Question

by using quadratic formula solve for x : 12abx²-(9a²-8b²)x-6ab=0​

Answer 1

Answer :

$\boxed{\bf x=\frac{3a}{4b} ,\frac{-2b}{3a} }$

Step-by-step explanation :

➤ Quadratic Polynomials :

   ✯ It is a polynomial of degree 2

   ✯ General form :

             ax² + bx + c  = 0

              $\boxed{\bf x=\frac{-b\pm\sqrt{b^2-4ac} }{2a} }$

                           

   ✯ Determinant, D = b² - 4ac

   ✯ Based on the value of Determinant, we can define the nature of roots.

           D > 0 ; real and unequal roots

           D = 0 ; real and equal roots

           D < 0 ; no real roots i.e., imaginary

   ✯ Relationship between zeroes and coefficients :

             ✩ Sum of zeroes = -b/a

             ✩ Product of zeroes = c/a

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Given polynomial,

12abx² - (9a² - 8b²)x - 6ab = 0

     $\bf x=\frac{-[-(9a^2-8b^2)]\pm\sqrt{[-(9a^2-8b^2)]^2-4(12ab)(-6ab)} }{2(12ab)} \\\\\\\ x=\frac{(9a^2-8b^2)\pm\sqrt{(9a^2-8b^2)^2+288a^2b^2} }{24ab} \\\\\\ x= \frac{(9a^2-8b^2)\pm\sqrt{81a^4+64b^4-2(9a^2)(8b^2)+288a^2b^2} }{24ab} \\\\\\ x=\frac{(9a^2-8b^2)\pm\sqrt{81a^4+64b^4-144a^2b^2+288a^2b^2} }{24ab} \\\\\\x=\frac{(9a^2-8b^2)\pm\sqrt{81a^4+64b^4+144a^2b^2} }{24ab}$

     $\bf x=\frac{(9a^2-8b^2)\pm\sqrt{(9a^2+8b^2)^2} }{24ab} \\\\\\ x=\frac{(9a^2-8b^2)\pm(9a^2+8b^2)}{24ab} \\\\\\ \implies x=\frac{(9a^2-8b^2)+9a^2+8b^2}{24ab} \\\\ \implies x =\frac{18a^2}{24ab} \\\\ \implies x=\frac{3a}{4b} \\\\ (or) \\\\ \implies x=\frac{(9a^2-8b^2)-(9a^2+8b^2)}{24ab} \\\\ \implies x=\frac{9a^2-8b^2-9a^2-8b^2}{24ab} \\\\ \implies x=\frac{-16b^2}{24ab} \\\\ \implies x=\frac{-2b}{3a}$

∴ x = 3a/4b , -2b/3a