Question

1. What is the solution of n² – 49 = 0? (1 point)

Answer 1

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

Answer 2

1. The solution of the equation is the third option.
2. The equation has no real solution because the answers are both imaginary numbers.
3. The side length of a square with an area of 144x² is 12x.
5. The value of z so that 9 and -9 are both solutions is equal to 22.