We have to factorize 25 - x 2 = 0 x 2 = 25 x 2 - 25 = 0 x 2 - 5 2 = 0 Now, a 2 - b 2 = a - b a + b x - 5 x + 5 = 0 By zero product rule x - 5 = 0 or x + 5 = 0 x = + 5 , - 5

Solve x 2 + 2 x 3 - 8 3 = 0

We have to factorize 3 x 2 + 2 x - 8 = 0 Now, 3 x 2 + 2 x - 8 = 0 3 x 2 + 6 x - 4 x - 8 = 0 3 x ( x + 2 ) - 4 ( x + 2 ) = 0 3 x - 4 x + 2 = 0 By zero product rule 3 x - 4 = 0 or x + 2 = 0 x = + 4 3 , - 2

Solve x 2 + 2 = 9 x 2

We have to factorize x 2 - 9 x 2 + 2 = 0 Taking L.C.M of denominators, we get 2 x 2 - 9 x + 4 = 0 × 2 2 x 2 - 8 x - x + 4 = 0 2 x x - 4 - 1 ( x - 4 ) = 0 2 x - 1 x - 4 = 0 By zero product rule 2 x - 1 = 0 or x - 4 = 0 x = + 1 2 , 4 .

Solve x 2 - 11 x 5 = 12 5

We will factorize x 2 - 11 x 5 = 12 5 x 2 - 11 x 5 - 12 5 = 0 Taking L.C.M of denominators and solving further 5 x 2 - 11 x - 12 = 0 5 x 2 - 15 x + 4 x - 12 = 0 5 x x - 3 - 4 ( x - 3 ) = 0 5 x - 4 x - 3 = 0 By zero product rule 5 x - 4 = 0 or x - 3 = 0 x = 4 5 , 3 .

Solve x + 15 2 x = 11 2

We will factorize x + 15 2 x = 11 2 Taking L.C.M and solving further, we get 2 x 2 + 15 2 x = 11 2 2 x 2 + 15 - 11 x = 0 2 x 2 - 11 x + 15 = 0 2 x 2 - 6 x - 5 x + 15 = 0 2 x x - 3 - 5 ( x - 3 ) = 0 2 x - 5 x - 3 = 0 By zero product rule 2 x - 5 = 0 or x - 3 = 0 x = 5 2 , 3 .

We will factorize x + 16 3 = 4 x Taking L.C.M and solving further 3 x + 16 3 = 4 x We have to cross multiply and bring it to the form. Then we can factorize and solve. 3 x 2 + 16 x - 12 = 0 3 x 2 + 18 x - 2 x - 12 = 0 3 x x + 6 - 2 ( x + 6 ) = 0 3 x - 2 x + 6 = 0 By zero product rule 3 x - 2 = 0 or x + 6 = 0 x = 2 3 , - 6 .