Math  /  Algebra

QuestionFor the pair of polynomials given, select all the points of intersection of their graphs. g(x)=(x+7)(x5)h(x)=x5\begin{array}{l} g(x)=(x+7)(x-5) \\ h(x)=x-5 \end{array} A. (8,13)(-8,-13) B. (7,0)(-7,0) C. (5,10)(-5,-10) D. (6,11)(-6,-11) E. (5,0)(5,0)

Studdy Solution

STEP 1

1. We are given two polynomials: g(x)=(x+7)(x5) g(x) = (x+7)(x-5) and h(x)=x5 h(x) = x-5 .
2. We need to find the points where the graphs of g(x) g(x) and h(x) h(x) intersect.
3. A point of intersection occurs where g(x)=h(x) g(x) = h(x) .

STEP 2

1. Set g(x) g(x) equal to h(x) h(x) and solve for x x .
2. Verify which of the given points satisfy both g(x) g(x) and h(x) h(x) .

STEP 3

Set g(x)=h(x) g(x) = h(x) :
(x+7)(x5)=x5 (x+7)(x-5) = x-5

STEP 4

Simplify the equation by expanding the left-hand side:
x2+7x5x35=x5 x^2 + 7x - 5x - 35 = x - 5 x2+2x35=x5 x^2 + 2x - 35 = x - 5

STEP 5

Rearrange the equation to bring all terms to one side:
x2+2x35x+5=0 x^2 + 2x - 35 - x + 5 = 0 x2+x30=0 x^2 + x - 30 = 0

STEP 6

Solve the quadratic equation x2+x30=0 x^2 + x - 30 = 0 using the quadratic formula:
x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1 a = 1 , b=1 b = 1 , c=30 c = -30 .
x=1±1241(30)21 x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} x=1±1+1202 x = \frac{-1 \pm \sqrt{1 + 120}}{2} x=1±1212 x = \frac{-1 \pm \sqrt{121}}{2} x=1±112 x = \frac{-1 \pm 11}{2}

STEP 7

Calculate the two possible solutions for x x :
x=1+112=5 x = \frac{-1 + 11}{2} = 5 x=1112=6 x = \frac{-1 - 11}{2} = -6

STEP 8

Verify the points of intersection by substituting x=5 x = 5 and x=6 x = -6 into h(x) h(x) and checking against the given options.
For x=5 x = 5 : h(5)=55=0 h(5) = 5 - 5 = 0 Point: (5,0) (5, 0)
For x=6 x = -6 : h(6)=65=11 h(-6) = -6 - 5 = -11 Point: (6,11) (-6, -11)
The points of intersection are: E. (5,0) (5, 0) D. (6,11) (-6, -11)

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