Respuesta :

number one is x=9,-9
number two is x=0
number three is x=12,-12
number four is 5i, -5i

For this case we have:

Question 1:

We want to know the solution of[tex]x ^ 2-81 = 0[/tex]

Adding 81 to both sides of the quadratic equation we have:

[tex]x ^ 2-81 + 81 = 81\\x ^ 2 = 81[/tex]

Applying square root on both sides of the equation:

[tex]\sqrt {x ^ 2} = \pm \sqrt {81}\\x = \pm 9[/tex]

So, we have two solutions:

[tex]x_ {1} = + 9\\x_ {2} = - 9[/tex]

Answer:

[tex]x_ {1} = + 9\\x_ {2} = - 9[/tex]

Question 2:

In this case, we want to solve the following quadratic equation:

[tex]2x ^ 2-26 = 0[/tex]

Adding 26 to both sides of the quadratic equation we have:

[tex]2x ^ 2-26 + 26 = 26\\2x ^ 2 = 26[/tex]

Dividing between 2 on both sides of the equation:

[tex]\frac {2x ^ 2} {2} = \frac {26} {2}\\x ^ 2 = 13[/tex]

Applying square root on both sides of the equation:

[tex]\sqrt {x ^ 2} = \pm \sqrt {13}[/tex]

[tex]x = \pm \sqrt {13}[/tex]

So, we have two solutions:

[tex]x_ {1} = + \sqrt {13}\\x_ {2} = - \sqrt {13}[/tex]

Answer:

[tex]x_ {1} = + \sqrt {13}\\x_ {2} = - \sqrt {13}[/tex]

Question 3:

For this case, we have a quadratic function of the form [tex]y = f (x)[/tex], where [tex]f (x) = x ^ 2-144[/tex]. They ask us to find the roots. So:

[tex]x ^ 2-144 = 0[/tex]

Adding 144 to both sides of the quadratic equation we have:

[tex]x ^ 2-144 + 144 = 144\\x ^ 2 = 144[/tex]

Applying square root on both sides of the equation:

[tex]\sqrt {x ^ 2} = \pm \sqrt {144}\\x = \pm 12[/tex]

So, we have two solutions:

[tex]x_ {1} = + 12\\x_ {2} = - 12[/tex]

Answer:

[tex]x_ {1} = + 12\\x_ {2} = - 12[/tex]

Question 4:

For this case we have a quadratic function of the form [tex]y = f (x)[/tex], where [tex]f (x) = x ^ 2 + 25[/tex]

Antoine says he has no solution. We must verify:

[tex]x ^ 2 + 25 = 0[/tex]

Subtracting 25 from both sides of the equation:

[tex]x ^ 2 + 25-25 = -25\\x ^ 2 = -25[/tex]

Applying square root on both sides of the equation:

[tex]\sqrt {x ^ 2} = \pm \sqrt {-25}[/tex]

By definition:

[tex]i = \sqrt {-1}\\i ^ 2 = -1[/tex]

So:

[tex]x = \pm \sqrt {25i ^ 2}\\x = \pm5i[/tex]

So, we have two solutions:

[tex]x_ {1} = + 5i\\x_ {2} = - 5i[/tex]

Answer:

[tex]x_ {1} = + 5i\\x_ {2} = - 5i[/tex]

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