if x and y are the roots of x²-4x+5 then write equation for 3x and 3y
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Answer:
Solving Equations Algebraically
Grapher
Calculat zero. In other words, if a*b = 0, then either a = 0, or b = 0, or both. For more on factoring polynomials, see the review section P.3 (p.26) of the text.
Example 1.
2x2 - 5x - 12 = 0.
(2x + 3)(x - 4) = 0.
2x + 3 = 0 or x - 4 = 0.
x = -3/2, or x = 4.
Square Root Principle
If x2 = k, then x = ± sqrt(k).
Example 2.
x2 - 9 = 0.
x2 = 9.
x = 3, or x = -3.
Example 3.

Example 4.
x2 + 7 = 0.
x2 = -7.
x = ± .
Note that  =  = , so the solutions are
x = ± , two complex numbers.
Completing the Square
The idea behind completing the square is to rewrite the equation in a form that allows us to apply the square root principle.
Example 5.
x2 +6x - 1 = 0.
x2 +6x = 1.
x2 +6x + 9 = 1 + 9.
The 9 added to both sides came from squaring half the coefficient of x, (6/2)2 = 9. The reason for choosing this value is that now the left hand side of the equation is the square of a binomial (two term polynomial). That is why this procedure is called completing the square. [ The interested reader can see that this is true by considering (x + a)2 = x2 + 2ax + a2. To get "a" one need only divide the x-coefficient by 2. Thus, to complete the square for x2 + 2ax, one has to add a2.]
(x + 3)2 = 10.
Now we may apply the square root principle and then solve for x.
x = -3 ± sqrt(10).
Example 6.
2x2 + 6x - 5 = 0.
2x2 + 6x = 5.
The method of completing the square demonstrated in the previous example only works if the leading coefficient (coefficient of x2) is 1. In this example the leading coefficient is 2, but we can change that by dividing both sides of the equation by 2.
x2 + 3x = 5/2.
Now that the leading coefficient is 1, we take the coefficient of x, which is now 3, divide it by 2 and square, (3/2)2 = 9/4. This is the constant that we add to both sides to complete the square.
x2 + 3x + 9/4 = 5/2 + 9/4.
The left hand side is the square of (x + 3/2). [ Verify this!]
(x + 3/2)2 = 19/4.
Now we use the square root principle and solve for x.
x + 3/2 = ± sqrt(19/4) = ± sqrt(19)/2.
x = -3/2 ± sqrt(19)/2 = (-3 ± sqrt(19))/2
So far we have discussed three techniques for solving quadratic equations. Which is best? That depends on the problem and your personal preference. An equation that is in the right form to apply the square root principle may be rearranged and solved by factoring as we see in the next example.
Example 7.
x2 = 16.
x2 - 16 = 0.
(x + 4)(x - 4) = 0.
x = -4, or x = 4.
In some cases the equation can be solved by factoring, but the factorization is not obvious.
The method of completing the square will always work, even if the solutions are complex numbers, in which case we will take the square root of a negative number. Furthermore, the steps necessary to complete the square are always the same, so they can be applied to the general quadratic equation
ax2 + bx + c = 0.
The result of completing the square on this general equation is a formula for the solutions of the equation called the Quadratic Formula.
Quadratic Formula
The solutions for the equation ax2 + bx + c = 0 are

We are saying that completing the square always works, and we have completed the square in the general case, where we have a,b, and c instead of numbers. So, to find the solutions for any quadratic equation, we write it in the standard form to find the values of a, b, and c, then substitute these values into the Quadratic Formula.
One consequence is that you never have to complete the square to find the solutions for a quadratic equation. However, the process of completing the square is important for other reasons, so you still need to know how to do it!
Examples using the Quadratic Formula:
Example 8.
2x2 + 6x - 5 = 0.
In this case, a = 2, b = 6, c = -5. Substituting these values in the Quadratic Formula yields

Notice that we solved this equation earlier by completing the square.
Note: There are two real solutions. In terms of graphs, there are two intercepts for the graph of the function f(x) = 2x2 + 6x - 5.

Example 9.
4x2 + 4x + 1 = 0
In this example a = 4, b = 4, and c = 1.

There are two things to notice about this example.
There is only one solution. In terms of graphs, this means there is only one x-intercept.

The solution simplified so that there is no square root involved. This means that the equation could have been solved by factoring. (All quadratic equations can be solved by factoring! What I mean is it could have been solved easily by factoring.)
4x2 + 4x + 1 = 0.
(2x + 1)2 = 0.
x = -1/2.
Example 10.
x2 + x + 1 = 0
a = 1, b = 1, c = 1

Note: There are no real solutions. In terms of graphs, there are no intercepts for the graph of the function f(x) = x2 + x + 1. Thus, the solutions are complex because the graph of y = x2 + x + 1 has no x-intercepts.