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Factoring Quadratics

Factoring Quadratics 1.11 - Solution

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We have been given a right triangle with an area of 35 m2.35 \text{ m}^2.

Recall that area of a right triangle is half the product of its legs. With this, we can write an equation to model the area of the given triangle. (x11)(x8)2=35\begin{gathered} \dfrac{(x-11)(x-8)}{2}=35 \end{gathered} To find the dimensions of the triangle, we should solve the equation above. Let's start by writing it in standard form.
(x11)(x8)2=35\dfrac{(x-11)(x-8)}{2}=35
Write in slope-intercept form
(x11)(x8)=70(x-11)(x-8)=70
x(x8)11(x8)=70x(x-8)-11(x-8)=70
x28x11(x8)=70x^2-8x-11(x-8)=70
x28x11x+88=70x^2-8x-11x+88=70
x219x+88=70x^2-19x+88=70
x219x+18=0x^2-19x+18=0
Next, we will factor the resulting equation.
x219x+18=0x^2-19x+18=0
Factor
x218xx+18=0x^2-18x-x+18=0
x(x18)x+18=0x(x-18)-x+18=0
x(x18)(x18)=0x(x-18)-(x-18)=0
(x18)(x1)=0(x-18)(x-1)=0
Now, we can apply the Zero Product Property to solve it.
(x18)(x1)=0(x-18)(x-1)=0
x18=0(I)x1=0(II)\begin{array}{lc}x-18=0 & \text{(I)}\\ x-1=0 & \text{(II)}\end{array}
x=18x1=0\begin{array}{l}x=18 \\ x-1=0 \end{array}
x1=18x2=1\begin{array}{l}x_1=18 \\ x_2=1 \end{array}
Because the legs of a right triangle cannot have negative values, the value of xx is 18.18. From here, we can determine the dimensions as shown below. Leg I: 1811=17 mLeg II: 1818=10 m\begin{aligned} \textbf{Leg I: }& {\color{#0000FF}{18}}-11=\phantom{1}7 \text{ m}\\ \textbf{Leg II: }& {\color{#0000FF}{18}}-\phantom{1}8=10 \text{ m} \end{aligned}