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How do I factorise difficult quadratic equations?

A quadratic equation such as 6x 2 + x - 35 = 0 may appear much harder to solve than a simpler equation which students might be more used to, such as x 2 + 5x + 6 = 0. In reality, there are a couple of method...
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Answered by Freddie G. Maths tutor
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A stone is thrown upwards with a speed of v metres per second. The stone reaches a maximum height of h metres. h is directly proportional to v^2. When the stone is thrown at 10m/s, max height is 5m. Work out the maximum height reached when v = 24.

(Question 20 in AQA calculator paper from November 2017)This question is really wordy, so the first thing we want to do is condense all the information we are given into more manageable equations! The first ...
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Answered by Olivia H. Maths tutor
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Write x^2+6x-7 in the form (x+a)^2+b where a and b are integers

Complete the square.We want a quadratic we can simplify.Halve the linear term coefficient (6) and square it.Add it to the (x 2 +6x) term and subtract it from the 7.x 2 +6x+(6/2) 2 -7-(6/2) 2 As you can see t...
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Answered by George B. Maths tutor
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Write as a single fraction in it's simplest form: 2/(y+3)-1/(y-6)

The aim is to find a common denominator, we first therefore cross multiply to get:(2(y-6)-(y+3))/((y+3)(y-6))After expanding the relevant brackets we are left with:(2y-12-y-3)/((y+3)(y-6))Which is simply:(y-...
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Answered by James F. Maths tutor
7372 Views

A curve, C, has equation y =(2x-3)^5. A point, P, lies on C at (w,-32). Find the value of w and the equation of the tangent of C at point, P in the form y =mx+c.

To find the value of w, let x = w and y = -32. Substitute these values into the equation of the curve, C: y = (2x-3)^5 => -32 = (2(w) - 3 )^5. Note: the symbol, =>, means "implies that." Fift...
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Answered by Lewis M. Maths tutor
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