Problem 22
Solve each quadratic equation by completeling the square.
x^2 -6x-3 = 0
MY answer: x = 3 +- 2 radical (3)
Problem 28
2x^2+10x+11=0
My answer: x = -2.5 +- 0.5 radical (3)
Problem 38
Find two consecutive positive integers such that the sum of their square is 85.
My answer: the integers are 7 and 6
To solve each quadratic equation by completing the square, follow these steps:
Problem 22:
1. Write the equation in the form x^2 + bx + c = 0.
- Our equation is already in this form: x^2 - 6x - 3 = 0.
2. Take half of the coefficient of x, square it, and add it to both sides of the equation.
- The coefficient of x is -6. Half of -6 is -3, and (-3)^2 = 9.
- Add 9 to both sides of the equation: x^2 - 6x - 3 + 9 = 0 + 9, which simplifies to x^2 - 6x + 6 = 9.
3. Rewrite the equation as a perfect square trinomial by factoring the left side of the equation.
- To factor x^2 - 6x + 6, we need to find two numbers whose product is 6 and whose sum is -6.
- Since -3 * -3 = 9 and -3 + -3 = -6, we can rewrite the equation as (x - 3)^2 = 9.
4. Take the square root of both sides of the equation and solve for x.
- Taking the square root of (x - 3)^2 = 9 gives us x - 3 = ±3.
- Adding 3 to both sides yields x = 3 ± 3.
5. Simplify the solution.
- x = 3 + 3 simplifies to x = 6.
- x = 3 - 3 simplifies to x = 0.
Therefore, the solutions to x^2 - 6x - 3 = 0 are x = 6 and x = 0.
Problem 28:
1. Write the equation in the form x^2 + bx + c = 0.
- Our equation is already in this form: 2x^2 + 10x + 11 = 0.
2. Take half of the coefficient of x, square it, and add it to both sides of the equation.
- The coefficient of x is 10. Half of 10 is 5, and (5)^2 = 25.
- Add 25 to both sides of the equation: 2x^2 + 10x + 11 + 25 = 0 + 25, which simplifies to 2x^2 + 10x + 36 = 25.
3. Rewrite the equation as a perfect square trinomial by factoring the left side of the equation.
- To factor 2x^2 + 10x + 36, we need to find two numbers whose product is 2 * 36 = 72 and whose sum is 10.
- These numbers are 6 and 12, so we can rewrite the equation as (x + 6)(x + 6) = 25.
4. Take the square root of both sides of the equation and solve for x.
- Taking the square root of (x + 6)(x + 6) = 25 gives us x + 6 = ±5.
- Subtracting 6 from both sides yields x = -6 ± 5.
5. Simplify the solution.
- x = -6 + 5 simplifies to x = -1.
- x = -6 - 5 simplifies to x = -11.
Therefore, the solutions to 2x^2 + 10x + 11 = 0 are x = -1 and x = -11.
Problem 38:
To find two consecutive positive integers whose sum of squares is 85, follow these steps:
1. Represent the consecutive integers as x and x + 1.
- Let the first integer be x.
- The second integer will be one more than x, so it is x + 1.
2. Write the equation based on the given information.
- The sum of squares of the consecutive integers is 85, so we have the equation x^2 + (x + 1)^2 = 85.
3. Expand and simplify the equation.
- x^2 + (x + 1)^2 = 85 becomes x^2 + x^2 + 2x + 1 = 85.
- Combining like terms gives us 2x^2 + 2x + 1 = 85.
- Rearranging the equation gives us 2x^2 + 2x - 84 = 0.
4. Solve the quadratic equation by factoring or using the quadratic formula.
- This equation cannot be easily factored, so we can use the quadratic formula: x = (-b ± √(b^2 - 4ac))/(2a).
- In this case, a = 2, b = 2, and c = -84.
- Plugging these values into the quadratic formula yields x = (-2 ± √(2^2 - 4(2)(-84)))/(2(2)).
- Simplifying further gives x = (-2 ± √(4 + 672))/4.
- The discriminant, 4 + 672, is equal to 676, which is a perfect square.
- Taking the square root of the discriminant gives ±26, so the solutions are x = (-2 + 26)/4 and x = (-2 - 26)/4.
5. Simplify the solutions.
- x = (24)/4 simplifies to x = 6.
- x = (-28)/4 simplifies to x = -7.
Therefore, the two consecutive positive integers whose sum of squares is 85 are 6 and 7.