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How do you solve the equation ax bxc?
The expression you presented is not an equation. Do you mean ax2 + bx = c? Do you mean to solve it for x? I'm assuming that's the case, but you need to be more clear on your question. To solve for x then, the technique to use is called completing the square: ax2 + bx = c Multiply both sides by a: a2x2 + abx = ac Add the square of b/2 to both sides: a2x2 + abx + (b/2)2 = ac + (b/2)2 We now have a perfect square on the left, simplify: (ax + b/2)2 = ac + b2 / 4 (ax + b/2)2 = (4ac + b2) / 4 And now solve for x: ax + b/2 = ±[(4ac + b2) / 4]1/2 ax + b/2 = ± √(4ac + b2) / 2 ax = [-b ± √(4ac + b2)] / 2 x = [-b ± √(4ac + b2)] / 2a
How do you solve for x in ax-x equals c?
ax - x = c Factor 'x' Hence x(a - 1) = c (NB '1' is ONE , not 'I'. ) Divide both sides by 'a-1' Hence x = c/(a-1) Solved for 'x'.
How do i solve rank of a determinant?
Matrix inverses and determinants, square and nonsingular, the equations AX = I and XA = I have the same solution, X. This solution is called the inverse of A.
If x is between a and c how are ax and ac related?
If a,b, and c are positive a < x < b means ax < cx If a,b , and c are negative a < x < b means ax > cx
What To solve an equation of the form ax you need to divide by the?
Co variable
