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# Find the value of p which satisfies the equation $$\sqrt P-\frac 6 p = 1$$

### Question

Find the value of p which satisfies the equation $$\sqrt P-\frac 6 p = 1$$

A) 4

B) -4

C) 9

D) -9

### Explanation:

$$\sqrt P-\frac 6{\sqrt p}=1$$
Multiply through by $$\sqrt P$$
$$P - 6 = \sqrt P$$
Square both side $$(P - 6)^{2} = (\sqrt P)^{2}$$
$$P^{2} - 12P + 36 = P$$
$$P^{2} - 12P - P + 36 = 0$$
$$P = 9$$ or $$4$$
Check to see if 9 or 4 satisfied the equation
$$\sqrt P-\frac 6{\sqrt P}=1$$
When $$P = 9$$
$$\sqrt 9-\frac 6{\sqrt P}=1$$
$$3-\frac 6 3=1$$
$$3 - 2 = 1$$
$$1 = 1$$
Hence the value p = 9 satisfied the equation when p = 4
$$\sqrt 4-\frac 6{\sqrt 4}=1$$
$$2-\frac 6 2=1$$
$$2 - 3 = 1$$
$$-1 \neq 1$$
Hence the value p = 4 does not satisfy the equation $${\therefore} p = 9$$

## Dicussion (1)

• $$\sqrt P-\frac 6{\sqrt p}=1$$
Multiply through by $$\sqrt P$$
$$P - 6 = \sqrt P$$
Square both side $$(P - 6)^{2} = (\sqrt P)^{2}$$
$$P^{2} - 12P + 36 = P$$
$$P^{2} - 12P - P + 36 = 0$$
$$P = 9$$ or $$4$$
Check to see if 9 or 4 satisfied the equation
$$\sqrt P-\frac 6{\sqrt P}=1$$
When $$P = 9$$
$$\sqrt 9-\frac 6{\sqrt P}=1$$
$$3-\frac 6 3=1$$
$$3 - 2 = 1$$
$$1 = 1$$
Hence the value p = 9 satisfied the equation when p = 4
$$\sqrt 4-\frac 6{\sqrt 4}=1$$
$$2-\frac 6 2=1$$
$$2 - 3 = 1$$
$$-1 \neq 1$$
Hence the value p = 4 does not satisfy the equation $${\therefore} p = 9$$