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\[\sqrt{x ^{2}/2+11}=x-1\] solve??
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\[\sqrt{\frac{x^2}{2} + 11 }= x-1\]Like that?
yeah
square both side (x^2 /2) +11= (x-1)^2 (x^2 /2) +11 = x^2 -2x +1 2((x^2 /2) +11) = 2x^2 -4x +2 <-- multiply 2 on both sides to remove 1/2 before x^2 on LHS x^2+22 =2x^2 -4x +2 x^2 -4x -20=0 by quadratic eqt, we can solve for x
before x^2 on left hand side?
yes
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coeff. of x^2 on LHS is 1/2
okay, got it. thanks!
the answer is 2+2 \[\sqrt{6}\]
\[2+2\sqrt{6}\]
how do you get that?
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x=(−b±sqrt(b^2−4ac))/2a
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