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Let C : x^2-4x+2k be a parabola, with vertex m. By taking derivatives or otherwise discuss, as k varies, the coordinates of m and, accordingly, the number of solutions of the equation x^2-4x+2k=0. Illustrate your work with graphs

Write y=x²-4x+2k. And m:= (x_m, y_m) for the coordinates of our vertex. We deduce that x_m is exactly the value of x for which y'=2x-4=0, because m is a minimum p...

Answered by Massimiliano V. • Maths tutor

2584 Views

Solve the simultaneous equation, 3x + y = 8 and x + 3y = 12, to find a value for x and y.

Re arrange one of the equation to get a single variant answer. So, x = 12 - 3y. Substitute this into the other equation, so 3 (12-3y) + y = 8. Expand this equation to form 36 - 9y + y = 8. Collect the y t...

Answered by Grace R. • Maths tutor

5484 Views

Show that (x + 4)(x + 5)(x + 6) can be written in the form ax3 + bx2 + cx + d where a, b, c and d are positive integers.

=(x+4)(x²+6x+5x+30)=(x+4)(x²+11x+30)=x³+11x²+30x+4x²+44x+120=x³+15x²+74x+120

Answered by Jiawen Z. • Maths tutor

2360 Views

Solve for x and y using these simultaneous equations: 2x + 3y = 21, y = 2x - 1

Substitute y = 2x - 1 into the first equation.

2x + 3(2x -1) = 21

2x + 6x - 3 = 21

8x = 24

x = 3

Therefore y = 2(3) - 1 = 5

Answered by Alexandra L. • Maths tutor

2079 Views

factorise x^3 + 3x^2 - 13x - 15

For convience we can call the polynomial f(x). Using the fact that the product of the roots of a polynomial equals the constant term, we know that the product of the roots is -15. We can therefore guess a...

Answered by Ben I. • Maths tutor

4916 Views