Answer:
i) x^2 - 40x + a = 0 is shown to be true based on the relationship between the perimeter, length, width, and area of the rectangle.
ii) the solutions for the equation x^2 - 40x + 300 = 0 are x = 30 and x = 10.
iii) x = 34.14 and x = 5.86
Explanation:
i) Let's denote the length of the rectangle as "L" and the width as "W". The perimeter of a rectangle is given by the formula:
Perimeter = 2 * (Length + Width)
Given that the perimeter is 80 cm, we can write this as an equation:
80 = 2 * (L + W)
Simplifying, we get:
40 = L + W
Now, the area of a rectangle is given by the formula:
Area = Length * Width
Given that the area is "a" square cm, we can write this as an equation:
a = L * W
We are asked to show that the equation x^2 - 40x + a = 0 holds true. Since we have expressed L and W in terms of 40, let's substitute these values into the equation for the area:
a = (40 - W) * W
a = 40W - W^2
Now, let's substitute the value of W from the perimeter equation (W = 40 - L):
a = 40(40 - L) - (40 - L)^2
a = 1600 - 40L - (1600 - 80L + L^2)
a = 1600 - 40L - 1600 + 80L - L^2
a = -L^2 + 40L
Now, we want to show that x^2 - 40x + a = 0. Comparing the two equations:
x^2 - 40x + a = x^2 - 40L + (-L^2 + 40L)
x^2 - 40x + a = x^2 - L^2
Since we know that 40 = L + W, we have shown that x^2 - 40x + a = x^2 - L^2, which means that the given equation holds true.
ii) Given the equation x^2 - 40x + a = 0 and the value of a as 300, we can substitute this value into the equation:
x^2 - 40x + 300 = 0
To solve this equation by factoring, we need to find two numbers that multiply to 300 and add up to -40 (the coefficient of the x term). Let's find those numbers:
We're looking for two numbers, let's call them "p" and "q", such that:
p * q = 300
p + q = -40
The numbers that satisfy these conditions are -30 and -10, because:
-30 * -10 = 300
-30 + (-10) = -40
Now, we can rewrite the middle term (-40x) in the original equation using these numbers:
x^2 - 30x - 10x + 300 = 0
Now, we can factor by grouping:
x(x - 30) - 10(x - 30) = 0
Notice that we have a common factor of (x - 30):
(x - 30)(x - 10) = 0
Now we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero:
x - 30 = 0 or x - 10 = 0
Solving for "x" in each case:
x = 30 or x = 10
iii) To solve the quadratic equation x^2 - 40x + a = 0 when a = 200, we can directly substitute the value of "a" into the equation:
x^2 - 40x + 200 = 0
Now we have a quadratic equation in standard form. To solve it, we can use the quadratic formula:
For an equation of the form ax^2 + bx + c = 0, the quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = -40, and c = 200. Substituting these values into the formula:
x = (-(-40) ± √((-40)^2 - 4 * 1 * 200)) / (2 * 1)
x = (40 ± √(1600 - 800)) / 2
x = (40 ± √800) / 2
x = (40 ± 20√2) / 2
x = 20 ± 10√2
So, the solutions for the equation x^2 - 40x + 200 = 0 are:
x = 20 + 10√2
x = 20 - 10√2