Factor left-hand side to get:
x²+5x-6=0
(x+6)(x-1)=0
Now equate each factor to zero and find x.
x+6=0 => x=-6
x-1=0 => x=1
From these solutions, find the positive solution.
The following quadratic equation has the integer solutions
x^2+5x-6=0
what is the positive solution of the equation
2 answers
to get the solution/roots, we can use quadratic formula:
x = [-b +- sqrt(b^2 - 4ac)]/(2a)
where
a = numerical coefficient of x^2
b = numerical coefficient of x
c = the constant
in the problem, x^2+5x-6 = 0, thus
a = 1, b = 5, c = -6
substituting:
x = [-b +- sqrt(b^2 - 4ac)]/(2a)
since we need the positive root, we choose the plus sign (we replace the plus/minus sign (+-) by plus sign (+) only):
x = [-5 + sqrt(5^2 - 4(1)(-6))]/(2(1))
x = [-5 + sqrt(25 + 24)]/2
x = [-5 + sqrt(49)]/2
x = (-5 + 7)/2
x = 1
hope this helps~ :)
x = [-b +- sqrt(b^2 - 4ac)]/(2a)
where
a = numerical coefficient of x^2
b = numerical coefficient of x
c = the constant
in the problem, x^2+5x-6 = 0, thus
a = 1, b = 5, c = -6
substituting:
x = [-b +- sqrt(b^2 - 4ac)]/(2a)
since we need the positive root, we choose the plus sign (we replace the plus/minus sign (+-) by plus sign (+) only):
x = [-5 + sqrt(5^2 - 4(1)(-6))]/(2(1))
x = [-5 + sqrt(25 + 24)]/2
x = [-5 + sqrt(49)]/2
x = (-5 + 7)/2
x = 1
hope this helps~ :)
1 answer