Question

solve a2b2x2-(4b4-3a4)x-12a2b2=0 using quadratic formula​

Answer 1

Given is the following quadratic equation:-

          $\sf \rightarrow \quad (a^2 b^2 )x^2 - (4b^4 - 3a^4)x - 12a^2 2b^2=0$

We have to solve this quadratic equation using the quadratic which is also known as the Shri Dhanacharya formula:-

         $\sf \rightarrow \quad x= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where,

• a = coefficient of x²
• b = coefficient of x
• c = constant term

After comparing the given quadratic equation with the standard form of a quadratic equation i.e., ax² + bx + c = 0, we get

• a° = a²b²
• b° = -(4b⁴ - 3a⁴)
• c° = -12a²b²

Putting the following values in the formula, we get

$\sf \rightarrow \quad x = \dfrac{-(4b^4 - 3a^4) \pm \sqrt{\{-(4b^4 - 3a^4)\}^2 - 4(a^2b^2)(-12a^2b^2)}}{2a^2b^2}$

$\sf \rightarrow \quad x = \dfrac{-4b^4 + 3a^4 \pm \sqrt{16b^8 + 9a^8 - 24a^4b^4 - 4(-12a^4b^4)}}{2a^2b^2}$

$\sf \rightarrow \quad x = \dfrac{-4b^4 + 3a^4 \pm \sqrt{16b^8 + 9a^8+24a^4b^4}}{2a^2b^2}$

$\sf \rightarrow \quad x = \dfrac{-4b^4 + 3a^4 \pm \sqrt{(4b^4+3a^4)^2}}{2a^2b^2}$

$\sf \rightarrow \quad x = \dfrac{-4b^4 + 3a^4 \pm (4b^4 + 3a^4)}{2a^2b^2}$

Here, We have two cases, Let's proceed one by one:-

Case 1:

$\sf \rightarrow \quad x = \dfrac{\cancel{-4b^4} + 3a^4 \cancel{+ 4b^4} + 3a^4}{2a^2b^2}$

$\sf \rightarrow \quad x = \dfrac{ 3a^4+3a^4}{2a^2b^2}$

$\sf \rightarrow \quad \bf x = \dfrac{3a^2}{b^2}$

Case 2:

$\sf \rightarrow \quad x = \dfrac{-4b^4 + 3a^4 - (4b^4 + 3a^4)}{2a^2b^2}$

$\sf \rightarrow \quad x = \dfrac{-4b^4 \cancel{+ 3a^4} - 4b^4 \cancel{- 3a^4}}{2a^2b^2}$

$\sf \rightarrow \quad x = \dfrac{-4b^4 - 4b^4}{2a^2b^2}$

$\bf \rightarrow \quad x = \dfrac{-4b^2 }{a^2}$

$\sf So, \; the \; values\; of \;x\; are\; \bold{\dfrac{3a^2}{b^2}} \; and \; \bold{-\dfrac{4b^2}{a^2}}$