Answer:
1. \(n=\pm7\)
2. \(x=\pm7i[tex]
3.[tex]\pm 12x =a\)
4. 22
Step-by-step explanation:
1. \(n^2 - 49 = 0\)
\(n^2 =49\)
\(n= \sqrt{49}\)
\(n=\pm7\)
2. \(x^2+64 = 0\)
\(x^2 =-64\)
\(x= \sqrt{49}\)
\(x=\pm7i[tex]
3. Area of square = [tex]a^{2}\)
Where a is the side
We are given an area o square = \(144x^{2}\)
So, \(144x^{2}=a^{2}\)
\(\sqrt{144x^{2}}=a\)
\(\pm 12x =a\)
4. We are given that the solution of quadratic equation -9 and 9
So, equation becomes:
\((x+9)(x-9)=0\)
\(x^2- 81=0\)
So, in the given equation \(x^2 + z = 103\)
z should be the number from which if we subtract 103 so we get 81
Substitute z = 22
\(x^2 + 22 = 103\)
\(x^2 + 22 -103=0\)
\(x^2 -81=0\)
Thus the value of z is 22
