eek i forgot to type anything in the thread - lol here goes. 1. Simplify - (3xy³)² 2. Solve equation - 2x²-5x-4=0 3. P is inversely proportional to the square root of Q If P=12 when Q=49 find an expression for P in the terms of Q. if you can help please do!
5 / 2 = 2 5 % 2 = 1 5 / 2 * 2 = 4 5 % 2 = 1 ((5 / 2) * 2) + (5 % 2) = 5 correct me if I missed something.
1) 9x² * y^6 3) P= A Q^0.5 12 = A 7 P= 84Q Can you check question two as i get roots which arent at gcse level.
2) dotn think it factorises cleanly at first glance, use the quadratic equation (x=-b+- squareroot(b^2-4ac) /2a) think thats how it goes 3) P=-k * squareroot(Q) sub P&Q to find k
1) 9 * x^2 * y^6 (y^n)^a = y^n*a 2) -(-5) ± √ 5^2 - 4x2x-4 ................2 x 2 5 ± √25 -(-32) ..........4 5 + √57 5 - √57 .......4....................................4 5 + 7.549834435 5 - 7.549834435 ,,...........4...................................4 12.7549834435 -2.7.549834435 ..........4..................................4 x = -0.6374586088 or 3.137458609 Round as required. 3) P = 84 * (1 / √q) Hope that helps
ahh ah, i see i made a small error in my other post it should be... P=k * (1/root(Q)) so sub for P&Q to find k
2) Don't know if you mean -2x²-5x-4=0 or 2x²-5x-4=0 anyway using the quadradic equation (-b ± sqrt(b²-4ac))/2a: a = (-)2 b = -5 c = -4 plug'em in et voila (-2x²-5x-4=0): x1 = 11/2 x2 = 8 or if it should be 2x²-5x-4=0 x1 = 21/2 x2 = 8 I learnt that formula in 10th grade igcse as well...
-b ± squareroot (b²-4ac) 2a The method is actually given to you in the gcse paper, with how to find the area of sphere ect, therefore u dont need to memorise that method, just how to use it to solve quadratics.
