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Find the positive square root of 4 x 2⅘+ 12 x 2⅗+ 37 x 2⅖ + 42 x 2⅕ + 49. full explanation i will make you brainlliest​

Answer»

ution:-

LET \sf{x=\sqrt[5]{2} }.

Then \sf{4x^4+12x^3+37x^2+42x+49} will have a square root.

Let the square root be \sf{2x^2+ax+b}.

After squaring, it needs to equal to the above.

\sf{\to{(2x^2+ax+b)^2=4x^4+12x^3+37x^2+42x+49}}

Let's apply POLYNOMIAL expansion here.

\sf{\to{4x^4+a^2x^2+b^2+2(2ax^3+abx+2bx^2)}}

\sf{\to{4x^4+4ax^3+a^2x^2+4bx^2+2abx+b^2}}

Comparing the COEFFICIENTS

  • \sf{4ax^3=12x^3} [a=3]
  • \sf{2\times3\times bx=42x} [b=7]

Therefore, it is a COMPLETE square.

\sf{(2x^2+3x+7)^2}

After substituting the value

\sf{2\times 2^\frac{2}{5} +3\times 2^\frac{1}{5} +7} is the positive value.



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