Question

find the roots of the following equation by applying quadratic formula​

Answer 1

Answer:

x = (a-b)/6 and x = (a+b)/6

Step-by-step explanation:

Answer 2

the roots of the equations,

$36 {x}^{2} - 12ax + ( {a}^{2} + {b}^{2}) = 0$

$x = \frac{ - bx + - \sqrt{ {b}^{2} - 4ac } }{2a}$

$x = \frac{12a + - \sqrt{ {(12a)}^{2} - 4 \times 36 \times ( {a}^{2} - {b}^{2} ) } }{2 \times 36} \\ x = \frac{12a + - \sqrt{ {(12a)}^{2} - 4 \times 36 \times ( {a}^{2} - {b}^{2} ) } }{72}$

$x = \frac{a + - \sqrt{ {a}^{2} - ( {a}^{2} - {b}^{2} ) } }{6} \\ x = \frac{a + - b }{6}$

Take +ve sign,

$x = \frac{a + b }{6}$

Take -ve sign,

$x = \frac{a - b }{6}$