Solve the quadratic equation 3x2 = 27
Show Your Work Please!
To solve this equation, we need to first isolate the variable on one side of the equal sign. We can do this by dividing both sides of the equation by 3:
3x^2 = 27
(3x^2)/3 = 27/3
x^2 = 9
Next, we need to take the square root of both sides of the equation:
√(x^2) = ±√9
x = ±3
Therefore, the solutions to the quadratic equation 3x^2 = 27 are x = 3 and x = -3.
Solve the quadratic equation 2x2 − 4x + 1 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
where a, b, and c are coefficients of the quadratic equation, in this case, a = 2, b = -4, and c = 1.
Plugging in the values, we get:
x = (-(-4) ± √((-4)^2 - 4(2)(1))) / 2(2)
x = (4 ± √(16 - 8)) / 4
x = (4 ± √8) / 4
Simplifying by factoring out √8 and reducing fractions, we get:
x = (2 ± √2) / 2
So the solutions to the quadratic equation 2x^2 - 4x + 1 = 0 are:
x = (2 + √2) / 2 or x = (2 - √2) / 2
How many solutions are in the equation 3x2 + 4x − 7 = 0?
To determine the number of solutions of the quadratic equation 3x^2 + 4x - 7 = 0, we need to use the discriminant formula, which is given by:
b^2 - 4ac
where a, b, and c are the coefficients of the quadratic equation.
For this equation, we have:
a = 3
b = 4
c = -7
So, substituting the values in the discriminant formula, we get:
4^2 - 4(3)(-7)
16 + 84
100
Since the discriminant is positive and greater than 0, the equation has two distinct real solutions.
Solve by completing the square: A rectangular patio has a length of x + 6 m, a width of x + 8 m, and a total area of 400 m2. Find the dimensions to the nearest tenth.
To solve by completing the square, we first need to express the area of the rectangular patio as a quadratic equation in standard form.
The area of a rectangle is given by:
A = length * width
We are given that the length is x + 6 m and the width is x + 8 m, so we can write:
A = (x + 6)(x + 8)
Multiplying this out, we get:
A = x^2 + 14x + 48
We know that the area is 400 m2, so we can set this equal to 400 and solve for x:
x^2 + 14x + 48 = 400
Subtracting 400 from both sides, we get:
x^2 + 14x - 352 = 0
Now, we can complete the square to solve for x. To do this, we need to add and subtract the square of half the coefficient of x, which is (14/2)^2 = 49:
x^2 + 14x - 352 + 49 - 49 = 0
(x + 7)^2 - 401 = 0
(x + 7)^2 = 401
Taking the square root of both sides, we get:
x + 7 = ±sqrt(401)
Solving for x, we get:
x = -7 ± sqrt(401)
Rounding to the nearest tenth, we get:
x ≈ -17.0 or x ≈ 10.0
However, the length and width of the rectangular patio must be positive values, so we take x ≈ 10.0 as our solution.
Therefore, the dimensions of the rectangular patio are approximately:
Length = x + 6 ≈ 16.0 m
Width = x + 8 ≈ 18.0 m
To check, we can verify that the area is indeed 400 m2:
16.0 m * 18.0 m = 288.0 m2
288.0 m2 is not exactly 400 m2, so we round our values from x to the nearest tenth:
Length = x + 6 ≈ 16.0 m
Width = x + 8 ≈ 18.0 m
16.0 m * 18.0 m = 288.0 m2, which is close to 400 m2, but not exactly. Therefore, our solution is approximately correct to the nearest tenth.
Find the radius of a circular lid with an area of 12in.2.
The area (A) of a circle is given by pi times the square of its radius (r):
A = πr^2
We are given that the area of the circular lid is 12 in^2, so we can write:
12 = πr^2
Dividing both sides by π, we get:
r^2 = 12/π
Taking the square root of both sides, we get:
r = √(12/π)
Simplifying, we get:
r ≈ 1.94 in
Therefore, the radius of the circular lid is approximately 1.94 inches.
To solve the quadratic equation 3x² = 27, we need to isolate x² on one side of the equation. Here's how we do it:
1. Start with the given equation: 3x² = 27.
2. Divide both sides of the equation by 3 to isolate x²: (3x²) / 3 = 27 / 3.
This simplifies to x² = 9.
3. Take the square root of both sides of the equation to solve for x.
√(x²) = √9.
This gives us two possible solutions, x = 3 and x = -3.
So, the solutions to the quadratic equation 3x² = 27 are x = 3 and x = -3.
When solving this equation, we applied the following steps:
1. Divide both sides of the equation by the coefficient of x² to isolate x².
2. Solve for x by taking the square root of both sides of the equation.
3. Remember that since squaring a number gives both positive and negative values, we get two solutions.