The difference between the solutions of the equation 5x^2−x−k=0 is 1.8. Find the solutions.
Using the quadratic formula:
x = (1 ± √(1 - 4(5)(-k) )/1-
= (1 + √(1+20k))/10 or x = (1 - √(1+20k))/10
then (1 + √(1+20k))/10 - (1 - √(1+20k))/10 = 1.8
(1 + √(1+20k)) - (1 - √(1+20k)) = 18
2√(1+20k) = 18
√(1+20k) = 9
square both sides, easy after that
a = 5
b = -1
c = -k
x1 = [1 +sqrt (1+20k) ] /10
x2 = [1 -sqrt (1+20k) ] /10
x1-x2 = 1.8
1.8*10 = [1 +sqrt (1+20k) ] [1 -sqrt (1+20k) ]
but we know that a^2-b^2 = (a+b)(a-b)
18 = 1 - (1+20k)
17 = -1 -20 k
18 = -20 k
check my arithmetic and continue to get x1 and x2
whoops, made algebra error, go with Reiny solution
To find the solutions of the equation 5x^2 − x − k = 0, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Comparing this to our equation, we have:
a = 5
b = -1
c = -k
We are given that the difference between the solutions of the equation is 1.8. Let's call the solutions x1 and x2. We can express this difference as:
x2 - x1 = 1.8
Substituting into the quadratic formula:
(-b ± √(b^2 - 4ac)) / (2a) = x2 - x1
(-(-1) ± √((-1)^2 - 4 * 5 * (-k))) / (2 * 5) = x2 - x1
Simplifying further:
(1 ± √(1 + 20k)) / 10 = 1.8
Now, we can solve for k:
1 ± √(1 + 20k) = 18
± √(1 + 20k) = 17
1 + 20k = 289 or 1 + 20k = 289
20k = 288 or 20k = -288
k = 14.4 or k = -14.4
Now that we have determined the value of k, we can substitute it back into the quadratic formula to find the solutions for x:
For k = 14.4:
x1 = (-(-1) - √((-1)^2 - 4 * 5 * (-14.4))) / (2 * 5)
= (1 - √(1 + 288)) / 10
x2 = (-(-1) + √((-1)^2 - 4 * 5 * (-14.4))) / (2 * 5)
= (1 + √(1 + 288)) / 10
And for k = -14.4:
x1 = (-(-1) - √((-1)^2 - 4 * 5 * 14.4)) / (2 * 5)
= (1 - √(1 - 288)) / 10
x2 = (-(-1) + √((-1)^2 - 4 * 5 * 14.4)) / (2 * 5)
= (1 + √(1 - 288)) / 10
Therefore, the solutions of the equation are:
For k = 14.4: x1 = (1 - √289) / 10 and x2 = (1 + √289) / 10
For k = -14.4: x1 = (1 - √287) / 10 and x2 = (1 + √287) / 10