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x^2-9x+20=0

x 2 -9x+20=0x 2 -5x-4x+20=0x(x-5)-4(x-5)=0(x-4)(x-5)=0x-4=0x=4x-5=0x=5
GL
Answered by Georgia L. • Maths tutor
3699 Views

Solve these simultaneously to find values for a and b: 6a + b = 16 and 5a - 2b = 19

In order to tackle questions like this with two letters of unknown value, first what we try to do is eliminate one of the variables completely from an equation. If we call 6a + b = 16 eqn 1 and 5a - 2b = 19 ...
MP
Answered by Malvika P. • Maths tutor
5367 Views

Show that (x + 1)(x + 2)(x + 3) can be written in the form ax3 + bx2 + cx + d where a, b, c and d are positive integers.

(x+1)(x+2) = ( x^2 + 3x + 2) - multiplying out the first 2 terms(x^2 + 3x + 2)(x + 3) = x^3 + 3x^2 + 2x + 3x^2 + 9x + 6 - multiplying the product of the first two terms by the last termx^3 + 6x^2 + 11x + 6 -...
RK
Answered by Rachel K. • Maths tutor
7387 Views

Integrate exp(2x)cos(8x) by parts

Let u=exp(2x) and v'=cos(8x)From these you can obtain u' and vu=2exp(2x) and v=1/8 sin(8x)Formula: integral(uv'dx)=uv-integral(vu'dx)=1/8 exp(2x)sin(8x)-integral(1/4 sin(8x)exp(2x))=1/8exp(2x)sin(8x)+1/16cos...
CD
Answered by Chloe D. • Maths tutor
4017 Views

Solve the equation 8x^6 + 7x^3 -1 = 0

The first thing to recognise is this is a quadratic in disguise, therefore we can rewrite the equation in terms of a new variable y. Where y=x 3 The equation then becomes 8y 2 +7y-1=0 . We then factorise thi...
KP
Answered by Kelan P. • Maths tutor
8358 Views