Question

root5 = p + root q......Find p and q​

Answer 1

Answer:

p=0p=0

q=\frac{1}{11}q=

11

1

Step-by-step explanation:

We have,

\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=p-7\sqrt{5}q

7−

5

7+

5

−

7+

5

7−

5

=p−7

5

q

⇒ \frac{(7+\sqrt{5})^2-(7-\sqrt{5})^2}{(7)^2-(\sqrt{5})^2}=p-7\sqrt{5}q

(7)

2

−(

5

)

2

(7+

5

)

2

−(7−

5

)

2

=p−7

5

q

⇒ $$\frac{49+5+14\sqrt{5}-49-5-14\sqrt{5}}{49-5}=p-7\sqrt{5}q$$

( Since, (a±b)² = a² ± 2ab + b² )

⇒ $$\frac{28\sqrt{5}}{44}=p-7\sqrt{5}q$$

⇒ $$\frac{7\sqrt{5}}{11}=p-7\sqrt{5}q$$

⇒ $$0-7\sqrt{5}(-\frac{1}{11})=p-7\sqrt{5}q$$

By comparing,

We get,

$$p=0$$

$$q=-\frac{1}{11}$$

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