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.
Pick on option
The solution to the quadratic equation 2x^2 - 5x = 6 is approximately x = 1.03.
How many real solutions does the following quadratic equation have? 4x^2+x+3=0
two real solutions two real solutions one real solution one real solution no real solutions no real solutions three real solutions
To determine the number of real solutions for the quadratic equation 4x^2 + x + 3 = 0, we can use the discriminant formula. The discriminant (D) is given by D = b^2 - 4ac, where a, b, and c are the coefficients of the equation.
In this case, a = 4, b = 1, and c = 3. Substituting these values into the discriminant formula, we have:
D = (1)^2 - 4(4)(3)
D = 1 - 48
D = -47
Since the discriminant is negative (D < 0), the quadratic equation 4x^2 + x + 3 = 0 has no real solutions. Therefore, the answer is: no real solutions.