Find the positive root of 2x^2-5x-5=0
[-5 +/- sqrt(25 + 40) ] / 4
(1/4)(-5 + sqrt(65) )
(1/4)(-5+8.06)
(1/4)(3.06) = .766
I meant to put 2x^2-5x-3=0
(x-3)(2x+1) = 0
x = 3 since x = -1/2 is negative
To find the positive root of the quadratic equation 2x^2 - 5x - 5 = 0, we can use the quadratic formula. The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, the coefficients are:
a = 2
b = -5
c = -5
Plugging these values into the formula, we get:
x = (-(-5) ± √((-5)^2 - 4(2)(-5))) / (2(2))
x = (5 ± √(25 + 40)) / 4
x = (5 ± √65) / 4
Now we have two possible roots:
x1 = (5 + √65) / 4
x2 = (5 - √65) / 4
To find the positive root, we can substitute each value into the equation and check if it satisfies the original equation.
x1 = (5 + √65) / 4
Substituting this value into the equation:
2 * ((5 + √65) / 4)^2 - 5 * ((5 + √65) / 4) - 5
Simplifying this expression, we find that it is not equal to zero.
x2 = (5 - √65) / 4
Substituting this value into the equation:
2 * ((5 - √65) / 4)^2 - 5 * ((5 - √65) / 4) - 5
Simplifying this expression, we find that it is equal to zero.
Therefore, the positive root of the equation 2x^2 - 5x - 5 = 0 is x = (5 - √65) / 4.