Solve x^2=4(x-3)^2

Start by look at the right hand side of the equation, first expand (x-3)2 which is also the same as (x-3)(x-3). This comes out to equal (x2-6x+9). Next multiply (x2 -6x+9) by 4 which was left outside of the bracket before it was expanded. This comes out to equal (4x2-24x+36).Now you are left with (x2=4x2-24x+36), minus the (x2) from the left hand side (to group all of the x values onto one side of the equation) to get (0=3x2-24x+36), divide through by 3 (due to 3 being a common factor through the equation) to simplify the equation. Next factorise the quadratic equation, which comes to being (x-2)(x-6)=0. Therefore the solutions of the equations are x=2 and x=6. To prove the solutions are correct, substitute each of the values separately and for the solution to be correct it should end up 0=0.

LR
Answered by Louis R. Maths tutor

3999 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve the simultaneous equations 4x + 5y = 13 and 3x - 2y = 27.


If m=a^x, n=a^y, and a^2 =(m^y n^x)^z Show xyz=1.


How to solve a simple simultaneous equation


Solve the simultaneous equations, 8x + 4y = 4 and 6x - 18y = 16


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences