Question

The square root of 4 - 4c + 2b + c2
- bc + b2
/4 is
A) b/2 - c + 2
B) b + c - 2
C) b/2 + c - 2
D) b - c + 2

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{Expression is}$

$\mathsf{4-4c+2b+c^2-bc+\dfrac{b^2}{4}}$

$\underline{\textsf{To find:}}$

$\textsf{Square root of}$

$\mathsf{4-4c+2b+c^2-bc+\dfrac{b^2}{4}}$

$\underline{\textsf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{4-4c+2b+c^2-bc+\dfrac{b^2}{4}}$

$\textsf{Using the identity,}$

$\boxed{\mathsf{a^2-2ab+b^2=(a-b)^2}}$

$\mathsf{=4-4c+2b+\left(\dfrac{b}{2}-c\right)^2}$

$\mathsf{=4+4\left(\dfrac{b}{2}-c\right)+\left(\dfrac{b}{2}-c\right)^2}$

$\mathsf{=\left(\dfrac{b}{2}-c\right)^2+2\left(\dfrac{b}{2}-c\right)(2)+4}$

$\mathsf{=\left(\left(\dfrac{b}{2}-c)+2\right)^2}$

$\mathsf{=\left(\dfrac{b}{2}-c+2\right)^2}$

$\implies\mathsf{4-4c+2b+c^2-bc+\dfrac{b^2}{4}=\left(\dfrac{b}{2}-c+2\right)^2}$

$\textsf{Taking square root on bothsides, we get}$

$\mathsf{\sqrt{4-4c+2b+c^2-bc+\dfrac{b^2}{4}}=\dfrac{b}{2}-c+2}$

$\therefore\textsf{option (A) is correct}$

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Answer 2

Answer:

Expression is

\mathsf{4-4c+2b+c^2-bc+\dfrac{b^2}{4}}4−4c+2b+c

2

−bc+

4

b

2

\underline{\textsf{To find:}}

To find:

\textsf{Square root of}Square root of

\mathsf{4-4c+2b+c^2-bc+\dfrac{b^2}{4}}4−4c+2b+c

2

−bc+

4

b

2

\underline{\textsf{Solution:}}

Solution:

\mathsf{Consider,}Consider,

\mathsf{4-4c+2b+c^2-bc+\dfrac{b^2}{4}}4−4c+2b+c

2

−bc+

4

b

2

\textsf{Using the identity,}Using the identity,

\boxed{\mathsf{a^2-2ab+b^2=(a-b)^2}}

a

2

−2ab+b

2

=(a−b)

2

\mathsf{=4-4c+2b+\left(\dfrac{b}{2}-c\right)^2}=4−4c+2b+(

2

b

−c)

2

\mathsf{=4+4\left(\dfrac{b}{2}-c\right)+\left(\dfrac{b}{2}-c\right)^2}=4+4(

2

b

−c)+(

2

b

−c)

2

\mathsf{=\left(\dfrac{b}{2}-c\right)^2+2\left(\dfrac{b}{2}-c\right)(2)+4}=(

2

b

−c)

2

+2(

2

b

−c)(2)+4

\mathsf{=\left(\left(\dfrac{b}{2}-c)+2\right)^2}

\mathsf{=\left(\dfrac{b}{2}-c+2\right)^2}=(

2

b

−c+2)

2

\implies\mathsf{4-4c+2b+c^2-bc+\dfrac{b^2}{4}=\left(\dfrac{b}{2}-c+2\right)^2}⟹4−4c+2b+c

2

−bc+

4

b

2

=(

2

b

−c+2)

2

\textsf{Taking square root on bothsides, we get}Taking square root on bothsides, we get

\mathsf{\sqrt{4-4c+2b+c^2-bc+\dfrac{b^2}{4}}=\dfrac{b}{2}-c+2}

4−4c+2b+c

2

−bc+

4

b

2

=

2

b

−c+2

\therefore\textsf{option (A) is correct}∴option (A) is correct

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