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Nana76 [90]
3 years ago
9

Solve x2 − 12x + 5 = 0 using the completing-the-square method. x = six plus or minus the square root of five x = negative six pl

us or minus the square root of five x = six plus or minus the square root of thirty one x = negative six plus or minus the square root of thirty one
Mathematics
1 answer:
Andru [333]3 years ago
5 0

we are given

x^2-12x+5=0

we have to solve it by completing square method

step-1: Move 5 on right side

x^2-12x+5-5=0-5

x^2-12x=-5

step-2: Break middle term

x^2-2*6*x=-5

step-3: Add 6^2 both sides

x^2-2*6*x+(6)^2=-5+(6)^2

(x-6)^2=31

step-3: Solve for x

\sqrt{(x-6)^2} =\sqrt{31}

we wil get two values

First value is

x-6=\sqrt{31}

add both sides 6

x-6+6=6+\sqrt{31}

x=6+\sqrt{31}

Second value is

x-6=-\sqrt{31}

add both sides 6

x-6+6=6-\sqrt{31}

x=6-\sqrt{31}

so, solutions are

x=6+\sqrt{31}

x=6-\sqrt{31}.............Answer

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1. not continuous, as the function definitions deliver different function values at x=1 when approaching this x from the left and from the right side.

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2 = a + b

3.

0 = 2a + b

4.

a = -2

b = 4

Step-by-step explanation:

the function is continuous at a specific point or value of x, if the f(x) = y functional value is the same coming from the left and the right side at that point.

1. that means that for x=1

3 - x = ax² + bx

so,

3 - 1 = a×1² + b×1 = a + b

2 = a + b

we have to use a=2 and b=3

2 = 2 + 3 = 5

2 is not equal 5, so the assumed equality is false, so the function is not continuous there.

2. point 1 gave us already the working relationship between a and b.

2 = a + b

only if that is true, is the function continuous at x=1.

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5x - 10 = ax² + bx

5×2 - 10 = a×2² + b×2 = 4a + 2b

10 - 10 = 4a + 2b

0 = 4a + 2b

0 = 2a + b

4. to find a and b to be continuous at both locations x=1 and x=2 both expressions in a and b must apply.

so, they establish a system of 2 equations with 2 variables.

2 = a + b

0 = 2a + b

a = 2 - b

0 = 2×(2-b) + b = 4 - 2b + b = 4 - b

b = 4

therefore

a = 2 - 4 = -2

5. I cannot draw a graph here.

just use now the function

3 - x, x < 1

‐2x² +4x, 1 <= x < 2

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