Question

For what value of a the quadratic equation 9 x square - 3a x + 1 is equal to zero has equal roots

Answer 1

Answer:-

a = ± 2

Given :-

$9x^2 -3ax + 1$

To find :-

The value of a for which the quadratic equation has equal roots.

Solution:-

For equal roots we have :-

$\huge \boxed {D = 0}$

Where,

$D = b^2 -4ac$

we have,

$a = 9 , b = -3a , c = 1$

Put the given value,

→$(-3a) ^2 -4\times 9 \times 1 = 0$

→$9a^2 -36 = 0$

→$9a^2 = 36$

→$a^2 = \dfrac{36}{9}$

→$a^2 = 4$

→$a = \sqrt{4}$

→$a = \pm{2}$

hence,

The value of a for which the quadratic equation have real roots is a = ± 2.

Answer 2

GIVEN :

Quadratic Equation : 9x² - 3ax + 1

The quadratic equation has equal roots.

We know that,

Discriminant = 0 when equation has equal roots.

Discriminant Formula = b² - 4ac

From the above equation,

a = 9 b = -3a c = 1

b² - 4ac = 0

(-3a)² - 4(9)(1) = 0

=> 9a² - 36 = 0

=> 9a² = 36

=> a² = 36/9

=> a² = 4

=> a = √4

=> a = ±2

Therefore, the value of a is ±2.