Question

I need help in finding numerical value please

Answer 1

Answer:

Step-by-step explanation:

$\sqrt{18}=\sqrt{2*3*3}=3\sqrt{2}$

$\sqrt{12}=\sqrt{2*2*3}=2\sqrt{3}$

$A=\sqrt{2}+\sqrt{18}\\\\=\sqrt{2}+3\sqrt{2}\\\\=(1+3)\sqrt{2}\\\\=4\sqrt{2}$

$B=\sqrt{3}+\sqrt{12}=\sqrt{3}+2\sqrt{3}\\\\ = (1+2)\sqrt{3}\\\\ = 3\sqrt{3}$

$\sqrt{B^{2}+A^{2}-2*B*A}=\sqrt{(B-A)^{2}}\\\\=B-A\\\\=3\sqrt{3}-4\sqrt{2}$

= 3* 1.732 - 4 *1.4142

= 5.196 - 5.6568

= -0.4608

Answer 2

Exact Answer:   $4\sqrt{2} - 3\sqrt{3}$

Approximate Answer:  0.46070182678574

Round the approximate value however you need to.

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Work Shown:

$C = \sqrt{ B^2 + A^2 - 2BA}\\\\C = \sqrt{ B^2 - 2BA + A^2}\\\\C = \sqrt{ (B-A)^2 }\\\\C = -(B-A) \ \ \text{ see note below}\\\\C = -B+A\\\\C = A-B\\\\C = (A) - (B)$

$C = (\sqrt{2}+\sqrt{18}) - (\sqrt{3}+\sqrt{12})\\\\C = \sqrt{2}+\sqrt{18} - \sqrt{3}-\sqrt{12}\\\\C = \sqrt{2}+\sqrt{9*2} - \sqrt{3}-\sqrt{4*3}\\\\C = \sqrt{2}+\sqrt{9}*\sqrt{2} - \sqrt{3}-\sqrt{4}*\sqrt{3}\\\\C = \sqrt{2}+3\sqrt{2} - \sqrt{3}-2\sqrt{3}\\\\C = 4\sqrt{2} - 3\sqrt{3} \ \ \text{ ..... exact value}\\\\C \approx 0.46070182678574 \ \ \text{ ..... approximate value}$

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Note: The rule used here is

If x is positive then $\sqrt{x^2} = x$

Or if x is negative, then $\sqrt{x^2} = -x$

Since B-A = -0.4607 approximately, we will use the second version of the rule above.