Question

for what value of a 30 a x square - 6 X + 1 equal to zero has no real roots​

Answer 1

GIVEN :

For what value of 'a'  $30ax^2 - 6x + 1=0$ has no real roots​

TO FIND :

For what value of 'a'  $30ax^2 - 6x + 1=0$ has no real roots​

SOLUTION :

Given quadratic equation is $30ax^2 - 6x + 1=0$

For a quadratic equation given in the form below :

$ax^2 +bx + c=0$

The equation will have  the conditions

• Two distinct real roots when

     [tex]Discriminant = b^2 - 4 ac > 0 [/tex]

• Two equal real roots when

      $Discriminant = b^2- 4 a c = 0$

• No real roots when

      [tex]Discriminant =  b^2- 4 a c < 0 [/tex]

Since, when the quadratic equation will have no real roots then, Discriminant of this equation will be less than zero .

i.e., [tex]Discriminant =  b^2- 4 a c < 0 [/tex]

$Discriminant = (-6)^2 - 4 (30a)(1) < 0$  (where a=30a , b=-6 and c=1)

 $(-6)^2 - 4 (30a)(1) < 0$

 36 - 120a < 0

36 < 120a

 120 a > 36

 $a >\frac{36}{120}$

∴ $a >\frac{3}{10}$

∴ For value  $a >\frac{3}{10}$ , given quadratic equation $30ax^2 - 6x + 1=0$ will have no real roots.

Answer 2

Given:

A equation :- $30ax^{2} -6x+1 =0$ has no real roots.

To Find:

Value of a ?

Solution:

Since, given equation is -

$30ax^{2} -6x+1 =0$

Since, it has no real roots, therefore

we have, D ≤ 0

      Since, $D= B^{2} - 4AC$ here, A=30a , B=-6 and C= 1

Therefore,

            $(-6)^{2} - 4 (30a)(1) \leq 0\\36-120 a\leq 0\\36\leq 120 a\\3/10\leq a\\a\geq 3/10$

Hence, $a\geq 3/10$.