Question

For positive real numbers a and b, which is not true? a)√ab=√a√bb)(a+√b)(a-√b)=a²-b² c)√a/b=√a/√b d)(√a+√b) (√a-√b)=a+b​

Answer 1

Answer:

option 4

Step-by-step explanation:

option four

Answer 2

We need to recall the following rules of a square root.

• $\sqrt{x} \sqrt{x} =x$
• $\sqrt{x} \sqrt{y} =\sqrt{xy}$
• $\sqrt{\frac{x}{y} } =\frac{\sqrt{x} }{\sqrt{y} }$

Given:

$a$ and $b$ are the positive real numbers.

Option a) $\sqrt{ab}=\sqrt{a} \sqrt{b}$

From the product rule of square root, we get

$\sqrt{ab}=\sqrt{a} \sqrt{b}$

Hence, this equation is true.

Option b) $(a+\sqrt{b})(a-\sqrt{b})=a^2-b^2$

Let's consider,

$(a+\sqrt{b})(a-\sqrt{b})=a^2-a\sqrt{b}+a\sqrt{b}-b^2$

$(a+\sqrt{b})(a-\sqrt{b})=a^2-b^2$

Hence, this equation is true.

Option c) $\sqrt{\frac{a}{b} } =\frac{\sqrt{a} }{\sqrt{b} }$

From the quotient rule of square root, we get

$\sqrt{\frac{a}{b} } =\frac{\sqrt{a} }{\sqrt{b} }$

Hence, this equation is true.

Option d)  $(\sqrt{a} +\sqrt{b})(\sqrt{a} -\sqrt{b})=a+b$

Let's consider,

$(\sqrt{a} +\sqrt{b})(\sqrt{a} -\sqrt{b})=\sqrt{a}\sqrt{a}-\sqrt{a}\sqrt{b}+\sqrt{b}\sqrt{a}-\sqrt{b}\sqrt{b}$

$(\sqrt{a} +\sqrt{b})(\sqrt{a} -\sqrt{b})=a-\sqrt{ab}+\sqrt{ab}-b$

$(\sqrt{a} +\sqrt{b})(\sqrt{a} -\sqrt{b})=a-b$

Hence, this equation is not true.

Thus, the correct option is d) $(\sqrt{a} +\sqrt{b})(\sqrt{a} -\sqrt{b})=a+b$