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For this question I would consider different methods for finding the roots of quadratics and remind the student that an exam could ask for a particular method to be used (as on one of the recent GCSE maths p...

Firstly I would state the substitution rule, letting 5x Cos(3x)=w(x), and differentiating with respect to w(x). This gives y'(x)=2 w'(x) (w(x)).I would then demonstrate the product rule stating where u and v...

You first calculate the delta of the equation: delta = b^2-4 a c = 16 - 16 = 0 => the solution is double.x1,2 = (-b±sqrt(delta))/2/a = 4/2/4 = 1/2 which is real.so the solutions are 1/2 and 1/2.

(The full answer produced answer is annotated working out, but since this text box won't let me submit pictures, I'll do my best to transcribe)First, we gather the two fractions by using a common denominator...

If a quadractic equation ax^2 + bx + c = 0 has no real solutions, this means that the discriminant is less than 0, i.e. b^2-4ac<0. Let's put our equation in this form: x^2 + kx + 8 = k rearranges to x^2 +...