Question

(p+2q)^2=16
Find the values ofp and q

Answer 1

Step-by-step explanation:

Solve simultaneously by adding the two equations.

p^2 + q^2 = 16

p^2 - q^2 = 8

——————–

2(p^2) = 24

From this we can work out p:

p^2 = 12

p = sqrt(12)

p = 2(sqrt(3))

This is roughly 3.46.

Therefore from the first equation we can sub in to get:

(2(sqrt(3)))^2 + q^2 = 16

12 + q^2 = 16

q^2 = 16–12

q^2 = 4

q = +/- 2

As q can be positive or negative, we will get two possible solutions.

Assuming q = +2, first of all:

(2(sqrt(3)))^3 + 2^3

= (2^3)(sqrt(3)^3) + 2^3

= (8)(3(sqrt(3))) + 8

= 24(sqrt(3)) + 8

This could also be expressed as (8)(3(sqrt(3)) + 1).

If you use a calculator, this equals 49.5692193816530518016. It’s not always an attractive answer.

As for q = -2:

(2(sqrt(3)))^3 + (-2)^3

The first term = 24(sqrt(3)), as shown previously. Therefore:

= 24(sqrt(3)) - 8

= 33.5692193816530518016

Answer 2

Answer:

Step-by-step explanation:

$p^2+4q^2+4pq=16$............1

$(p+2q)^2=16$

$(p+2q)=\sqrt16$

$(p+2q)=4$........2

p=4-2q.......2

substituting 2 in 1,

p^2+q^2+2pq=16

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

4q^2-16q+q^2+8q-4q^2=0

8q+q^2=0

q(8+q)=0

q=-8

-8 in 2,

p=4-2(-8)

=4+16

=20