Solving for Repeated Solution in Quadratic Equation

The given equation is $12x^2 - ax + 7 = ax^2 + 9x + 3$.

To find the value of '$a$', we first rearrange the equation into the standard quadratic form $Ax^2 + Bx + C = 0$.

• Combine like terms: $12x^2 - ax^2 - ax - 9x + 7 - 3 = 0$
• Group terms: $(12 - a)x^2 + (-a - 9)x + 4 = 0$
• Standard form: $(12 - a)x^2 - (a + 9)x + 4 = 0$

For a quadratic equation to have only one (repeated) solution, its discriminant ($\Delta$) must be zero. The discriminant is calculated as $\Delta = B^2 - 4AC$. Here, $A = (12 - a)$, $B = -(a + 9)$, and $C = 4$.

Calculating Discriminant for 'a'

Set the discriminant to zero:

$ \Delta = B^2 - 4AC = 0 $

Substitute the coefficients:

$ (-(a + 9))^2 - 4(12 - a)(4) = 0 $

Simplify the equation:

• $(a + 9)^2 - 16(12 - a) = 0$
• $(a^2 + 18a + 81) - (192 - 16a) = 0$
• $a^2 + 18a + 81 - 192 + 16a = 0$
• $a^2 + 34a - 111 = 0$

Finding Positive Integral Solution for 'a'

Solve the quadratic equation for '$a$': $a^2 + 34a - 111 = 0$.

We can factor this equation. We need two numbers that multiply to -111 and add up to 34. These numbers are 37 and -3.

$ (a + 37)(a - 3) = 0 $

This gives two possible values for '$a$':

• $a + 37 = 0 \implies a = -37$
• $a - 3 = 0 \implies a = 3$

The question asks for the positive integral solution. Therefore, the value of '$a$' is 3.