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

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The discussion focuses on several AS level math problems, including factorization, finding points of intersection, and solving inequalities. For the first problem, the correct approach to factorizing 6x² - 5x - 6 involves using the completed square form and factoring out 6. The second problem requires substituting one equation into another to find the intersection points, with a hint provided for rearranging the first equation. The third problem discusses conditions for a line to intersect a curve at exactly one point, emphasizing the need for a single solution in the quadratic formula. Lastly, the fifth problem addresses solving an inequality by analyzing the signs of the factors involved. Overall, the thread highlights common challenges faced by students studying math independently.
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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
 
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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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