find value of (2x+x*2^(1/2))^2 when x=4(2^(1/2) -1)

asked by guest
on Nov 17, 2024 at 8:13 am



You asked:

Calculate the value of (2x+x⋅212)2{\left( 2 x + x \cdot {2}^{\frac{1}{2}} \right)}^{2} when x=4(212−1)x = 4 \left({2}^{\frac{1}{2}} - 1\right).

MathBot Answer:

(2x+x⋅212)2=(−8+2(−4+42)+82)2{\left( 2 x + x \cdot {2}^{\frac{1}{2}} \right)}^{2} = \left(-8 + \sqrt{2} \left(-4 + 4 \sqrt{2}\right) + 8 \sqrt{2}\right)^{2}

(212x+2x)2=(2⋅4(212−1)+212⋅4(212−1))2=(2⋅4(112−1)+4(112−1)12)2=(2⋅4(2−1)+2⋅4(2−1))2=(2(42−4)+2(42−4))2=(2(42−4)+82−8)2=(82−8+8−42)2=(42)2=(−8+2(−4+42)+82)2\begin{aligned}\left(2^{\frac{1}{2}} x + 2 x\right)^{2}&=\left(2 \cdot 4 \left(2^{\frac{1}{2}} - 1\right) + 2^{\frac{1}{2}} \cdot 4 \left(2^{\frac{1}{2}} - 1\right)\right)^{2}\\&=\left(2 \cdot 4 \left(\frac{1}{\sqrt{\frac{1}{2}}} - 1\right) + \frac{4 \left(\frac{1}{\sqrt{\frac{1}{2}}} - 1\right)}{\sqrt{\frac{1}{2}}}\right)^{2}\\&=\left(2 \cdot 4 \left(\sqrt{2} - 1\right) + \sqrt{2} \cdot 4 \left(\sqrt{2} - 1\right)\right)^{2}\\&=\left(2 \left(4 \sqrt{2} - 4\right) + \sqrt{2} \left(4 \sqrt{2} - 4\right)\right)^{2}\\&=\left(\sqrt{2} \left(4 \sqrt{2} - 4\right) + 8 \sqrt{2} - 8\right)^{2}\\&=\left(8 \sqrt{2} - 8 + 8 - 4 \sqrt{2}\right)^{2}\\&=\left(4 \sqrt{2}\right)^{2}\\&=\left(-8 + \sqrt{2} \left(-4 + 4 \sqrt{2}\right) + 8 \sqrt{2}\right)^{2}\end{aligned}


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