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Solve x^2 = 4 (x-3)^2 - MyTutor First expand the bracket (x-3) 2 separately to give x 2 - 6x +9 Next multiply by the 4 outside the brackets to give 4x 2 -24x +36 Now all the terms have been expanded you can collect the like …
How do I integrate cos^2x with respect to x? - MyTutor Finally, we multiply this by 1/2 (the constant we took outside the integral before) to give us a final result of (sin2x)/4 + x/2. Of course, don't forget the +C assuming this is an indefinite …
write showing all working the following algebraic expression as a ... write showing all working the following algebraic expression as a single fraction in its simplest form: 4- [ (x+3)/ ( (x^2 +5x +6)/ (x-2))]
Solve 4 (x-5)=18 - MyTutor There are 2 ways to solve this equation:Method 1:Expand the brackets by multiplying everything inside the brackets by 4 to make 4x-20=18Then +20 to both sides making 4x=38Then divide …
Solve 8 (4^x ) – 9 (2^x ) + 1 = 0 - MyTutor At first this equation seems tricky, but we can perform a clever substitution to simplify it. We notice that if let y = 2^x, then we can rewrite this as: 8 (y^2) -...
From factorising a^2-b^2 hence or otherwise simplify fully (x^2 This question is a GCSE Higher tier style question. First the student should recognise that a2-b2= (a-b) (a+b). Using this to solve (x2+4)2- (x2-2)2 - Where a = (x2...
Solve x^2 - 6x + 8 < 0 - MyTutor Here, our x^2 term is positive so we will draw a U shaped curve -> Don't draw the axes on yet!The question is asking us to solve for one or two values of x. The x^2 - 6x + 8 equation is often …
How do I factorise and solve 2x^2+4x-6=0? - MyTutor The 2 in front of the x^2 can cause confusion. In this case, the equation can be divided through by 2, to produce x^2+2x-3=0. From here it can be factorised by working out what adds to make '2' …
Solve x^2 = 4 (x – 3)^2 - MyTutor To solve this equation we need to firast expand it. x2=4 (x-3)2 To do this we first expand the brackets. Remeber anything squared means times it by itself (x-3)2...
Solve the simultaneous equations “x^2+y^2=4” and “x=2-y”. What … The full solutions to the simultaneous equations are therefore: “x=2 and y=0” or “x=0 and y=2”. To answer the second part of the question, we should recognise that the circle with radius 2 is …