Respuesta :
The length of the rectangle is 11 units
The width of the rectangle is 5 units.
Explanation:
The equation of the trinomial is [tex]r^{2}-6 r-55[/tex]
We need to find the length and the width of the rectangle.
The length and width of the rectangle can be determined by factoring the trinomial [tex]r^{2}-6 r-55=0[/tex]
Factoring using the quadratic formula, we have,
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
where [tex]a=1, b=-6[/tex] and [tex]c=-55[/tex]
Substituting these in the above formula, we have,
[tex]r=\frac{-(-6) \pm \sqrt{(-6)^{2}-4 \cdot 1(-55)}}{2 \cdot 1}[/tex]
[tex]r=\frac{6\pm \sqrt{36+220}}{2}[/tex]
[tex]r=\frac{6\pm \sqrt{256}}{2}[/tex]
[tex]r=\frac{6\pm 16}{2}[/tex]
Hence, the two roots of the trinomial are
[tex]r=\frac{6+ 16}{2}[/tex] and [tex]r=\frac{6- 16}{2}[/tex]
Solving [tex]r=\frac{6+ 16}{2}[/tex] , we get,
[tex]r=\frac{22}{2}[/tex]
[tex]r=11[/tex]
Solving [tex]r=\frac{6- 16}{2}[/tex] , we get,
[tex]r=\frac{-10}{2}[/tex]
[tex]r=-5[/tex]
Thus, the solutions of the trinomial are 11 and -5
Since, the length of the rectangle is larger than the width, the length of the rectangle is 11 units.
Also, the width of the rectangle cannot be negative.
Thus, ignoring the negative sign, we have, [tex]r=5[/tex]
Hence, the width of the rectangle is 5 units.