What is The solution set of 2k^2=21k-49 ?
this an be arranged to the quadratic
2k^2 - 21k + 49 = 0
(2k-7)(k-7) = 0
k = 7/2 or k-7
Thanks! how do u determine that they are both -7 though? Doesn't it have to equal 21? or doesn't that 21 matter?
why don't you expand (2k-7)(k-7 to see what you get?
oh it makes sense now thanks!
what is Expand:-4(2m - 7x + 9)
To find the solution set of 2k^2 = 21k - 49, we can rearrange the equation into a quadratic equation and solve for k.
Step 1: Move all terms to one side of the equation:
2k^2 - 21k + 49 = 0
Step 2: The equation is now in the form ax^2 + bx + c = 0, where a = 2, b = -21, and c = 49. We can use the quadratic formula to solve for k:
k = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values, we get:
k = (21 ± √((-21)^2 - 4 * 2 * 49)) / (2 * 2)
Step 3: Simplify the expression under the square root:
k = (21 ± √(441 - 392)) / 4
k = (21 ± √49) / 4
k = (21 ± 7) / 4
Step 4: Solve for k:
For the positive square root:
k = (21 + 7) / 4
k = 28 / 4
k = 7
For the negative square root:
k = (21 - 7) / 4
k = 14 / 4
k = 3.5
Therefore, the solution set for the equation 2k^2 = 21k - 49 is {7, 3.5}.