
a=1/81

Like? Yes (3) | No (3)  

Sanjeewani Indurkar   11 years AGO

for a=1/81, sqrt(b^2-4ac)=0 for given quadratic equation. So least value of "a" to get roots as real value will be 1/81

Like? Yes (4) | No (5)  

babu rao ganpat rao apte   11 years AGO

hey,,hello.. every one..

by real we mean b^2-4ac should be a +ve value...

thus..b=4...a=1...c=log(a)...

least value of b^2-4ac will be zero...which is only possible when u know..a=1/81....

so this was the answer.... enjoy urself.... :) :)

Like? Yes (2) | No (5)  

Rats   11 years AGO

To roots of equation to be real.
log(base3)a=-4
So, a=1/81

Like? Yes (1) | No (6)  

G REDDI SAI AKHIL   11 years AGO

X^2-4X-log(base3)1/81=0
X^2-4X-log(base3)3^-4=0
X^2-4X+4log(base3)3=0
X^2-4X+4=0
x^2-2x-2x+4=0
(x-2)(x-2)=0
x=2,x=2

Like? Yes (1) | No (7)