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Solve x^2+6x+5=0 for x

x 2 +6x+5=(x+5)*(x+1)=0 therefore x+5=0 or x+1=0, hence x=-5 and x=-1
WR
Answered by William R. • Maths tutor
3572 Views

Integrate y=x^2 between the limits x=3 and x=1

Integrate y=x^2 which is 1/3 x^3 Subsitute the limits, (1/3 (3)^3)-(1/3 (1)^3) 27/3 - 1/3 = 26/3
CW
Answered by Caleb W. • Maths tutor
4175 Views

Given that y=((3x+1)^2)*cos(3x), find dy/dx.

As why is in the for y=uv where u and v are funtions of x, dy/dx=u'v+v'u (where ' implies the derivative) u=(3x+1) 2 , v=cos(3x) therefore using the chain rule u'=2 3 (3x+1)=18x+6 and v'=-3sin(3x). Using thi...
WR
Answered by William R. • Maths tutor
4126 Views

x^2 - 10x + 33 ≡ (x - a)^2 + b. Work out the value of a and b.

x 2 - 10x + 33 ≡ (x - a ) 2 + b Work out the value of a and b . Our aim here is to write the expression on the left in the same form as the one on the right so that we can compare the two. Therefore, conside...
JC
Answered by Joshua C. • Maths tutor
12156 Views

How do you solve algebraic fractions with quadratics?

First you need to remove the fractions from each side, take the equation: (x+1)/(x+3) = (2x-1)/(3x-1) now multiply by x+3 to give x+1 = (2x-1)(x+3)/(3x-1) now multiply by (3x-1) to give (3x-1)(x+1)=(2x-1)(x+...
ER
Answered by Eleanor R. • Maths tutor
3810 Views