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Solve x^2 - 5x - 14 = 0

This is an example of a quadratic equation which we can solve on of two ways as it takes the form ax^2 + bx - c = 0a = 1, b = -5, c = -14 (1) Factorisation which we can do ourselves OR (2) Use the quadrat...

JB
Answered by Jessica B. Maths tutor
7433 Views

Solve algebraically: 1) 6a + b = 16, 2) 5a - 2b = 19

Firstly multiply the first equation by 2 resulting in the equation 12a + 2b = 32. Following this, add the first equation to the second equation, this will result in the 2b cancelling out with just 17a = 5...

AS
Answered by Andrew S. Maths tutor
3039 Views

Find the equation of the normal to the curve at the point (1, -1 ): 10yx^2 + 6x - 2y + 3 = x^3

Firstly, an expression for dy/dx needs to be found to allow us to find the gradient of the normal. As a normal is a straight line, the equation y-y1=m(x-x1) can be used to find its e...

TD
Answered by Tutor125302 D. Maths tutor
4645 Views

How to solve a simultaneous equation?

In simultaneous equations, at this level, you are often given two equations both containing an x and a y. For example:Equation1: y + 2x = 3Equation2: 4y + 2x = 8.The first step in approaching this is to t...

SB
Answered by Sophie B. Maths tutor
3400 Views

Express x^2-4x+9 in the form (x-p)^2+q where p and q are integers

The first step would be to expand the second equation:(x-p)^2+qx^2-px-px+p^2+q

this simplifies to x^2-2px+p^2+q

After this you examine the two equations and identify their similariti...

MN
Answered by Mark N. Maths tutor
13173 Views

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