Question

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 1

Solution:-

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.