Question

If m and n are natural numbers such that √7+√48=√m + √n. find m2+ n2

Answer 1

Answer:

Step-by-step explanation:

√(7+√48) = √(4+3+√(16.3) ) =√(2² + (√3)² + 4√3)) =√(2² + (√3)² + 2.2.√3))

= √(2 + √3)² = 2 + √3

Therefore from (1),

2 + √3 = √m + √n

There is a theorem on surd which states that if √(a+√b) = √x + √y , then √(a-√b) = √x-√y . Application of the theorem to the above equation leads us to the following two linear algebraic equations in the two unknown variables m and n.

2 + √3 = √m + √n

2 - √3 = √m - √n

Add and subtract the above two equations to obtain

2 = √m and √n = √3 These give on squaring,

m = 4 and n = 3 Substituting the above values for m and n,

m² + n² = 4² + 3² = 16 + 9 = 25

Answer 2

Hey

$\sqrt{m} + \sqrt{n} = \sqrt{7} + \sqrt{48} \\ \\on \: comparison \\ = > \sqrt{m} = \sqrt{7} \: and \: \sqrt{n} = \sqrt{48} \\ \\ = > m = 7 \: and \: n = 48 \\ \\ now, \\ m {}^{2} + n {}^{2} = 7 {}^{2} + 48 {}^{2} = 49 + 2304 = 2353$