Complete the square to solve the following quadratic equation: x^2+2x−8=0 .
x=2, x=−4 x equals 2 , x equals negative 4 x=−2, x=4 x equals negative 2 , x equals 4 x=−10, x=8 x equals negative 10 , x equals 8 x=10, x=−8
To complete the square, we need to rewrite the equation in the form (x + p)^2 = q.
Starting with x^2 + 2x - 8 = 0, we can find p by taking half of the coefficient of x and squaring it:
p = 2/2 = 1
p^2 = 1^2 = 1
Now we add and subtract p^2 inside the parentheses:
x^2 + 2x + 1 - 1 - 8 = 0
(x + 1)^2 - 9 = 0
The equation is now in the form (x + p)^2 = q, where p = 1 and q = -9. We can solve for x by taking the square root of both sides:
(x + 1)^2 = 9
x + 1 = ±√9
x + 1 = ±3
Now solve for x:
x = -1 + 3 = 2
x = -1 - 3 = -4
Therefore, the solution is x = 2 or x = -4.
To complete the square and solve the quadratic equation x^2 + 2x - 8 = 0, follow these steps:
Step 1: Move the constant term to the right side of the equation:
x^2 + 2x = 8
Step 2: Divide the coefficient of x by 2, square it, and add it to both sides of the equation:
x^2 + 2x + (2/2)^2 = 8 + (2/2)^2
x^2 + 2x + 1 = 9
Step 3: Factor the perfect square trinomial on the left side of the equation:
(x + 1)^2 = 9
Step 4: Take the square root of both sides of the equation:
x + 1 = ±√9
Step 5: Simplify the square root of 9 to get two possible solutions:
x + 1 = ±3
Step 6: Solve for x by subtracting 1 from both sides of the equation:
x = -1 ± 3
Step 7: Simplify to get the final solutions:
x = 2 or x = -4
Therefore, the quadratic equation x^2 + 2x - 8 = 0 can be solved by completing the square, and the solutions are x = 2 and x = -4.
To solve the quadratic equation x^2 + 2x - 8 = 0 by completing the square, follow these steps:
Step 1: Move the constant term (-8) to the other side of the equation, so your equation becomes x^2 + 2x = 8.
Step 2: Take half of the coefficient of x (which is 2) and square it: (2/2)^2 = 1.
Step 3: Add the result from Step 2 to both sides of the equation: x^2 + 2x + 1 = 9.
Step 4: Rewrite the left side of the equation as a perfect square: (x + 1)^2 = 9.
Step 5: Take the square root of both sides of the equation: x + 1 = ± √9.
Step 6: Solve for x by separating the equation into two cases:
- Case 1: x + 1 = √9.
- Subtract 1 from both sides: x = √9 - 1.
- Simplify: x = 3 - 1 = 2.
- Case 2: x + 1 = -√9.
- Subtract 1 from both sides: x = -√9 - 1.
- Simplify: x = -3 - 1 = -4.
Therefore, the solutions to the quadratic equation x^2 + 2x - 8 = 0 are x = 2 and x = -4.
Complete the square to solve the following quadratic equation: x^2−22x+1=15 .
x=−11+315−−√, −11−315−−√ x equals negative 11 plus 3 Start Root 15 End Root comma negative 11 minus 3 Start Root 15 End Root x=−11+14−−√, −11−14−−√ x equals negative 11 plus Start Root 14 End Root comma negative 11 minus Start Root 14 End Root x=11+14−−√, 11−14−−√ x equals 11 plus Start Root 14 End Root comma 11 minus Start Root 14 End Root x=11+315−−√, 11−315−−√
To complete the square and solve the quadratic equation x^2 - 22x + 1 = 15, follow these steps:
Step 1: Move the constant term (15) to the right side of the equation:
x^2 - 22x = 15 - 1
x^2 - 22x = 14
Step 2: Take half of the coefficient of x (-22/2 = -11) and square it:
(-11)^2 = 121
Step 3: Add the result from Step 2 to both sides of the equation:
x^2 - 22x + 121 = 14 + 121
x^2 - 22x + 121 = 135
Step 4: Rewrite the left side of the equation as a perfect square:
(x - 11)^2 = 135
Step 5: Take the square root of both sides of the equation:
x - 11 = ± √135
Step 6: Simplify the square root of 135 to get two possible solutions:
x - 11 = ± √(9 * 15)
x - 11 = ± √9 * √15
x - 11 = ± 3 * √15
Step 7: Solve for x by adding 11 to both sides of the equation:
x = 11 ± 3 * √15
Therefore, the quadratic equation x^2 - 22x + 1 = 15 can be solved by completing the square, and the solutions are x = -11 + 3√15, -11 - 3√15.
Using the quadratic formula, find the solution to x^2−6x−16=0.
x=8, x=−8 x equals 8 , x equals negative 8 x=8 x equals 8 x=8, x=−2 x equals 8 , x equals negative 2 x=−2
To solve the quadratic equation x^2 - 6x - 16 = 0 using the quadratic formula, we have:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the equation x^2 - 6x - 16 = 0, the coefficients are:
a = 1
b = -6
c = -16
Substituting these values into the quadratic formula, we get:
x = (-(-6) ± √((-6)^2 - 4(1)(-16))) / (2(1))
x = (6 ± √(36 + 64)) / 2
x = (6 ± √100) / 2
x = (6 ± 10) / 2
This gives us two possible solutions:
x = (6 + 10) / 2 = 16 / 2 = 8
x = (6 - 10) / 2 = -4 / 2 = -2
Therefore, the solutions to the quadratic equation x^2 - 6x - 16 = 0 are x = 8 and x = -2.
Which of the following is a solution to the quadratic equation 2x^2−5x=6 ? Assume that the solution has been rounded to the nearest hundredth, if applicable.
x=0.89 x equals 0.89 x=1.03 x equals 1.03 x=−0.89 x equals negative 0.89 x=−1.03 x equals negative 1.03
To find the solution to the quadratic equation 2x^2 - 5x = 6, we can rearrange the equation to be in the form ax^2 + bx + c = 0:
2x^2 - 5x - 6 = 0
Now we can use the quadratic formula to find the solution:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -5, and c = -6. Substituting these values into the formula:
x = (-(-5) ± √((-5)^2 - 4(2)(-6))) / (2(2))
x = (5 ± √(25 + 48)) / 4
x = (5 ± √73) / 4
The solutions are x = (5 + √73) / 4 and x = (5 - √73) / 4.
Now we can evaluate these solutions and round them to the nearest hundredth:
x ≈ (5 + √73) / 4 ≈ 1.03
x ≈ (5 - √73) / 4 ≈ -1.03
Therefore, the solution to the quadratic equation 2x^2 - 5x = 6 is approximately x = 1.03 and x = -1.03.