Find the value of 2x²-3xy when x=√2+3 and y=√2-2

asked by guest
on Nov 25, 2024 at 4:37 am



You asked:

Calculate the value of 2x2−3xy2 {x}^{2} - 3 x y when x=2+3x = \sqrt{2} + 3 and y=2−2y = \sqrt{2} - 2.

MathBot Answer:

2x2−3xy=−3(−2+2)(2+3)+2(2+3)22 {x}^{2} - 3 x y = - 3 \left(-2 + \sqrt{2}\right) \left(\sqrt{2} + 3\right) + 2 \left(\sqrt{2} + 3\right)^{2}

2x2−3xy=2(2+3)2−3(2−2)(2+3)=−3(2−2)(2+3)+122+22=12−32+122+22=92+34=−3(−2+2)(2+3)+2(2+3)2\begin{aligned}2 x^{2} - 3 x y&=2 \left(\sqrt{2} + 3\right)^{2} - 3 \left(\sqrt{2} - 2\right) \left(\sqrt{2} + 3\right)\\&=- 3 \left(\sqrt{2} - 2\right) \left(\sqrt{2} + 3\right) + 12 \sqrt{2} + 22\\&=12 - 3 \sqrt{2} + 12 \sqrt{2} + 22\\&=9 \sqrt{2} + 34\\&=- 3 \left(-2 + \sqrt{2}\right) \left(\sqrt{2} + 3\right) + 2 \left(\sqrt{2} + 3\right)^{2}\end{aligned}


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