Solve the equation 2y^(1/2) -7y^(1/4) +3 = 0

2y1/2 -7y1/4+ 3 = 0 We need to use a substitution to obtain a quadratic.Let y1/4 = x (use the y with the smallest fractional power as your substitution)From this, we can see that y1/2 = x2 (using the laws of indices: (ya)b = yab )We substitute this in and obtain an equation in terms of x. The right hand side will stay the same as this is just equal to 0.The equation becomes:2x2 -7x + 3 = 0 We can now solve this by factorizing, (2x - 1)(x -3) = 0 we now get our solutions:(2x - 1) = 0 rearranging for x we get: x = 1/2(x - 3) = 0 x = 3 Sub our values for x into the original substitution y1/4 = x We can rearrange this substitution for y:(y1/4)4 = (x)4y = x4 Now y = (1/2)4 = 1/16and y = (3)4 = 81 so the solutions of the equation are 1/16 and 81.

SL

Related Maths A Level answers

All answers ▸

Differentiate y=x^3


Find the binomial expansion of (-8+4x)^(2/3) up to and including the term in x^2.


Given that y = 4x^3 – 5/(x^2) , x =/= 0, find in its simplest form dy/dx.


Write cosx - 3sinx in the form Rcos(x + a)