WebDec 24, 2016 · Let #f(x) = x^3+x^2+kx-15# By the remainder theorem, the remainder when #f(x)# is divided by #(x-2)# will be #f(2)# So we have: #3 = f(2) = 8+4+2k-15 = 2k-3# Add #3# to both ends to get: #6 = 2k# Divide both sides by #2# and transpose to get: #k = 3# WebYou seem to be assuming that k2 + 2k − 1 is equal to (k −1)2, which is not true. Your equation k2 + 2k − 1 = 0 is correct, and it has roots k = −1± 2 ... More Items Examples Quadratic equation x2 − 4x − 5 = 0 Trigonometry 4sinθ cosθ = 2sinθ Linear equation y = 3x + 4 Arithmetic 699 ∗533 Matrix [ 2 5 3 4][ 2 −1 0 1 3 5] Simultaneous equation
Solve x^2-2kx+k^2+k-5=0 Microsoft Math Solver
Web$\begingroup$ We just wrote an equation, specifically, $2x^3+x^2+kx+6 = (x+3)P(x)$. Without doing more work, $P(x)$ could be anything, but that's ok. When you plug in $x = … WebThe values of k are precisely the solutions of Δ > 0 that's k 2 − 1 > 0, which gives k < − 1 ∨ k > 1. The quadratic formula uses b 2 − 4 a c. For there to be 2 distinct real roots, b 2 − 4 a c > … ipad model md510ll/a software upgrade
Solve (K+2)x^2-Kx+6=0 Microsoft Math Solver
Web2 x 3 + x 2 + k x + 6 = ( x + 3) q ( x) + 0 This is an equality and must work for ALL values of x, including x = -3 : 2 ( − 3) 3 + ( − 3) 2 + k ( − 3) + 6 = ( − 3 + 3) q ( − 3) 2 ( − 3) 3 + ( − 3) 2 + k ( − 3) + 6 = 0 k can be solved You may refer factor theorem for more info Share Cite Follow answered Aug 17, 2014 at 7:59 AgentS 12k 3 33 66 WebSince x-2 is supposed to be a factor, then we should get a remainder of zero when we divide. The remainder we got, 10-2k, must then be zero: 10-2k = 0 Adding 2k: 10 = 2k Dividing by 2: 5 = k This is the value for k that makes x-2 a factor of f (x). WebSolve the equation kx x 2 + 6 = 0 for the value of x if roots of this equation are real and equal. Login Study Materials NCERT Solutions NCERT Solutions For Class 12 NCERT … ipad mobile printer bluetooth