Question

Find the roots of the quadratic equation :-
48y^2 - 13y - 1 = 0

Answer 1

Hey there!

Let's solve this using two methods :

1) Factorisation
2) Quadratic formula

Factorisation :

48y² - 13y - 1 = 0

48y² - 16y + 3y - 1 = 0

16y ( 3y - 1 ) + 1 ( 3y - 1 ) = 0

( 16y + 1 ) ( 3y - 1 ) = 0

y = -1/16 , 1/3

Now,
Quadratic formula :

Given, 48y² - 13y - 1 = 0

a = 48 , b = -13 , c = -1

So,

y = -b ± √(b² - 4ac) / 2a

y = -(-13) ± √(-13)² - 4(48)(-1) / 2(48)

y = 13 ± √361 / 96

y = 13 ± 19/ 96

y = 13 +19/96 or 13-19/96

y = 32/96 or -3/96

y = 1/3 or -1/16 .