The cube of the number p is 16 times the number- Find p where- -0 and -4-A- 16B- 5C- 4D- 14

The correct option is C 4We are given that p^3 = 16p ⇒ p^3 – 16p = 0 ⇒ p(p^2 16) = 0 ⇒ Either p = 0 or (p^2 – 16 = 0) But p ≠ 0 (Given) ⇒ p^2 – 16 = 0 ⇒ ...

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Standard VIII

Mathematics

Cube Numbers

The cube of t...

Question

The cube of the number p is 16 times the number. Find p where, p $\neq$ 0 and p $\neq$ -4.

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Solution

The correct option is D 4
We are given that
$p^{3} = 16 p$
$\Rightarrow p^{3} - 16 p = 0$
$\Rightarrow p (p^{2} - 16) = 0$
$\Rightarrow E i t h e r p = 0 o r (p^{2} - 16 = 0)$
$B u t p \neq 0 (G i v e n)$
$\Rightarrow p^{2} - 16 = 0$
$\Rightarrow p^{2} = 16$
$\Rightarrow p = \pm 4$
$B u t p \neq - 4$
$∴ p = 4$ .

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The cube of the number p is 16 times the number. Find p where, p ≠ 0 and p ≠ -4.

The correct option is D 4We are given that p3=16p ⇒p3–16p=0 ⇒p(p2−16)=0 ⇒Either p=0 or (p2–16=0) But p≠0 (Given) ⇒p2–16=0 ⇒p2=16 ⇒p=±4 But p≠−4 ∴p=4.

The cube of the number p is 16 times the number. Find p where, p $\neq$ 0 and p $\neq$ -4.