Question

Medical Physics has two main
categories​

Answer 1

Answer:

$Aɴsᴡᴇʀ</p><p></p><p>\begin{gathered} \scriptsize \sf{ We \: have \: , \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \scriptsize \sf{x ^{2} + 2 \sqrt{2x} - 6 = 0 } \\ \scriptsize \sf{ \implies \: x^{2} + 3 \sqrt{2} x - \sqrt{2x} - 6 = 0 } \: \: \: \: \: \: \: \: \: \\ \scriptsize \sf{ \implies \: x(x + 3 \sqrt{2}) - \sqrt{2}(x + 3 \sqrt{2}) = 0 } \\ \scriptsize \sf{ \implies \: (x + 3 \sqrt{2})(x - \sqrt{2}) = 0 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \scriptsize \sf{ \implies \: x + 3 \sqrt{2} = 0 \: or , \: x - \sqrt{2} = 0 } \: \: \: \: \: \\ \scriptsize \sf{ \implies \: x = - 3 \sqrt{2} \: or \: , \: x = \sqrt{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }\end{gathered}Wehave,x2+22x−6=0⟹x2+32x−2x−6=0⟹x(x+32)−2(x+32)=0⟹(x+32)(x−2)=0⟹x+32=0or,x−2=0⟹x=−32or,x=2 \scriptsize \sf{Thus , x = - 3 \sqrt{2} \: and \: x = \sqrt{2} \: are \: two \: roots \: of \: the \: given \: equation . }Thus,x=−32andx=2aretworootsofthegivenequation.</p><p></p><p>━━━━━━━━━━━━━━━━━━━━━━━━━━━━</p><p></p><p>$

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