This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
You're on a roll — here's how to solve the equation t^4 + 4t^3 - 6t^2 - 4t + 1 = 0: Step 1: Divide the entire equation by t^2. Since t=0 would lead to 1=0, t ≠ 0. We can safely divide by t^2. (t^4)/(t^2) + (4t^3)/(t^2) - (6t^2)/(t^2) - (4t)/(t^2) + (1)/(t^2) = 0 t^2 + 4t - 6 - (4)/(t) + (1)/(t^2) = 0 Step 2: Group terms with similar coefficients. (t^2 + (1)/(t^2)) + 4(t - (1)/(t)) - 6 = 0 Step 3: Make a substitution. Let y = t - (1)/(t). Then y^2 = (t - (1)/(t))^2 = t^2 - 2(t)((1)/(t)) + (1)/(t^2) = t^2 - 2 + (1)/(t^2). So, t^2 + (1)/(t^2) = y^2 + 2. Step 4: Substitute y into the equation and solve the resulting quadratic equation for y. (y^2 + 2) + 4y - 6 = 0 y^2 + 4y - 4 = 0 Using the quadratic formula y = -b ± sqrt(b^2 - 4ac)2a: y = -4 ± sqrt(4^2 - 4(1)(-4))2(1) y = -4 ± sqrt(16 + 16)2 y = -4 ± sqrt(32)2 y = -4 ± 4sqrt(2)2 y = -2 ± 2sqrt(2) Step 5: Substitute back y = t - (1)/(t) and solve for t. Case 1: y = -2 + 2sqrt(2) t - (1)/(t) = -2 + 2sqrt(2) Multiply by t: t^2 - 1 = (-2 + 2sqrt(2))t t^2 - (-2 + 2sqrt(2))t - 1 = 0 Using the quadratic formula for t: t = -(-2 + 2sqrt(2)) ± sqrt((-2 + 22))^2 - 4(1)(-1)2(1) t = 2 - 2sqrt(2) ± sqrt((4 - 82) + 8) + 42 t = 2 - 2sqrt(2) ± sqrt(12 - 82)2 To simplify sqrt(12 - 82), we look for (a-bsqrt(2))^2 = a^2 + 2b^2 - 2absqrt(2). Comparing 12 - 8sqrt(2) with a^2+2b^2 - 2absqrt(2), we have 2ab=8 ab=4 and a^2+2b^2=12. If a=2, b=2, then a^2+2b^2 = 4+2(4) = 12. So sqrt(12 - 82) = sqrt((2-22))^2 = |2-2sqrt(2)| = 2sqrt(2)-2. t = 2 - 2sqrt(2) ± (2sqrt(2)-2)2 Two solutions from this case: t_1 = 2 - 2sqrt(2) + (2sqrt(2)-2)2 = (0)/(2) = 0 This solution is extraneous because we divided by t^2 earlier, assuming t ≠ 0. t_2 = 2 - 2sqrt(2) - (2sqrt(2)-2)2 = 2 - 2sqrt(2) - 2sqrt(2) + 22 = 4 - 4sqrt(2)2 = 2 - 2sqrt(2) Case 2: y = -2 - 2sqrt(2) t - (1)/(t) = -2 - 2sqrt(2) Multiply by t: t^2 - 1 = (-2 - 2sqrt(2))t t^2 - (-2 - 2sqrt(2))t - 1 = 0 Using the quadratic formula for t: t = -(-2 - 2sqrt(2)) ± sqrt((-2 - 22))^2 - 4(1)(-1)2(1) t = 2 + 2sqrt(2) ± sqrt((4 + 82) + 8) + 42 t = 2 + 2sqrt(2) ± sqrt(12 + 82)2 To simplify sqrt(12 + 82), we look for (a+bsqrt(2))^2 = a^2 + 2b^2 + 2absqrt(2). Comparing 12 + 8sqrt(2) with a^2+2b^2 + 2absqrt(2), we have 2ab=8 ab=4 and a^2+2b^2=12. If a=2, b=2, then a^2+2b^2 = 4+2(4) = 12. So sqrt(12 + 82) = sqrt((2+22))^2 = 2+2sqrt(2). t = 2 + 2sqrt(2) ± (2 + 2sqrt(2))2 Two solutions from this case: t_3 = 2 + 2sqrt(2) + (2 + 2sqrt(2))2 = 4 + 4sqrt(2)2 = 2 + 2sqrt(2) t_4 = 2 + 2sqrt(2) - (2 + 2sqrt(2))2 = (0)/(2) = 0 This solution is also extraneous. The valid solutions for t are: t = 2 - 2sqrt(2), 2 + 2sqrt(2) Last free one today — make it count tomorrow, or type /upgrade for unlimited.