Notice that this is a biquadratic equation because it only has even powers of \(x\) . If we substitute \(u = x^2\) , then the equation becomes \[3u^2 + 10u - 8 = 0.\] We can solve this quadratic equation for \(u\) using factoring or the quadratic formula. Using the quadratic formula, we get \[u=\frac{-10\pm\sqrt{100+4\times 3 \times 8}}{2\times 6}=\frac{-10\pm \sqrt{196}}{6}=\frac{-10\pm 14}{6}\] which means that \(u = \frac{2}{3}\) or \(u = -4\) .

Since \(u = x^2\) , we have \[x^2 = \frac{2}{3} \quad \text{or} \quad x^2 = -4.\] Since the square of a real number cannot be negative, the equation \(x^2=-4\) has no real solutions. Taking the square root of both sides of the first equation gives us \[x = \pm\sqrt{\frac{2}{3}} = \pm \frac{\sqrt{6}}{3}\] Therefore, the real solutions of \(3x^4 + 10x^2 - 8 = 0\) are \(x=\frac{\sqrt{6}}{3}\) and \(x=-\frac{\sqrt{6}}{3}\) .

To eliminate the negative exponent, we multiply both sides of the equation by \(x^3\) , obtaining \[x^6 + 3x^3 - 10 = 0.\] This equation is quadratic in \(x^3\) , so if we let \(u=x^3\) we obtain: \[u^2 + 3u - 10 = 0\] This can be solved by factoring. To do this, we are looking for two numbers that multiply to \(-10\) and add up to \(3\) . These two numbers are \(5\) and \(-2\) , so we have: \[(u+5)(u-2)=0\] Thus, either \(u+5=0\) or \(u-2=0\) . Solving for u we obtain \[u=-5 \quad \text{or} \quad u=2.\] Since \(u = x^3\) , we have \[x^3 = 2 \quad \text{or} \quad x^3 = -5.\] Taking the cube root of each side gives us \[x = \sqrt[3]{2} \quad \text{or} \quad x = \sqrt[3]{-5} = -\sqrt[3]{5}.\] 

Notice that the expression \(x^2 + 3x\) appears in both factors on the left-hand side. If we let \(u = x^2 + 3x\) , we obtain \[(u + 4)(u + 5) = 6.\] Expanding the product, we have \[u^2 + 9u + 20 = 6,\] which gives us \[u^2 + 9u + 14 = 0.\] To factor the quadratic expression, we need to find two numbers that multiply to 14 and add up to 9. Those numbers are 7 and 2, so we have \[(u + 2)(u + 7) = 0,\] Therefore, \(u+2=0\) or \(u+7=0\) which means \[u = -2 \quad \text{or} \quad u = -7.\] Since \(u = x^2 + 3x\) , we have \[x^2 + 3x = -2 \quad \text{or} \quad x^2 + 3x = -7.\] Solving the first quadratic equation: \[x^2 + 3x + 2 = 0.\] To factor the expression we look for two numbers that multiply to 2 and add up to 3. Those numbers are 1 and 2, so we have \[(x+1)(x+2)=0,\] which has roots \(x=-1\) and \(x=-2\) . Solving the second quadratic equation: \[x^2 + 3x + 7 = 0\] using the quadratic formula: \[x = \frac{-3 \pm \sqrt{3^2 - 4(1)(7)}}{2(1)} = \frac{-3 \pm \sqrt{-19}}{2}.\] This means that this equation has no real solutions. Therefore, the solutions of the given equations are \(x=-1\) and \(x=-2\) .

