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2002 Winter diet paper 2 Q.10 - HSN forum # 2002 Winter diet paper 2 Q.10

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### #1gary

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Posted 19 May 2005 - 02:11 PM

I know you use the discriminant for this but I am completely lost with the caculation please help.

### #2aldo

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Posted 19 May 2005 - 02:38 PM

You are given the equation of the circle & the line.

Re-arrange the equation of the line so you have y= k - 2x.

Substitute this value of y into the equation for the circle.

A line is a tangent to a circle where b - 4ac = 0, so you want to arrange the above equation ang get the following:

5x - (2+4k)x + (k -4) = 0

Now using the discriminant, b - 4ac = 0, and factorising you should get that k = 7 OR k = -3 and since you are asked to find the value of k where k > 0 you can conclude that the value for k is k = 7.

That make sense???

### #3oddbins

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Posted 19 May 2005 - 02:39 PM

I'm stuck on this question to. What is the discriminant?

### #4dfx

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Posted 19 May 2005 - 02:46 PM

5x sq - (2+4k)x + (k sq -4) = 0

At this piont, suppose your equation was:

5x sq - (1 + 2k) 2x ....

Wouldn't it make more sense since you've taken out the common factor of 2x, not just x ? Or does the coefficient of x outside the bracket ALWAYS have to be 1?

### #5aldo

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Posted 19 May 2005 - 02:51 PM

A quadratic is an expression of the form:

ax + bx + c

If you try to calculate the discriminant of a quadratic you must have in the above form.

a is the coefficient of x b is the coefficient of x

c is a constant term

Therefore, your discriminant value would be incorrect as it is the coefficient of the x term on its own.

That make sense??

### #6dfx

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Posted 19 May 2005 - 02:53 PM

Yeah, that makes perfect sense. Thanks! #### 1 user(s) are reading this topic

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