Question

What is the following product? Assume y greater-than-or-equal-to 0

3 StartRoot 10 EndRoot (y squared StartRoot 4 EndRoot + StartRoot 8 y EndRoot)

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

What is the following product? Assume y ≥ 0

$\sf{3 \sqrt{10}( {y}^{2} \sqrt{4} + \sqrt{8y} ) = }$

$\sf{a. \: \: \: 6 {y}^{2} \sqrt{10} + 12 \sqrt{5y} }$

$\sf{b. \: \: \: 6 \sqrt{10} + 12 \sqrt{5y} }$

$\sf{c. \: \: \: 6 {y}^{2} \sqrt{10} + 4 \sqrt{5y} }$

$\sf{d. \: \: \: 3 {y}^{2} \sqrt{10} + 12 \sqrt{5y} }$

EVALUATION

Here the given expression is

$\sf{3 \sqrt{10}( {y}^{2} \sqrt{4} + \sqrt{8y} )}$

We simplify it as below

$\sf{3 \sqrt{10}( {y}^{2} \sqrt{4} + \sqrt{8y} )}$

= $\sf{3 \sqrt{10}( {y}^{2} \sqrt{2 \times 2} + \sqrt{2 \times 2 \times 2 \times y} )}$

$\sf{ = 3 \sqrt{10}(2 {y}^{2} + 2 \sqrt{2 \times y} )}$

$\sf{ = 3 \sqrt{10}(2 {y}^{2} + 2 \sqrt{2 y} )}$

$\sf{ = 3 \sqrt{10} \times 2 {y}^{2} +3 \sqrt{10} \times 2 \sqrt{2 y} }$

$\sf{ = 6 \sqrt{10} {y}^{2} +3 \times 2 \sqrt{10 \times 2 \times y} }$

$\sf{ = 6 \sqrt{10} \: {y}^{2} +3 \times 2 \sqrt{2 \times 5 \times 2 \times y} }$

$\sf{ = 6 \sqrt{10} \: {y}^{2} +3 \times 2 \times 2\sqrt{ 5 \times y} }$

$\sf{ = 6 \sqrt{10} \: {y}^{2} +12\sqrt{ 5y} }$

$\sf{ = 6 {y}^{2} \sqrt{10} + 12 \sqrt{5y} }$

FINAL ANSWER

Hence the correct option is

$\sf{a. \: \: \: 6 {y}^{2} \sqrt{10} + 12 \sqrt{5y} }$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. If x=√3a+2b + √3a-2b / √3a+2b - √3a-2b

prove that bx²-3ax+b=0

https://brainly.in/question/19664646

2. the order of the surd 7√8

https://brainly.in/question/31962770