Question

If 9p^2+16q^2=25 and 3p+4q=1

Answer 1

Answer:

$9p {}^{2} + 16q {}^{2} = 25 \\ (3p + 4q) {}^{2} - 2 \times 3 p \times 4q = 25\\ 1 {}^{2} - 24pq = 25 \\ 24pq = - 24 \\ pq = - 1$

$3p + 4q = 1 \\ 3p + 4 \times ( \frac{ - 1}{p} ) = 1 \\ 3p {}^{2} - p - 4 = 0 \\ 3p {}^{2} - 4p + 3p - 4 = 0 \\ (3p - 4)(p + 1) = 0 \\ \: either \: \: (3p - 4) = 0 = > p = \frac{4}{3} \\ or \: \: (p + 1) = 0 = > p = - 1$

and the value of q is

$when \: \: p = \frac{4}{3} \: \: then \: q = \frac{ - 3}{4} \\ \\ when \: \: p = - 1 \: \: then \: \: q = 1$

......xD......,✌️✌️✌️✌️✌️✌️

Answer 2

Answer:

Step-by-step explanation:

9p^2 + 16^2=25

(3p)^2 + (4p)^2 =25 ---(¡)

3p + 4q = 1

4q = 1-3p ---(¡¡)

Now put value of 4q in equation (¡)

(3p)^2 + (1-3p)^2=25

9p^2 + (1)^2 + (3p)^2 - 6p =25

9p^2 + 1 + 9p^2 - 6p =25

18p^2 -6p = 24

18p^2-6p -24 = 0

6(3p^2-1p -4) = 0

3p^2 -1p - 4= 0/6

3p^2 -1p -4 =0

3p^2+3p-4p -4 =0

3p(p+1)-4(p+1)=0

(3p-4) (p+1) =0

3p-4 =0

3p =4

p =4/3 ---(¡¡¡)

p+1=0

P=-1 ---(¡v)

Now put value of both p in equation(ii)

4q = 1-3p

4q = 1-4/3

4q = -1/3

q = -1/12---(v)

4q = 1 - (-1)

4q = 2

q=2/4

q=1/2