Question

A4. Do as directed.

a. A leader on the court must be able to facilitate plays.
(Pick out the Modal Auxiliary and state its function)

b. A shooting guard must have high accuracy.
(Identify whether the underlined word is a Gerund or Present Participle)
​

Answer 1

Answer:

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Question:−

Simplify the problem.

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Solution:−

$\sqrt{a + b + 2x + 2 \sqrt{ab + (a + b)x + {x}^{2} } }$

Now, let

y = $\sqrt{ab + (a + b)x + {x}^{2} }$y=

Therefore,

$\sqrt{a + b + 2x + 2 \sqrt{ab + (a + b)x + {x}^{2} } }$

= $\sqrt{a + b + 2x + 2 + 2 \sqrt{y} }$=

Now, we will factorise y.

$\sqrt{ab + (a + b)x + {x}^{2} }$

= $\sqrt{ab + ax + bx + {x}^{2} }$=

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

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

Now,

= $\sqrt{a + b + 2x + 2 + 2 \sqrt{y} }$=

= $\small\sqrt{(x + a) + (x + b) + 2 \sqrt{(x + a)(x + b)} }$=

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

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

= $\sqrt{x + a} + \sqrt{x + b}$=

$\star\:\:\bf\large\underline\blue{Answer:-}$⋆

$\sqrt{x + a } + \sqrt{x + b}$