Step 1. Recognize that you probably won't be asked to solve very many 4th order equations that don't have some trick associated with them. In this case, the equation is recognizably quadratic in x2.
Step 2. Actually, or virtually, substitute y=x2 to get
4y2-16y+15=0
Step 3. Factor as (2y-3)(2y-5)=0 (or (2x2-3)(2x2-5)=0)
Step 4. Recognize the solutions
y=3/2, y=5/2
Step 5. Reverse step 2 to get
x2=3/2
x2=5/2
Step 6. Solve each of these
x = +sqrt[3/2] = +sqrt[3]/sqrt[2]
x = -sqrt[3/2] = -sqrt[3]/sqrt[2]
x = +sqrt[5/2] = +sqrt[5]/sqrt[2]
x = -sqrt[5/2] = -sqrt[5]/sqrt[2]
_____
When asked to solve equations of this sort, you are usually being asked to find all solutions. This means that you must consider both positive and negative square roots of "b" when solving x2=b. A 4th order equation will have four (4) solutions. For word problems or other problems that place restrictions on the allowed solutions, some solutions may turn out to be irrelevant.
Often, a quick look at the coefficients of a quadratic will tell you if it is easily factored or not. In this case, the factors of the coefficient of x4 (4) are going to be either (1*4) or (2*2). Likewise, the factors of 15 (coefficient of x0) are going to be (1*15) or (3*5). You have to guess which pairs might be combined to give the 16 that is the middle coefficient. Here, we recognize that 2*5+2*3 will do it. The negative sign (on 16) means that we need to choose to factor 15 as (-3*-5) or 4 as (-2*-2). Usually we like to keep the leading coefficient positive.
If you don't immediately see the factors, you can always use the quadratic formula. Sometimes that is a bit more work, but it gets you to the same place.
y = (-b + sqrt[b2-4ac])/(2a) = (-(-16) + sqrt[(-16)2-4(4)(15)]/(2*4)
= (16 + sqrt[256-240])/8 = (16 + 4)/8 = 5/2 or 3/2
Step 2. Actually, or virtually, substitute y=x2 to get
4y2-16y+15=0
Step 3. Factor as (2y-3)(2y-5)=0 (or (2x2-3)(2x2-5)=0)
Step 4. Recognize the solutions
y=3/2, y=5/2
Step 5. Reverse step 2 to get
x2=3/2
x2=5/2
Step 6. Solve each of these
x = +sqrt[3/2] = +sqrt[3]/sqrt[2]
x = -sqrt[3/2] = -sqrt[3]/sqrt[2]
x = +sqrt[5/2] = +sqrt[5]/sqrt[2]
x = -sqrt[5/2] = -sqrt[5]/sqrt[2]
_____
When asked to solve equations of this sort, you are usually being asked to find all solutions. This means that you must consider both positive and negative square roots of "b" when solving x2=b. A 4th order equation will have four (4) solutions. For word problems or other problems that place restrictions on the allowed solutions, some solutions may turn out to be irrelevant.
Often, a quick look at the coefficients of a quadratic will tell you if it is easily factored or not. In this case, the factors of the coefficient of x4 (4) are going to be either (1*4) or (2*2). Likewise, the factors of 15 (coefficient of x0) are going to be (1*15) or (3*5). You have to guess which pairs might be combined to give the 16 that is the middle coefficient. Here, we recognize that 2*5+2*3 will do it. The negative sign (on 16) means that we need to choose to factor 15 as (-3*-5) or 4 as (-2*-2). Usually we like to keep the leading coefficient positive.
If you don't immediately see the factors, you can always use the quadratic formula. Sometimes that is a bit more work, but it gets you to the same place.
y = (-b + sqrt[b2-4ac])/(2a) = (-(-16) + sqrt[(-16)2-4(4)(15)]/(2*4)
= (16 + sqrt[256-240])/8 = (16 + 4)/8 = 5/2 or 3/2