For the equation 6x^2+x-2=0 determined whether the given quadratic eq has roots .if so find roots of equation
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6x^2+x-2=0
has two root
using quadratic formula
x=1/2 & -2/3
has two root
using quadratic formula
x=1/2 & -2/3
Answered by
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General form of quadratic equation is
ax² + bx + c = 0
Here, a = 6, b = 1 , c = -2
If its determinant is less than zero. The x values imaginary. That you learn in Complex number in higher classes.
FOR DETERMINANT:
b² - 4ac = (1)² - 4× 6 ×(-2)
b² - 4ac = 1 + 48 = 49 > 0
Therefore, this equation has two distinct real roots:
x = 【- b ± √(b² - 4ac)/ 2a
x = [ - 1 ±√49]/ 2×6
x = (- 1 + 7)/12 ; x = (-1 - 7)/12
x = (6/12) ; x =( -8/12)
x = 1/2 ; x= -2/ 3.
ax² + bx + c = 0
Here, a = 6, b = 1 , c = -2
If its determinant is less than zero. The x values imaginary. That you learn in Complex number in higher classes.
FOR DETERMINANT:
b² - 4ac = (1)² - 4× 6 ×(-2)
b² - 4ac = 1 + 48 = 49 > 0
Therefore, this equation has two distinct real roots:
x = 【- b ± √(b² - 4ac)/ 2a
x = [ - 1 ±√49]/ 2×6
x = (- 1 + 7)/12 ; x = (-1 - 7)/12
x = (6/12) ; x =( -8/12)
x = 1/2 ; x= -2/ 3.
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