Asked by x
Solve the following quadratic equation using square roots:
(x-5)^2= 49
simple
(x-5)^2= 49
simple
Answers
Answered by
x
Q2 Write x^2+12x+35=0 in its factored form.
simple
simple
Answered by
x
Solve the following quadratic equation:
x2 - 14x+49 = 0
x2 - 14x+49 = 0
Answered by
x
A basketball is shot from a height of 6 feet with an initial upward velocity of 20 feet per second. The height of the basketball can be represented by the equation
h = -16t^2 + 20t + 64 where t = time after the shot in seconds and h = the height in feet. When will the basketball hit the ground?
h = -16t^2 + 20t + 64 where t = time after the shot in seconds and h = the height in feet. When will the basketball hit the ground?
Answered by
GPT-4o mini
To solve the equation \((x - 5)^2 = 49\) using square roots, follow these steps:
1. **Take the square root of both sides**:
\[
x - 5 = \pm \sqrt{49}
\]
2. **Calculate the square root**:
\[
x - 5 = \pm 7
\]
3. **Solve for \(x\)** by considering both the positive and negative cases:
- Case 1:
\[
x - 5 = 7
\]
\[
x = 7 + 5 = 12
\]
- Case 2:
\[
x - 5 = -7
\]
\[
x = -7 + 5 = -2
\]
4. **Final solutions**:
\[
x = 12 \quad \text{and} \quad x = -2
\]
So, the solutions to the equation \((x - 5)^2 = 49\) are \(x = 12\) and \(x = -2\).
1. **Take the square root of both sides**:
\[
x - 5 = \pm \sqrt{49}
\]
2. **Calculate the square root**:
\[
x - 5 = \pm 7
\]
3. **Solve for \(x\)** by considering both the positive and negative cases:
- Case 1:
\[
x - 5 = 7
\]
\[
x = 7 + 5 = 12
\]
- Case 2:
\[
x - 5 = -7
\]
\[
x = -7 + 5 = -2
\]
4. **Final solutions**:
\[
x = 12 \quad \text{and} \quad x = -2
\]
So, the solutions to the equation \((x - 5)^2 = 49\) are \(x = 12\) and \(x = -2\).
Answered by
GPT-4o mini
To factor the quadratic equation \(x^2 + 12x + 35 = 0\), we need to find two numbers that multiply to \(35\) (the constant term) and add up to \(12\) (the coefficient of the linear term).
The two numbers that satisfy these conditions are \(5\) and \(7\) since:
- \(5 \times 7 = 35\)
- \(5 + 7 = 12\)
Now, we can express the quadratic in its factored form:
\[
x^2 + 12x + 35 = (x + 5)(x + 7) = 0
\]
So, the factored form of the equation is:
\[
(x + 5)(x + 7) = 0
\]
The two numbers that satisfy these conditions are \(5\) and \(7\) since:
- \(5 \times 7 = 35\)
- \(5 + 7 = 12\)
Now, we can express the quadratic in its factored form:
\[
x^2 + 12x + 35 = (x + 5)(x + 7) = 0
\]
So, the factored form of the equation is:
\[
(x + 5)(x + 7) = 0
\]
Answered by
GPT-4o mini
To solve the quadratic equation \(x^2 - 14x + 49 = 0\), we can factor it.
First, we notice that the equation can be expressed as a perfect square:
\[
x^2 - 14x + 49 = (x - 7)^2
\]
Thus, we rewrite the equation as:
\[
(x - 7)^2 = 0
\]
Now, we take the square root of both sides:
\[
x - 7 = 0
\]
Solving for \(x\) gives:
\[
x = 7
\]
The solution to the quadratic equation \(x^2 - 14x + 49 = 0\) is:
\[
\boxed{7}
\]
This is a double root, meaning \(x = 7\) is the only solution.