When multiplying the first and fourth factors together, we obtain \[(x+1)(x+4) = x^2 + 5x + 4\] while when we multiply the second and third factors we obtain \[(x+2)(x+3) = x^2 + 5x + 6.\] Letting \(u = x^2 + 5x\) the original equation can be written as \[(u + 4)(u + 6) = 120.\] Expanding the product gives us \[u^2 + 10u + 24 = 120,\] or \[u^2 + 10u - 96 = 0.\] To factor this, we need to find two numbers that multiply to -96 and add up to 10. Those numbers are 16 and -6 so we obtain: \[(u + 16)(u - 6) = 0,\] Therefore, \(u+16=0\) or \(u-6=0\) , which means that \(u = -16\) or \(u = 6\) . Since \(u = x^2 + 5x\) , we have \[x^2 + 5x = -16 \quad \text{or} \quad x^2 + 5x = 6.\] The second equation gives us \[x^2 + 5x - 6 = 0.\] To factor this, we are looking for two numbers that multiply to -6 and add up to 5, which are 6 and -1. So we have: \[(x+6)(x-1) = 0,\] which has roots \(x=-6\) or \(x=1\) . The first equation gives us \[x^2 + 5x + 16 = 0\] Using the quadratic formula, we have \[x = \frac{-5 \pm \sqrt{5^2 - 4(1)(16)}}{2(1)} = \frac{-5 \pm \sqrt{-39}}{2} = \frac{-5 \pm i\sqrt{39}}{2}\] Thus the roots are \(x=-6\) , \(x=1\) , and \(x=\frac{-5 \pm i\sqrt{39}}{2}\) .

To make the equation more convenient to solve, we will use the method of completing the square. By completing the square for the first two terms, we obtain \[(x^2 + 5x)^2 - 25x^2+31x^2 + 30x + 5 =0\] or \[(x^2 + 5x)^2 +6x^2 + 30x + 5 =0\] or \[(x^2 + 5x)^2 +6(x^2 + 5x) + 5 =0.\] Now we set \(u = x^2 + 5x\) so that the above equation becomes \[u^2+6u+5=0.\] This can be factored to obtain: \[(u+1)(u+5)=0.\] This means that \(u+1 = 0\) or \(u+5=0\) , so we have \[u=-1 \quad \text{or} \quad u=-5.\] Since \(u = x^2 + 5x\) , we obtain \[x^2 + 5x = -1 \quad \text{or} \quad x^2 + 5x = -5.\] Solving the first equation, we have \[x^2 + 5x + 1 = 0.\] Using the quadratic formula \[x = \frac{-5 \pm \sqrt{5^2 - 4(1)(1)}}{2(1)} = \frac{-5 \pm \sqrt{21}}{2}.\] Solving the second equation, we have \[x^2 + 5x + 5 = 0.\] Using the quadratic formula: \[x = \frac{-5 \pm \sqrt{5^2 - 4(1)(5)}}{2(1)} = \frac{-5 \pm \sqrt{5}}{2}.\] Thus the solutions of the given equation are \(x = \frac{-5 \pm \sqrt{21}}{2}\) and \(x = \frac{-5 \pm \sqrt{5}}{2}\) .

Notice that the second fraction is the reciprocal of the first. Let \(u = \frac{x^2 + 2x}{x^2 - 1}\) . Then the equation becomes \[8u + \frac{3}{u} - 11 = 0.\] Multiplying both sides by \(u\) , we obtain \[8u^2 - 11u + 3 = 0.\] Factoring this quadratic equation: \[8u^2 - 11u + 3 = \frac{1}{8} \left[(8u)^2 - 11(8u) + 24 \right]\] We look for two numbers whose sum is \(-11\) and whose product is \(24\) . These numbers must both be negative, as their sum is negative and their product is positive. The numbers are \(-8\) and \(-3\) . Therefore, \[\frac{1}{8} \left[(8u)^2 - 11(8u) + 24 \right] = \frac{1}{8} (8u - 8)(8u - 3) = (u - 1)\left(\frac{8u - 3}{8}\right).\] Thus, either \(u - 1 = 0\) , which means \(u = 1\) , or \(8u - 3 = 0\) , which means \(u = \frac{3}{8}\) .

