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For x!=0, multiply the equation by x to get x^4+5x^2+1=0. Then substitute t=x^2 where t>=0. So the equation has a form t^2+5t+1. Then find the discriminant and two roots. One of the roots t2<0 doesn...

Get a line a form y=-ax/b-c/b, then substitute into a cirle equation (x-p)^2 +(y-s)^2=r^2. Get a quadratic and find whether a discriminant is equal to zero. If it is then the line is tangent to a circle. ...

You start by dividing 420 by whichever numbers you know that it is divisible by.

420 is obviously divisible by 10 so you can start with 10 x 42.

Then, you factorise 10 and 10 = 2 x...

To solve this equation we want to find the value of one unknown, x. The solution will be x = ___. So first we need to put all the terms with x on one side of the =, and all the constants on the other side...

Pythagoras' theorem is: a^2+b^2=c^2 (a=short side, b=short side, c=longest side/hypotenuse, ^=squared). Now applying that to this question would mean that a=PQ, b=PR and c=QR. So we can use the figures gi...