Question

Solve the equation for all values of r. (22 – 16) (22 + 49) = 0 Please help

Answer 1

Correct expression: $(x^2-16)(x^2+49)=0$

Answer:

x =  4

x = -4

x = 7i or $7 \sqrt{-1}$

x = -7i or $(-7) \sqrt{-1}$

Step-by-step explanation:

$(x^2-16)(x^2+49)=0$

We do factorization of x^2 - 16

(x^2-16) = (x+4)(x-4)

(x+4)(x-4)(x^2+49) = 0

We require that any term in the multiplication to be zero to fulfil the requirement

so:

if x + 4 = 0 then (x+4)(x-4)(x^2+49) = 0

x = -4

if x - 4 = 0 then (x+4)(x-4)(x^2+49) = 0

x = 4

if (x^2+49) = 0 then (x+4)(x-4)(x^2+49) = 0

x^2 = -49

we have two roots:

If your course has already worked with complex number and the root of -1 then:

7i and -7i are solution as well

because 7i will be: $7^2 \sqrt{-1} ^2= 49 (-1) = -49$

and the same reasoning for -7i: $(-7)^2 \sqrt{-1} ^2= 49 (-1) = -49$