Therefore, we have two equations: \[u = \frac{x^2 + 2x}{x^2 - 1} = 1, \quad u = \frac{x^2 + 2x}{x^2 - 1} = \frac{3}{8}.\] 

From the first equation, we get \[x^2 + 2x = x^2 - 1\] or \[2x = -1\] which simplifies to \[x = -\frac{1}{2}. \tag{*}\] 

The second equation gives \[8(x^2 + 2x) = 3(x^2 - 1),\] which simplifies to \[8x^2 + 16x = 3x^2 - 3,\] yielding \[5x^2 + 16x + 3 = 0.\] Factoring: \[\frac{1}{5} \left[(5x)^2 + 16(5x) + 15 \right] = 0.\] We look for two numbers whose sum is \(16\) and whose product is \(15\) . The numbers are \(15\) and \(1\) . Thus, \[\begin{align} \frac{1}{5} \left[(5x)^2 + 16(5x) + 15 \right] &= \frac{1}{5}(5x + 15)(5x + 1) \\ &= (x + 3)\left(5x + \frac{1}{5}\right) = 0. \end{align}\] Thus, \[x = -3 \quad \text{or} \quad x = -\frac{1}{5}. \tag{**}\] 

Considering both (*) and (**), the solutions are \[x = -\frac{1}{2}, \quad x = -3, \quad x = -\frac{1}{5}.\] All the values of \(x\) thus found are solutions of the given equation since they cause none of its denominators to vanish.

This equation can be solved using a substitution to turn it into a quadratic equation. Since \(\frac{1}{3} = 2\cdot \frac{1}{6}\) , we can let \(u = x^{1/6}\) . This means that \(u^2 = (x^{1/6})^2 = x^{2/6} = x^{1/3}\) . Using this substitution, the equation becomes \[u^2 + u - 2 = 0.\] We can factor the quadratic expression in \(u\) as follows. We need to find two numbers that multiply to -2 and add to 1. Those two numbers are 2 and -1. This means \[(u + 2)(u - 1) = 0\] Therefore, either \(u+2=0\) or \(u-1=0\) , which gives us \[u = -2 \quad \text{or} \quad u = 1.\] Since \(u = x^{1/6}\) , we have \[x^{1/6} = -2 \quad \text{or} \quad x^{1/6} = 1.\] To find \(x\) , we raise both sides of each equation to the power of 6. For the first equation, \[(x^{1/6})^6 = (-2)^6\] which gives us \[x = 64.\] For the second equation \[(x^{1/6})^6 = 1^6\] which gives us \[x = 1.\] However, we need to check if these solutions satisfy the original equation, as these steps might introduce spurious solutions.

If we plug in \(x=64\) in the original equation we obtain: \[64^{1/3} + 64^{1/6} -2 = 4 + 2 - 2 = 4 \neq 0.\] So \(x=64\) is not a solution of the original equation.

If we plug in \(x=1\) we obtain: \[1^{1/3} + 1^{1/6} - 2 = 1 + 1 - 2 = 0.\] So \(x=1\) is a solution. Therefore, the only solution is \(x = 1\) .

To eliminate the denominators, we multiply each side by the least common multiple of all denominators: \(x,x-2,x^{2}-4\) . Because \(x^{2}-4=(x-2)(x+2)\) , the LCM of the denominators is \(x(x-2)(x+2)\) or \(x(x^{2}-4)\) \[\begin{align} x(x-2)(x+2)\left(\dfrac{5}{x}+\dfrac{1}{x-2}\right) & =x(x-2)(x+2)\dfrac{4}{x^{2}-4}\\ 5(x^{2}-4)+x(x+2) & =4x\\ 5x^{2}-20+x^{2}+2x & =4x\tag{expand LHS}\\ 6x^{2}-2x-20 & =0\tag{simplify} \end{align}\] Using the quadratic formula, we find the solutions of the above equation: \[x=\frac{2\pm\sqrt{4+240}}{12}=\frac{2\pm22}{12}\] so \(x=-5/3\) or \(x=2\) . Now we need to check if these values satisfy the original equations: 
Checking \(x=-5/3\) : \[\begin{align} \text{Left Hand Side (LHS)} & =\frac{5}{-5/3}+\frac{1}{-5/3-2}\\ & =-3-\frac{1}{11/3}\\ & =-3-\frac{3}{11}=\frac{-36}{11} \end{align}\] \[\begin{align} \text{Right hand side (RHS)} & =\frac{4}{(-5/3)^{2}-4}\\ & =\frac{4}{\frac{25}{9}-4}\\ & =\frac{4}{\frac{-11}{9}}=-\frac{36}{11} \end{align}\] Because LHS = RHS, \(x=-5/3\) is a solution. 
Checking \(x=2\) : \[\text{LHS}=\frac{5}{2}+\frac{1}{2-2}\text{ division by zero! undefined}\] \[\text{RHS}=\frac{4}{4-4}\text{ division by zero! undefined}\] so \(x=2\) is not acceptable and the only solution is \(x=-5/3\) .

