The number of values of x in the interval [0,3 ] satisfying the equation 2 ^2 x+5 x-3=0 is

Step-by-step solution

Solving 2 ^2x + 5 x - 3 = 0 The equation is quadratic in x . Substitute y = x , yielding 2y^2 + 5y - 3 = 0 . Factoring gives (2y - 1)(y + 3) = 0 . Solutions are y = 1 2 and y = -3 . Since the range of sine is [-1, 1] , x = -3 has no solution. For x = 1 2 , the general solution is x = n + (-1)^n 6 , where n Z . Within [0, 3 ] , valid values occur for n = 0, 1, 2, 3 , giving x = 6 , 5 6 , 13 6 , 17 6 . Thus, there are exactly four solutions. 4