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quadratics - value question

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quadratics - value question

by cgc » Thu Jan 21, 2010 10:22 pm
Having trouble solving this questions quickly.... Any suggestions?

If one root of the equation 2x(power2)+3x-k=0 is 6, what is the value of k?

A. 90
B. 42
C. 18
D. 10
E. -10

I went ahead and solved it this way...
1st - I created the quadratic formula below by dividing the equation by 2 so the x(power2) remains....(only way I know how)
x(power2)+3/2x-1/2k=0

2nd - I solved for the second value of x given that one value of x as 6 or 12/2 since the equation was just divided by 2. Therefore, -12/2 + 15/2 = 3/2
(x-12/2) (x+15/2)

3rd - I am am left to solve for k.
(15/2)(-12/2) = -90/2 = -45

4th - Therefore, 1/2k=-45...k=-90 or answer choice a.

Does anyone have a faster solution to this lenghty question?

Thanks,
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by ace_gre » Thu Jan 21, 2010 10:32 pm
Hi , I would directly substitute x= 6,

2(6)^2 + 3*6 - k =0
36 *2 + 18 = k
72 + 18 = 90 :)
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by sreak1089 » Thu Jan 21, 2010 10:44 pm
There are couple of ways to do this:

Method 1:

2*x^2 + 3x - k = 0

Lets assume roots of the above eqn are a & b and we know one of the roots. So, lets assume a = -6

Above eqn can be re-written as:
x^2 +(3/2)x - (k/2) = 0

We know, Sum of the roots = a+b = (3/2) = -6 + b = (3/2) => b = 6+3/2 = 15/2
We know that product of roots = a*b = (-k/2) = (-6*15/2) = -k/2 => k = (-6*15/2 * 2) = 90

Method 2:

We know that one of the roots = 6, hence substitute x in the original eqn.

2*(6^2) + 3*6) - k = 0
k = 72 + 18 = 90.

Cheers.
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by cgc » Fri Jan 22, 2010 10:21 am
the second method is defenately faster.

thanks,
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by sreak1089 » Fri Jan 22, 2010 7:36 pm
Obviously :)
cgc wrote:the second method is defenately faster.

thanks,
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