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PLEASE HELP ITS DUE TOMMOROW

The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all possible values of k.

User CPerkins
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1 Answer

4 votes

Answer:

- 3/2 ±sqrt(17)/2 = k

Explanation:

Area of a triangle is

A = 1/2 bh

1.5 = 1/2 b*h

b = distance between the x intercepts

h = y intercept

y=0.5(x−3)(x+k)

The x intercepts are 3, -k

Distance between them is (3--k) = 3+k so b = (3+k)

The y intercept is .5(-3 * k) = 1.5k so h = 1.5k

Substituting into the equation for A

1.5 = .5 (3+k) (1.5k)

1.5 = .75k(3+k)

Divide by .75 on each side

1.5/.75 = k(3+k)

2 = k(3+k)

2 = 3k+k^2

2 = k^2 +3k

We will need to complete the square

(3/2) ^2 =9/4 is what we need to add to each side

2+9/4 = k^2 +3k +9/4

8/4 + 9/4 = (k+3/2)^2

17/4 = (k+3/2)^2

Take the square root of each side

±sqrt(17/4) = sqrt( (k+3/2)^2)

±sqrt(17/4) = k+3/2

Subtract 3/2 from each side

- 3/2 ±sqrt(17/4) = k

We know the square root of (a/b) =sqrt(a)/sqrt(b)

- 3/2 ±sqrt(17)/sqrt(4) = k

- 3/2 ±sqrt(17)/2 = k

User Alder
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