i need to solve this equation by completing the square:
x^2-14x+1=0
i don't' know how to do this, nor do i know what complete the square means.
please help.
x^2-14x+1=0
take the constant to the other side
x^2-14x = -1
now you want to turn the left side into a perfect square,
take 1/2 of the x term coefficient, then square it, and add it to both sides
x^2 - 14x + 49 = -1 + 49
now the left side is a perfect square
(x-7)^2 = 48
take the square root of both sides
x-7 = ±√48
x = 7 ±4√3
So is that the same way you'd do 3x^2+9x-12=0?
yes, you could, but divide each term by 3 first, if it says "solve by completing the square"
on the other hand it factors quite nicely
3x^2+9x-12=0 divide by 3
x^2 + 3x - 4 = 0
(x+4)(x-1) = 0
x = -4 or x = 1
So what if you were factofing 2x^2-x-35=0?
To solve the given equation, you can use a technique called completing the square. Completing the square involves rearranging the equation into a perfect square trinomial form, which can then be easily solved.
1. Start with the quadratic equation: x^2 - 14x + 1 = 0.
2. To complete the square, focus on the terms involving x. Ignore the constant term (1) for now.
3. Divide the coefficient of x by 2 and square it: (-14/2)^2 = 49.
4. Add this value as a constant to both sides of the equation: x^2 - 14x + 49 + 1 = 49.
This step is done to make the left side of the equation a perfect square trinomial.
5. Simplify the equation: x^2 - 14x + 50 = 49.
6. Now, rewrite the equation as a perfect square trinomial by factoring the left side: (x - 7)^2 = 49.
7. Take the square root of both sides: x - 7 = ±√49.
8. Solve for x by adding or subtracting the square root of 49 from both sides:
x - 7 = ±7.
Two equations are derived:
x - 7 = 7,
x - 7 = -7.
9. Solve each equation separately:
For x - 7 = 7, add 7 to both sides: x = 14.
For x - 7 = -7, add 7 to both sides: x = 0.
So, the equation x^2 - 14x + 1 = 0 is solved by completing the square, giving x = 14 and x = 0 as the solutions.