What is degree of;
1. (x²-4x+13)(x-5)³=0
2.(x+1)⁵(x-1)²=0
3.(x²+4)(x-3)³=0
4.(x-√2)⁶(x+√2)⁶ x⁴=0
What is the real roots of an equation;
1. (x²-4x+13)(x-5)³=0
2.(x+1)⁵(x-1)²=0
3.(x²+4)(x-3)³=0
4.(x-√2)⁶(x+√2)⁶ x⁴=0
What is the number of real roots;
1. (x²-4x+13)(x-5)³=0
2.(x+1)⁵(x-1)²=0
3.(x²+4)(x-3)³=0
4.(x-√2)⁶(x+√2)⁶ x⁴=0
Pasagut mga idle...
Brainliestt ko Makatama
1. (x²-4x+13)(x-5)³=0
2.(x+1)⁵(x-1)²=0
3.(x²+4)(x-3)³=0
4.(x-√2)⁶(x+√2)⁶ x⁴=0
What is the real roots of an equation;
1. (x²-4x+13)(x-5)³=0
2.(x+1)⁵(x-1)²=0
3.(x²+4)(x-3)³=0
4.(x-√2)⁶(x+√2)⁶ x⁴=0
What is the number of real roots;
1. (x²-4x+13)(x-5)³=0
2.(x+1)⁵(x-1)²=0
3.(x²+4)(x-3)³=0
4.(x-√2)⁶(x+√2)⁶ x⁴=0
Pasagut mga idle...
Brainliestt ko Makatama
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Answer:
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.
Example 3.
y2 + √2
x + 2/x
πx2/2 + x
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.
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Step-by-step explanation: