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1. f(x) = x2 - 4x + 5

2. f(x) = x2 + 3x + 4

3. f(x) = -x2 + 2x - 3

4. f(x) = -x2 - 5x - 9

5. f(x) = 3x2 - x + 2

 May 23, 2017
 #1
avatar+7352 
+1

On these, we want to solve for x when f(x) = 0

 

1.

0 = x2 - 4x + 5

 

We can solve this for x using the quadratic formula with

a = 1, b = -4, and c = 5

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \\~\\ x = {-(-4) \pm \sqrt{(-4)^2-4(1)(5)} \over 2(1)} ={4 \pm \sqrt{16-20} \over 2} ={4\pm\sqrt{-4}\over2} =\frac{4\pm2i}{2} \\~\\ x=2\pm i\)

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2.

0 = x2 + 3x + 4

 

We can solve this for x using the quadratic formula with

a = 1, b = 3, and c = 4

 

Can you finish this one from here?

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3.

0 = -x2 + 2x - 3

 

We can solve this for x using the quadratic formula with

a = -1, b = 2, and c = -3

 

\(x = {-2 \pm \sqrt{2^2-4(-1)(-3)} \over 2(-1)} = {-2 \pm \sqrt{4-12} \over -2} = {-2 \pm \sqrt{-8} \over -2} = \frac{-2\pm2i\sqrt2}{-2} \\~\\ x=1\pm i\sqrt2\)

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4.

0 = -x2 - 5x - 9

 

We can solve this for x using the quadratic formula with

a = -1, b = -5, and c = -9

 

Can you finish this one from here?

__________________________________________________

 

5.

0 = 3x2 - x + 2

 

We can solve this for x using the quadratic formula with

a = 3, b = -1, and c = 2

 

\(x = {-(-1) \pm \sqrt{(-1)^2-4(3)(2)} \over 2(3)} ={1 \pm \sqrt{-23} \over 6} \\~\\ x={1 \pm i\sqrt{23} \over 6}\)

.
 May 23, 2017
edited by hectictar  May 23, 2017
 #2
avatar+2340 
+1

 

 

Finding the zeros of a function simply means to set the domain equal to zero and then solve. 

 

1. Let's do the first together:

 

\(x^2-4x+5=0\)

 

This equation is not factorable, so we have to use a different method other than factoring. Let's use the quadratic equation:

 

\(x = {-b\pm \sqrt{b^2-4ac} \over 2a}\)

 

a=1, b=-4, c=5

 

\(x = {-(-4) \pm \sqrt{(-4)^2-4(1)(5)} \over 2(1)}\)

 

Let's simplify this. First, I'll simplify what it is in the radical:
 

\(x = {4 \pm \sqrt{16-20} \over 2}\)

\(x = {4\pm2i\over 2}=2\pm i\)

 

The solution is 2+/- i.

2. Let's do the second one. Here is the original equation:

 

\(x^2+3x+4=0\)

 

This one, too, cannot be factored. I'll use the quadratic equation again:

 

\(x = {-3 \pm \sqrt{3^2-4(1)(4)} \over 2(1)}\)

 

I have simply substituted all the values into the equation. Now, solve for x:

 

\(x = {-3 \pm \sqrt{9-16} \over 2}\)

\(x = {-3 \pm \sqrt{-7} \over 2}\)

\(x = {-3\pm i\sqrt{7} \over2}\)

3. Here's the original equation:
 

\(-x^2+2x-3=0\)

 

I'll use the quadratic formula for the third time:

 

\(x = {-2 \pm \sqrt{2^2-4(-1)(-3)} \over 2(-1)}\)

\(x = {-2 \pm \sqrt{4-12} \over -2}\)

\(x = {-2\pm \sqrt{-8} \over -2}\)

\(x = {-2 \pm 2i\sqrt{2} \over -2}=1\pm i\sqrt{2}\)

 

4. Here's the original equation:

 

\(-x^2- 5x - 9=0\)

 

This is not factorable, so we must use the quadratic formula again. 

 

\(x = {5\pm \sqrt{-11} \over-2}\)

 

\(x = {-5 \pm i\sqrt{11} \over2}\)

 

5. This one is the same story as the previous ones:

 

\(x = {1\pm\sqrt{-23} \over6}\)

\(x=\frac{1\pm i\sqrt{23}}{6}\)

 May 23, 2017
edited by TheXSquaredFactor  May 23, 2017
 #3
avatar+98196 
+2

 

Whoa....hold on a minute, X^2.....These definitely have solutions.....!!!

 

They are not  real solutions, but they are solutions, nonetheless..........

 

 

 

cool cool cool

 May 23, 2017
 #4
avatar+2340 
0

Fine... you win...

 

Fixed for your liking!

TheXSquaredFactor  May 23, 2017
edited by TheXSquaredFactor  May 23, 2017

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