# 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 + 100b) 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...

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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.

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