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] Reply
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 Reply
Given :-
To Find :–
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]
[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]
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