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