Clearing of fractions (by multiplying each term by \((x+1)(x+2)(x+3)(x+4)\) ) and simplifying, we get: \[\begin{align} (x+1)(x+2)&(x+3)(x+4)\left(\frac{1}{x+1} + \frac{1}{x+2}\right) \\ &=(x+1)(x+2)(x+3)(x+4)\left( \frac{1}{x+3} + \frac{1}{x+4}\right) \end{align}\] \[\begin{align} (x+2)(x+3)&(x+4) + (x+1)(x+3)(x+4) \\ &= (x+1)(x+2)(x+4) + (x+1)(x+2)(x+3)\\ (x^2+5x+6)&(x+4) +(x^2+4x+3)(x+4) \\ & = (x^2+3x+2)(x+4) + (x^2+3x+2)(x+3) \\ \end{align}\] \[\begin{align} \text{LHS}=& (x+2)(x+3)(x+4)+(x+1)(x+3)(x+4) \\ =&(x+3)(x+4)[x+2+x+1]\\ =&(x^2+7x+12)[2x+3]\\ =&2x(x^2+7x+12)+3(x^2+7x+12)\\ =& 2x^3+14x^2+24x+3x^2+21x+36 \end{align}\] \[\begin{align} \text{RHS} &=(x+1)(x+2)(x+4)+(x+1)(x+2)(x+3)\\ &=(x+1)(x+2)[x+4+x+3]\\ &=(x^2+3x+2)[2x+7]\\ &=2x(x^2+3x+2)+7(x^2+3x+2)\\ &=2x^3+6x^2+4x+7x^2+21x+14\\ &=2x^3+13x^2+25x+14 \end{align}\] \[\begin{align} \text{LHS} &= \text{RHS}\\ 2x^3+17x^2+45x+36 &= 2x^3+13x^2+25x+14 \end{align}\] \[4x^2 + 20x + 22= 0\] \[2x^2 + 10x + 11 = 0\] 

Using the quadratic formula, we have: \[\begin{align} x &= \frac{-10 \pm \sqrt{10^2 - 4(2)(11)}}{2(2)} \\ &= \frac{-10 \pm \sqrt{100 - 88}}{4} \\ &= \frac{-10 \pm \sqrt{12}}{4} \\ &= \frac{-10 \pm 2\sqrt{3}}{4} \\ &= \frac{-5 \pm \sqrt{3}}{2} \end{align}\] Thus, the solutions are \(x = \frac{-5 + \sqrt{3}}{2}\) and \(x = \frac{-5 - \sqrt{3}}{2}\) .

Both of these values of \(x\) are valid roots of the given equation since none of the denominators ( \(x+1\) , \(x+2\) , \(x+3\) , \(x+4\) ) vanish when \(x\) takes these values.

We can factor the denominators to get \[\frac{x+3}{(x-1)(x+1)} + \frac{x-3}{x(x-1)} + \frac{x+2}{x(x+1)} = 0\] 

The lowest common denominator is \(x(x-1)(x+1)\) . Multiplying through by this, we obtain: \[\begin{align} x(x+3) + (x+1)(x-3) + (x-1)(x+2) &= 0 \\ x^2+3x+x^2-2x-3+x^2+x-2 &= 0\\ 3x^2 + 2x - 5 &= 0 \end{align}\] Factoring this quadratic, we get \[(3x+5)(x-1) = 0\] So \(x=1\) or \(x=-\frac{5}{3}\) . However, \(x=1\) makes the denominators of the first two terms zero. Therefore, \(x=1\) is not a solution. Hence, the only solution is \(x=-\frac{5}{3}\) .

