The integral value of a for which ax2 + ax + a = 2x2 - 3x - 6 has equal roots is 3 2 – 3 – 2

The given equation is ax² + ax + a = 2x² - 3x - 6.

First, we need to rearrange the equation to the standard quadratic form ax² + bx + c = 0.

So, we subtract 2x², add 3x and add 6 from both sides to get:

(a - 2)x² + (a + 3)x + (a + 6) = 0.

Now, for a quadratic equation to have equal roots, the discriminant (b² - 4ac) must be equal to zero.

So, we have (a + 3)² - 4*(a - 2)*(a + 6) = 0.

Solving this equation will give us the value of a.

Let's simplify the equation:

a² + 6a + 9 - 4a² + 8a + 24 = 0,

-3a² + 14a + 33 = 0.

Divide the equation by -1 to make the coefficient of a² positive:

3a² - 14a - 33 = 0.

Now, we can solve this quadratic equation for a using the quadratic formula a = [-b ± sqrt(b² - 4ac)] / 2a:

a = [14 ± sqrt((-14)² - 43(-33))] / 2*3,

a = [14 ± sqrt(196 + 396)] / 6,

a = [14 ± sqrt(592)] / 6,

a = [14 ± 2*sqrt(148)] / 6,

a = [7 ± sqrt(148)] / 3.

So, the integral value of a for which the given equation has equal roots is either 7 + sqrt(148) / 3 or 7 - sqrt(148) / 3. However, neither of these values is an integer, so there is no integral value of a for which the given equation has equal roots.