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reciprocals of the solutions

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by sanju09 » Sat Jun 19, 2010 12:22 am
What is the sum of the reciprocals of the solutions to the equation x^2 - (3/5) x = -11/3?
(A) -11/3
(B) 9/55
(C) 3/5
(D) 94/65
(E) 5/3
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by amising6 » Sat Jun 19, 2010 12:52 am
dude are you sure about question some value is wrong
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by mj78ind » Sat Jun 19, 2010 12:57 am
On simplification we get 15x^2 - 9x + 55 = 0 (bring the -11/3 to LHS and multiply throughout by 15).

Now we know if a and b are the roots of a quadratic, a + b = -middle term constant / first term constant and ab = last term constant / first term, thus (a+b)/ab = 1/a + 1/b = -middle term/ last term

=-(-9)/55 = 9/55

B

OA please?
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by sanju09 » Sat Jun 19, 2010 12:58 am
amising6 wrote:dude are you sure about question some value is wrong
such as?
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by sanju09 » Sat Jun 19, 2010 1:05 am
mj78ind wrote:On simplification we get 15x^2 - 9x + 55 = 0 (bring the -11/3 to LHS and multiply throughout by 15).

Now we know if a and b are the roots of a quadratic, a + b = -middle term constant / first term constant and ab = last term constant / first term, thus (a+b)/ab = 1/a + 1/b = -middle term/ last term

=-(-9)/55 = 9/55

B

OA please?
A quadratic in the form x^2 - S x + P = 0 has S as the sum and P as the product of its two solutions. That is if α and β were the two solutions, then the sum of their reciprocals will be (α + β) / (α β) = S/P. [spoiler]You got it right![/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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