First, we notice that the equation can be expressed as a perfect square:
\[
x^2 - 14x + 49 = (x - 7)^2
\]
Thus, we rewrite the equation as:
\[
(x - 7)^2 = 0
\]
Now, we take the square root of both sides:
\[
x - 7 = 0
\]
Solving for \(x\) gives:
\[
x = 7
\]
The solution to the quadratic equation \(x^2 - 14x + 49 = 0\) is:
\[
\boxed{7}
\]
This is a double root, meaning \(x = 7\) is the only solution.
Answered by
GPT-4o mini
To determine when the basketball will hit the ground, we need to set the height \(h\) to 0 in the equation:
\[
h = -16t^2 + 20t + 64
\]
Setting \(h\) to 0 gives us:
\[
0 = -16t^2 + 20t + 64
\]
We can rearrange the equation:
\[
-16t^2 + 20t + 64 = 0
\]
To make calculations easier, we can multiply the entire equation by -1:
\[
16t^2 - 20t - 64 = 0
\]
Next, we can either use the quadratic formula or factor it. Let's try factoring first. We look for two numbers that multiply to \(16 \cdot -64 = -1024\) and add to \(-20\).
However, it might be easier to use the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our case, \(a = 16\), \(b = -20\), and \(c = -64\).
Plugging in these values:
\[
t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4 \cdot 16 \cdot (-64)}}{2 \cdot 16}
\]
This simplifies to:
\[
t = \frac{20 \pm \sqrt{400 + 4096}}{32}
\]
Calculating the square root:
\[
t = \frac{20 \pm \sqrt{4496}}{32}
\]
Now, let's simplify \(\sqrt{4496}\):
\[
\sqrt{4496} \approx 67.07
\]
Substituting back gives:
\[
t = \frac{20 \pm 67.07}{32}
\]
Calculating the two possible values for \(t\):
1. **Using the positive root**:
\[
t = \frac{20 + 67.07}{32} \approx \frac{87.07}{32} \approx 2.72 \text{ seconds}
\]
2. **Using the negative root**:
\[
t = \frac{20 - 67.07}{32} \approx \frac{-47.07}{32} \approx -1.47 \text{ seconds (not possible)}
\]
Since time cannot be negative, we discard the second value.
Thus, the basketball will hit the ground approximately **2.72 seconds** after being shot.
So the final answer is:
\[
\boxed{2.72 \text{ seconds}}
\]
\[
h = -16t^2 + 20t + 64
\]
Setting \(h\) to 0 gives us:
\[
0 = -16t^2 + 20t + 64
\]
We can rearrange the equation:
\[
-16t^2 + 20t + 64 = 0
\]
To make calculations easier, we can multiply the entire equation by -1:
\[
16t^2 - 20t - 64 = 0
\]
Next, we can either use the quadratic formula or factor it. Let's try factoring first. We look for two numbers that multiply to \(16 \cdot -64 = -1024\) and add to \(-20\).
However, it might be easier to use the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our case, \(a = 16\), \(b = -20\), and \(c = -64\).
Plugging in these values:
\[
t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4 \cdot 16 \cdot (-64)}}{2 \cdot 16}
\]
This simplifies to:
\[
t = \frac{20 \pm \sqrt{400 + 4096}}{32}
\]
Calculating the square root:
\[
t = \frac{20 \pm \sqrt{4496}}{32}
\]
Now, let's simplify \(\sqrt{4496}\):
\[
\sqrt{4496} \approx 67.07
\]
Substituting back gives:
\[
t = \frac{20 \pm 67.07}{32}
\]
Calculating the two possible values for \(t\):
1. **Using the positive root**:
\[
t = \frac{20 + 67.07}{32} \approx \frac{87.07}{32} \approx 2.72 \text{ seconds}
\]
2. **Using the negative root**:
\[
t = \frac{20 - 67.07}{32} \approx \frac{-47.07}{32} \approx -1.47 \text{ seconds (not possible)}
\]
Since time cannot be negative, we discard the second value.
Thus, the basketball will hit the ground approximately **2.72 seconds** after being shot.
So the final answer is:
\[
\boxed{2.72 \text{ seconds}}
\]