+129907
2b^2 + b = 6
step 1.... divide both sides by 2
b^2 + (1/2)b = 3
step 2 .....take (1/2) of (1/2) = 1/4....square this = 1/16 add to both sides
b^2 + 1/2b + 1/16 = 3 + 1/16
step 3 ......factor the left side.....simplify the right side
(b + 1/4)^2 = 49/16
step 4 ......take positive/negative roots of both sides
b + 1/4 = ±√[ 49/16]
b + 1/4 = ±7/4
step 5......subtract 1/4 from both sides
b = 7/4 - 1/4 = 6/4 = 3/2 or b = -7/4 - 1/4 = -8/4 = -2
The mistake was in step 2.....1/4 was added to both sides, but it should have been (1/4)^2 = 1/16 !!
+129907
Here's 22
f(x) = -0.1x^2 + 3.2x -3.5
Let's find the roots....thus
-0.1x^2 + 3.2x -3.5 = 0 multiply both sides by -10
x^2 - 32x + 35 = 0 subtract 35 from both sides
x^2 - 32x = -35 take 1/2 of 32 = 16....square this = 256 ....add to both sides
x^2 - 32x + 256 = -35 + 256 factor and simplify
(x - 16)^2 = 221 take positive/negative roots of both sides
x - 16 = ±√221 add 16 to both sides
x = ±√221 + 16
So....the two roots are -√ 221 + 16 ≈ 1.13 and √ 221 + 16 ≈ 30.87
And the positive difference between these two is the width of the tunnel at the bottom =
30.87 - 1.13 ≈ 29.74 ft....so....the tunnel is wide enough at the bottom
The function is an upside-down parabola....the vertex occurs at the max height
To find the height...let's find the x coordinate of the vertex.....this is given by
-b / (2a) where b = 3.2 and a = -.1 ....so we have
-3.2 / [ 2(-.1) ] = -3.2 / -.2 = 32 / 2 = 16
So....putting this into the original function will give us the height....and we have
-.1(16)^2 + 3.2(16) - 3.5 =
-25.6 + 51.2 - 3.5 = 22.1 ft
However.......we need to find the width at 20 feet
So....set the function = 20 and solve
20 = -0.1x^2 + 3.2x -3.5 subtract 20 from both sides
Using the quadratic formula, we have that
x = ( -3.2 ±√ [ 3.2^2 - 4(-.1)(-23.5) ] ) / [ 2(-.1)] =
x = (-3.2 ±√ [ .84] ) / [ -.2 ] =
x ≈ 11.41 and x ≈ 20.58
So...the width at 20 feet equals the distance between these ≈ 9.17 feet
So....the tunnel is not wide enough at this point.....!!!!
