Bunch of AS level math problems(quadratics,indices etc)

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SUMMARY

This discussion focuses on solving various AS level math problems, specifically involving quadratics and inequalities. Key problems include factorizing the quadratic equation 6x² - 5x - 6 using the completed square method, finding the intersection of curves, and solving inequalities. The correct factorization for the quadratic is (3x + 2)(2x - 3), while the intersection points of the curves are (1, 1) and (-4, 3.5). Participants provide hints and methods for tackling these problems, emphasizing the importance of substitution and understanding the behavior of quadratic equations.

PREREQUISITES
  • Understanding of quadratic equations and their properties
  • Knowledge of the completed square method for factorization
  • Ability to solve systems of equations through substitution
  • Familiarity with inequalities and sign analysis
NEXT STEPS
  • Learn the completed square method for quadratic factorization
  • Study how to find intersections of curves using substitution
  • Explore the quadratic formula and its applications in determining unique solutions
  • Review sign charts and their use in solving inequalities
USEFUL FOR

Students preparing for AS level math exams, particularly those struggling with quadratics, inequalities, and curve intersections. This discussion is beneficial for anyone seeking to improve their problem-solving skills in algebra.

giddy
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Ok, so I have very little time to nail my AS level math papers and I got stuck at a couple places with a bunch of problems from various chapters. I've collected all my snags and I'm posting them together.
Problem 1
Use the completed square form to factorize 6x2 - 5x - 6

The attempt at a solution
(x + 5/2)2 - 25/4 - 6
{(x + 5/2)2 - (5/2)2} - 6
(by a2 - b2 = (a-b)(a +b))
[((x + 5/2) - (5/2))((x + 5/2) + (5/2)) ]- 6
[x(x + 10/2)] - 6 >>>>> Which is totally weird because the answer is supposed to be:
(3x +2)(2x-3)

Problem 2
Find the point of intersection of the foll. curves:
x + 2y = 3; x2 + xy = 2
I know I need to substitute one equation in the other. I managed all the other problems in the section except this one.
The expected answer is : (1,1) (-4,3.5)

Problem 3
Find values of k such that the straight line y = 2x + k meets the curve with equation x2 + 2xy + 2y2 = 5 exactly once.
I don't really know what to do, I tried substituting like the previous problem and came up with a complicated mess of an equation with k.

Problem 4
Show that the equation 2x+1 + 2x-1=160 can be written in the form 2.5 X 2x = 160. Find the value of x which satisfies the equation.
I'm a little embarrassed that i do now know how to really tackle this one!?:redface:

Problem 5
Solve the inequality : (x-1)(x-6)<0
There something in my book about a table of signs of the factors but I can't make any sense of it!

I'm studying math on my own and I'm a little absent minded sometimes and I miss the obvious when i can even understand complex things, so I'm sorry if any of the questions were silly.

Thanks
Gideon
 
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1) First step to make the completed square form is this 6(x2-5x/6)-6.

2) You'd need to show us your working to say if you did something wrong

3) If you substitute for y, you'll get a quadratic in x. If


x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}


because it meets the curve only once, means there is only one solution for 'x'. Looking at the formula above. How do you get it so that x is one value and not in the form x=(1±2)/2 or something like that?

4) This will help you am*an=am+n. Work backwards with this one.

5) What I'd do is see that the curve crosses the x-axis at 1 and 6. Draw the axis and see for what range of x, the curve lies for less than zero...or below the x-axis as it means.
 
giddy said:
Ok, so I have very little time to nail my AS level math papers and I got stuck at a couple places with a bunch of problems from various chapters. I've collected all my snags and I'm posting them together.
Problem 1
Use the completed square form to factorize 6x2 - 5x - 6

The attempt at a solution
(x + 5/2)2 - 25/4 - 6

This is not correct.

Start by factoring 6.

6(x2-(5/6)x-1)=
=6[(x-5/12)2-25/144-1]=
=6[(x-5/12)2-169/144]

Now use the fact that 169/144 = (c/d)2

Next use a2-b2=(a-b)(a+b)

Problem 2

Hint: for the first equation
2y=x-3
y=(x-3)/2

Now substitute for y in the second equation and solve for x :smile:

Problem 5

Ok, if you have a*b<0 what "sign" should the values of a, b have?

Hint: a<0 & b>0 or a>0 & b<0
 
Last edited:

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