Using the quadratic formula, I found the solution set is x=2,x=-9
11
Given the equation x2 - x - 20 = 0, we see that it can likely be factored easily. Through simple trial and error we find that: (x - 5)(x + 4) = 0. From here, we set both terms equal to zero to find all possible solutions. x - 5 = 0 x = 5 and x + 4 = 0 x = -4. Substituting 5 and -4 back into the original equation yield true results. Thus the solution set is x = -4, 5. More formal, the solution set is {-4, 5}. Note: There are other methods of solving this problem such as the quadratic formula or by graphing the function.
An open statement is a sentence that contains a variable , such as x. The solution set for an open sentence is the set of values that when substituted for the variable make a true statement. The members of the solution set are called solutions. Examples: x = 2. Solution set is {2} solution is 2. x2 - 5 = 4 Solution set is {-3, 3 } solutions are -3 and 3. x > 0 Solution set = {x " x > 0 } That is all positive numbers. Every positive number is a solution. There are some finer points that I did not mention such as the possibility of more than one variable and limitations on the values that allowed in the substitutions.
x - x2 - 9x + 14 = 0 ; whence, x2 + 8x = 14 , x2 + 8x + 16 = 30 , and x + 4 = ±√30 . Therefore, x = ±√30 - 4 .
x2 + 49 = 0 ∴ x2 = -49 ∴ x = 7i
I will assume you mean,X2 - 16 = 0X2 = 16take square root each sideX = (+/-) 4=========(-4, 0) and (4, 0)----------------------
X2 - X - 6 = 0what two factors of - 6 add up to - 1 ?(X + 2)(X - 3)============(- 2, 0 ) and (3, 0 )------------------------------solution set of points
Using the quadratic formula, I found the solution set is x=2,x=-9
6x2-24=0 6(x2-4)=0 x= {-2,2}
Yes. Consider x2 ≥ 0
Nope. Consider x2+0x+1=0. This means x2+1=0. This has two solutions, but they are complex numbers: +i and -i, where i is the squareroot of -1. How about x2+0x+0=0? This means x2=0. This has two solutions, sure, but they aren't distinct. In this case, x=0 for both solutions, so we just consider them one solution.
x2 + 6x = 16=> x2 + 6x - 16 = 0=> x2 + 8x -2x - 16 = 0=> (x+8)(x-2) = 0=> x = -8 or x = 2So, the solutions of the quadratic equation x2 + 6x = 16 are -8 and 2.
Given: x2 + 10x - 16 Let: x2 + 10x - 16 = 0 x2 + 10x - 16 = 0 ∴ x2 + 10x + 25 = 16 + 25 ∴ (x + 5)2 = 41 ∴ x = -5 ± √41 ∴ x2 + 10x - 16 = (x + 5 + √41)(x + 5 - √41)
x2 + 6x = 16x2 + 6x - 16 = 0(x + 8)(x - 2) = 0x + 8 = 0 or x - 2 = 0x = -8 or x = 2Since it is given that x > 0, then x = -8 is not a solution of the equation. Thus, the only solution is x = 2.
-22
x2 + 11x + 18 (x + 9)(x + 2) CHECK: x2 + 9x + 2x + 18 x2 + 11x + 18 SET EACH EQUAL TO ZERO: x + 9 = 0 x = -9 x + 2 = 0 x = -2 NOW YOU ARE DONE: Solution set: {-9, -2}