Question

Solve for x:a(a²+b²)x²+b²x-a​

Answer 1

x = a/(a² + b²) or x = -1/a

Explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

`x = (-b\pm√(b^2-4ac))/(2a) =(-b\pm√(D))/(2a)`

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

`\begin{array}{rcl}x & = & (-b\pm√(D))/(2a)\\\\ & = & \frac{-b^(2)\pm\sqrt{(b^(2)+2a^(2))^(2)}}{2a(a^(2) + b^(2))}\\\\ & = & (-b^(2)\pm(b^(2)+ 2a^(2)))/(2a(a^(2) + b^(2)))\\\\x = (-b^(2)+(b^(2) + 2a^(2)))/(2a(a^(2) + b^(2)))&\qquad& x =(-b^(2)-(b^(2) + 2a^(2)))/(2a(a^(2) + b^(2)))\\\\x =(-b^(2)+(b^(2) + 2a^(2)))/(2a(a^(2) + b^(2)))&\qquad& x =(-b^(2)-(b^(2) +2a^(2)))/(2a(a^(2) + b^(2)))\\\\\end{array}`

`\begin{array}{rcl}x = \large \boxed{\mathbf{(a)/(a^(2) + b^(2))}}&\qquad& x =(-b^(2)-(b^(2) +2a^(2)))/(2a(a^(2) + b^(2)))\\\\&\qquad& x =(-2b^(2)- 2a^(2))/(2a(a^(2) + b^(2)))\\\\&\qquad& x =(-2(a^(2)+ b^(2)))/(2a(a^(2) + b^(2)))\\\\&\qquad& x =\large \boxed{\mathbf{-(1)/(a)}}\\\\\end{array}`