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laiz [17]
4 years ago
14

If using the method of completing the square to solve the quadratic equation x^2+13x+15=0 which number would have to be added to

"complete the square"?
Mathematics
1 answer:
Ray Of Light [21]4 years ago
8 0

Answer:

The number to be added to complete the square is (13/2)^2

But the solved value for x are:

x = 11.71 or -1.28

Step-by-step explanation:

x ^ 2 + 13x + 15 = 0

x^2 + 13x = -15

(The coefficient of x^2 is 1, hence needless diving through)

Add the square of half the coefficient of x to both sides

x^2 + 13x + (13/2)^2 = -15 + ( 13/2)^2

(x + 13/2)^2 = -15 + 169/4

(x + 13/2)^2 = 109/4

Take the square of both sides

x + 13/2 = +-10.44/ 2

x = 10.44/2 + 13 /2

x = 23.44/2

x = 11.71

OR

x = 10.44/2 - 13/2

x = -2.56/2

x = -1.28

Therefore x = 11.71 or -1.28

But the number to be added to complete the square is (13/2)^2

Lets keep learning, maths is fun

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Answer:

See below:

Step-by-step explanation:

Problem 1:

Multiply Equation 1 by 4, keep Equation 2 the same.

x+y=8, multiply each term by 4:

4*x=4x, 4*y=4y, 8*4=32

so, the equivalent system is: 4x+4y=32 and x-y=2

Solve the system of equations:

x-y=2 becomes x=y+2

plug into 4x+4y=32 to solve for y

4(y+2)+4y=32---> 4y+8+4y=32--->8y=24---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 1 Answer:

Equivalent system: 4x+4y=32, x-y=2; solution: x=5, y=3

Problem 2:

Keep Equation 1 the same. Add 1 and 2.

To add an equation, add the left sides together, and then the rights.

so: x+y=8 + x-y=2 gives us: 2x=10

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plug x into x+y=8--->5+y=8--->y=3

Problem 2 answer:

Equivalent system: x+y=8, 2x=10; solution:x=5, y=3

Problem 3:

Subtract Equation 2 from 1, and keep 2 the same.

To subtract an equation, subtract the left sides, then the rights. We are subtracting 1<em> from </em>2, so its 2-1.

x-y=2 - x+y=8 gives us: -2y=-6

Solve for y by dividing by -2-->-2y/-2=-6/-2---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 3 answer:

Equivalent system: -2y=-6, x-y=2; solution: x=5, y=3

Problem 4:

Multiply the sum of Equation 1 and 2 by a factor of 3. Keep equation 2 the same.

First we add 1 and 2: (we did this earlier) ---> 2x=10 ---> now we multiply it all by 3---> 2x*(3)=10*(3)---> this gives us: 6x=30---> now divide by 6 to solve for x: 6x/6=30/6 gives us: x=5

Now, solve for y by plugging x into equation 2: x-y=2---> 5-y=2--->y=3

Problem 4 answer:

Equivalent system: 6x=30, x-y=2; solution: x=5, y=3

______

Quick Tip: One thing inherent of Equivalent systems is that they have the same set of solutions. Thus, we know the systems are equivalent when they have the same set of solutions for x and y. Moreover, you don't need to solve every time after you attempt to find an equivalent system, instead, just plug in the values found in problem 1 to each new set of equations to test if they are equivalent.

If we find x=5 and y=3 for x+y=8 and x-y=2, then all we have to do is plug them in to 6x=30 and -2y=-6 to see if they are equivalent.

6(5)=30 ---> true

-2(3)=-6 ---> true

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