For this case we have two quadratic equations:
[tex]x ^ 2 = 25[/tex]
We apply square root to both sides of the equation:
[tex]x = \pm \sqrt {25}\\x = \pm5[/tex]
Thus, we have two solutions:
[tex]x_ {1} = 5\\x_ {2} = - 5[/tex]
On the other hand we have:
[tex]3x ^ 2 = 48[/tex]
We divide by 3 on both sides of the equation:
[tex]x ^ 2 = \frac {48} {3}\\x ^ 2 = 16[/tex]
We apply square root to both sides of the equation:
[tex]x = \pm \sqrt {16}\\x = \pm4[/tex]
Thus, we have two solutions:
[tex]x_ {1} = 4\\x_ {2} = - 4[/tex]
Answer:
Equation 1:
[tex]x_ {1} = 5\\x_ {2} = - 5[/tex]
Equation 2:
[tex]x_ {1} = 4\\x_ {2} = - 4[/tex]
