If a^1/2 + b^1/2 – c^1/2 = 0, then the value of
(a + b-c)^2 is​

2 thoughts on “If a^1/2 + b^1/2 – c^1/2 = 0, then the value of<br />(a + b-c)^2 is​”

1. Given :-

   • [tex]\sf \;a^{\frac{1}{2}} + b^{\frac{1}{2}} – c^{\frac{1}{2}} = 0[/tex]

   To Find :–

   • Value of [tex] \mathsf{(a + b – c)^2 }[/tex]

   Solution :-

   We are given :-

   [tex]\mathsf\pink{{\: \: \: \: \: \: \: \:\;a^{\frac{1}{2}} + b^{\frac{1}{2}} – c^{\frac{1}{2}} = 0}}\\[/tex]

   [tex]\mathsf{\: \: \: \: \: \: \: \::\implies a^{\frac{1}{2}} + b^{\frac{1}{2}} = c^{\frac{1}{2}}}\\[/tex]

   ⠀⠀⠀⠀⠀[tex]\small\underline{\pmb{\sf Squaring \: on \: both \: sides:-}}\\[/tex]

   [tex]\implies \mathsf{\left(a^{\frac{1}{2}} + b^{\frac{1}{2}}\right)^2 = \left(c^{\frac{1}{2}}\right)^2}\\[/tex]

   • Here,left side expression is in form of Identity : (a+ b)² We know that (a + b )² = a² + b² + 2ab.

   [tex]\: \: \: \: \: \: \: \::\implies \mathsf{\left(a^{\frac{1}{2}}\right)^2 + \left(b^{\frac{1}{2}}\right)^2 + 2\left(a^{\frac{1}{2}}\right)\left(b^{\frac{1}{2}}\right) = \left(c^{\frac{1}{2}}\right)^2}\\[/tex]

   [tex]\;\;\textsf\green{{ \boxed{\mathsf{\left(p^{n}\right)^m = p^{mn}}}}}[/tex]

   [tex]\: \: \: \: \: \: \: \::\implies \mathsf{\left(a^{\frac{2}{2}}\right) + \left(b^{\frac{2}{2}}\right) + 2\left(a^{\frac{1}{2}}\right)\left(b^{\frac{1}{2}}\right) = \left(c^{\frac{2}{2}}\right)}\\[/tex]

   [tex]\: \: \: \: \: \: \: \::\implies \mathsf{a + b + 2\left(a^{\frac{1}{2}}\right)\left(b^{\frac{1}{2}}\right) = c}\\[/tex]

   [tex]\: \: \: \: \: \: \: \::\implies \mathsf{a + b – c = -2\left(a^{\frac{1}{2}}\right)\left(b^{\frac{1}{2}}\right)}\\[/tex]

   ⠀⠀⠀⠀⠀[tex]\small\underline{\pmb{\sf Squaring \: on \: both \: sides:-}}\\[/tex]

   [tex]\: \: \: \: \: \: \: \::\implies \mathsf{(a + b – c)^2 = (-1)^2(2)^2\left(a^{\frac{1}{2}}\right)^2\left(b^{\frac{1}{2}}\right)^2}\\[/tex]

   [tex]\: \: \: \: \: \: \: \::\implies \mathsf{(a + b – c)^2 = 4\left(a^{\frac{2}{2}}\right)\left(b^{\frac{2}{2}}\right)}\\[/tex]

   [tex]\: \: \: \: \: \: \: \:\pink{:\implies \mathsf{(a + b – c)^2 = 4ab}}\\\\[/tex]

   [tex]\therefore\:\underline{\textsf{ Value of (a + b – c)² is \textbf{4ab }}}.\\[/tex]

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2. Answer:

   √a+√b-√c=0

   or, √a+√b=√c

   Squaring both sides,

   (√a)²+2√a√b+(√b)²=(√c)²

   or, a+b+2√ab=c

   or, a+b-c=-2√ab

   Again squaring both sides,

   (a+b-c)²=(-2√ab)²

   or, (a+b-c)²=4ab

   ∴, The answer is c)4ab

   Step-by-step explanation:

   hope my answer help you

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