(a) Rewrite the equation as \[\sqrt{2x^{2}+4}=2x-2\] Then we square both sides: \[\begin{align} 2x^{2}+4 & =(2x-2)^{2}\\ 2x^{2}+4 & =4x^{2}-8x+4\\ 0 & =2x^{2}-8x\tag{simplify }\\ 0 & =2x(x-4)\tag{factor} \end{align}\] Therefore \(x=0\) or \(x=4\) . Substituting \(x=0\) in the original equation, we get \[\begin{align} \text{LHS} & =2+\sqrt{4}=4\\ \text{RHS} & =0 \end{align}\] Because LHS \(\neq\) RHS, \(x=0\) is an extraneous solution. 
Substituting \(x=4\) in the original equation, we get \[\begin{align} \text{LHS} & =2+\sqrt{2\cdot4^{2}+4}=2+6=8\\ \text{RHS} & =2\cdot4=8 \end{align}\] and LHS \(=\) RHS. So the only solution of this equation is \(x=4\) . 
(b) Rewrite the equation as \[-\sqrt{2x^{2}+4}=-(2+2x)\] or \[\begin{align} \sqrt{2x^{2}+4} & =2+2x\\ 2x^{2}+4 & =(2+2x)^{2}\tag{square both sides}\\ 2x^{2}+4 & =4+8x+4x^{2}\tag{expand}\\ 0 & =2x^{2}+8x\tag{simplify}\\ 0 & =2x(x+4) \end{align}\] which gives \(x=0\) or \(x=-4\) . Substituting \(x=0\) in the original equation, we get \[\begin{align} \text{LHS} & =2-\sqrt{4}=0,\\ \text{RHS} & =-2\cdot0=0 \end{align}\] Because LHS \(=\) RHS, \(x=0\) is a solution. Substituting \(x=-4\) in the original equation, we get \[\begin{align} \text{LHS} & =2-\sqrt{2\cdot(-4)^{2}+4}=2-6=-4\\ \text{RHS} & =-2\cdot(-4)=8 \end{align}\] Because LHS \(\neq\) RHS, \(x=-4\) is not a solution (it is an extraneous solution) and so \(x=0\) is the only solution.

(a) \[\begin{align} \sqrt{x+19} & =9-\sqrt{x+10}\tag{rewrite the equation}\\ x+19 & =81-18\sqrt{x+10}+x+10\tag{square both sides}\\ 18\sqrt{x+10} & =72\tag{simplify}\\ \sqrt{x+10} & =4\\ x+10 & =16\tag{square both sides}\\ x & =6 \end{align}\] Let’s check if \(x=6\) satisfies the original equation: \[\begin{align} \text{LHS} & =\sqrt{6+19}+\sqrt{6+10}=\sqrt{25}+\sqrt{16}=5+4\\ \text{RHS} & =9 \end{align}\] Because LHS \(=\) RHS, \(x=6\) is the solution. 
(b) \[\begin{align} \sqrt{x-2} & =3+\sqrt{2x+5}\tag{rewrite the equation }\\ x-2 & =9+6\sqrt{2x+6}+2x+5\tag{square both sides}\\ -6\sqrt{2x+6} & =x+16\tag{simplify}\\ 36(2x+6) & =x^{2}+32x+16\tag{square both sides}\\ x^{2}-40x+76 & =0\tag{simplify } \end{align}\] Now we can factor it by trial and error as \[x^{2}-40x+76=(x-2)(x-38)\] which gives \(x=2\) or \(x=38\) . Alternatively we can use the quadratic equation to find the roots \[\begin{align} x & =\frac{40\pm\sqrt{1600-4\cdot76}}{2}\\ & =\frac{40\pm\sqrt{1296}}{2}\\ & =\frac{40\pm36}{2}\\ & =20\pm18 \end{align}\] which gives \(x=2\) or \(x=38\) . We have to test these values to see if they satisfy the original equation. 
Substituting 2 for \(x\) : \[\begin{align} \text{LHS} & =\sqrt{2-2}-\sqrt{2\cdot2+5}=-3,\\ \text{RHS} & =3. \end{align}\] Because LHS \(\neq\) RHS, \(x=2\) is not a solution. 
Substituting 38 for \(x\) : \[\begin{align} \text{LHS} & =\sqrt{38-2}-\sqrt{2\cdot38+5}=6-9=-3,\\ \text{RHS} & =3. \end{align}\] Because LHS \(\neq\) RHS, \(x=38\) is not a solution. Therefore, this problem does not have any solutions.