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f(x)=x^2+12x+32=0, solve for x

Method 1: Quadratic formulaf(x)= ax2+bx+c, therefore a=1, b=12, c=32 - quadratic formula= (-b+/-(b2-4ac)1/2)/2ax=(-12+(144-128)1/2/2 = -4 or x=(-12-(144-128)

Answered by Maths tutor
3168 Views

Find the coordinates of the sationary points on the curve x^2 -xy+y^2=12

To find stationary points, we need to find dy/dx and set it equal to 0. Here we must use implicit differentiation: d/dx(x2) + d/dx(-xy) + d/dx(y2) = d/dx(12). Hence 2x - x(dy/dx) - y...

AB
Answered by Adam B. Maths tutor
9143 Views

f(x) = x^3+2x^2-x-2 . Solve for f(x) = 0

f(x) = x3 + 2x2 - x - 2 Use the factor theorem to test for f(1) = 0 f(1) = 13 + 2*12 - 1 - 2 = 0 Therefore x = 1 is a solution and (x-1) is a factor of f(x) Now...

ES
Answered by Elena S. Maths tutor
4403 Views

If y=2x+4x^3+3x^4 and z=(1+x)^2, find dy/dx and dz/dx.

Differentiation dy/dx = 2+12x^2+12x^3
Chain ruleu = 1+xz=u^2dz/dx = dz/du * du/dx = 2u * 1 = 2(1+x)

Answered by Maths tutor
3482 Views

solve this simulatneous equations (with clear algebraic working) : 5x-2y = 33 , 5x + 8y = 18


5x -2y = 33 (substract) 5x + 8y = 18
-10y = 15
y = -1.5
5x + 8(-1.5) = 18
5x = 30
x= 6

LE
Answered by Lara E. Maths tutor
4561 Views

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