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Please expand the brackets in the following equation to get a quadratic equation. Then, please show using the quadratic formula that the solutions to the equation are x=3 and x=5. Here is the starting equation: (x-3)(x-5)=0

(X-3)(x-5)=0Use F.O.I.LFirst x multiplied by x gives x²outer x multiplied by -5 gives -5xinner x multiplied by -3 gives -3xlast -5x-3 gives +15combining we get x²-8x+15=0the quadrati...

Answered by Jacob G. • Maths tutor

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Show that the funtion (x-3)(x^2+3x+1) has two stationary points and give the co-ordinates of these points

Stationary points are points where the gradient of a function is equal to 0. In this question the product rule can be used to find the gradient of the given function. The product rule is given by u'v+uv'...

Answered by Joe R. • Maths tutor

3136 Views

Find the nature of the turning points of the graph given by the equation x^4 +(8/3)*x^3 -2x^2 -8x +177 (6 marks)

(1 mark) Differentiate equation in the question: 4x³+8x²-4x-8(1 mark) Equate this to zero: (x-1)(x+1)(x+2)=0(1 mark) Find turning points (roots of above equation): x=1,-1,-2(1 mark) ...

Answered by Elizabeth B. • Maths tutor

3158 Views

show that y = (kx^2-1)/(kx^2+1) has exactly one stationary point when k is non-zero.

Stationary points are found by considering the points at which the gradient of the function equal zero. For the above, you need to employ the quotient rule, since both numerator and denominator are f(x), ...

Answered by Rob M. • Maths tutor

4856 Views

A ball is projected vertically upwards from the ground with speed 21 ms^–1. The ball moves freely under gravity once projected. What is the greatest height reached by the ball?

Set out information given in question, and taking the upward direction to be positive: s (displacement) = ?, u (initial speed) = 21ms^-1, v (final speed at maximum height) = 0ms^-1, a ...

Answered by Shruti S. • Maths tutor

8151 Views