Solve thsi word problem ?
The length of the hypotenuse of a right triangle is 1 cm more than triple that of the shorter leg. The length of the longer leg is 1 cm less than triple that of the shorter leg. Find the lengths of the three sides of the triangle.
Find the value(s) of k so that each parabola has only one x-intercept.
a) y = x^2 + kx + 100
b) y = -4x^2 + 28x + k
Shorter leg = x
Hypotenuse = 3x+1
Longer leg = 3x-1
We know x^2 + (3x-1)^2 = (3x+1)^2
Expanding we get
x^2 + 9x^2 - 6x + 1 = 9x^2 + 6x + 1
Putting into standard form we get
x^2 - 12x = 0
x(x-12) = 0
So either x = 0 or x = 12
x = 0 gives 0 for the longer leg which does not make sense, so our only answer is x = 12
Shorter leg = 12, Longer leg = 3(12)-1 = 35, and Hypotenuse = 3(12)+1=37.
We can verify that 12^2 + 35^2 = 37^2 to check our answer.
We get one intercept when `b^2-4ac = 0`
In the first problem a=1, b=k, and c=100
We get `b^2-4ac=k^2-4(1)(100)=0`
k^2 - 400 = 0
`k = +-sqrt(400)=+-20`
The second one a=-4, b=28 and c=k
We get `b^2-4ac=28^2-4(-4)(k) = 0`
784+16k = 0, solving we get `k = -784/16=-49` . So our answer is k=-49.