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Learning Task 4. Solve the problem by applying the sum and product of roots
of quadratic equations.

The perimeter of a rectangular metal plate is 36 dm and its area is 80
dm². Find its dimensions. (Relate the measures to the sum and product of a
quadratic equation.)



The perimeter of a rectangle is twice the sum of its length and width while
its area is the product of its length and width. Such that,

Perimeter = 2(L+w) and Area = L•w.​

Learning Task 4 Solve the problem by applying the sum and product of rootsof quadratic equationsThe perimeter of a rectangular metal plate is 36 dm and its area class=

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Answer:

Thus, the dimensions of the metal plate are 10 dm and 8 dm.

Step-by-step explanation:

For a quadratic equation:

[tex]x^2+bx+c=0[/tex]

The sum of the roots is -b and the product is c. Note the leading coefficient is 1.

We know the perimeter of the rectangular metal plate is 36 dm and its area is 80 dm^2. Being L and W its dimensions, then:

P=2(L+W)=36

A=L.W=80

Note both formulas are closely related to the roots of the quadratic equation, we only need to adjust the data for the perimeter to be exactly the sum of L+W and not double of it.

Thus we use the semi perimeter instead as P/2=L+W=18

The quadratic equation is, then:

[tex]x^2-18x+80=0[/tex]

Factoring by finding two numbers that add up to 18 and have a product of 80:

[tex](x-10)(x-8)=0[/tex]

The solutions to the equation are:

x=10, x=8

Thus, the dimensions of the metal plate are 10 dm and 8 dm.

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