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D01-41

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Re: D01-41  [#permalink]

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New post 06 Oct 2018, 09:38
I solved it to the point, a=-8.

Then, used the approach sum of roots = -b/a (if the equation is of the form ax2 +bx+c)

So the sum is 8/1= 8 =2x (since the roots are same). This is how I got the value of x=4

Then after that using the method of substitution I could get the value of b.

x(square)+ ax-b=0
(4*4)+4(-8)-b=0
16-32-b=0
-16-b=0
Hence b=-16

Please let me know if my approach is correct.
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Re: D01-41  [#permalink]

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New post 14 Oct 2018, 21:54
Bunuel wrote:
The equation \(x^2 + ax - b = 0\) has equal roots, and one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3. What is the value of b?

A. \(-\frac{1}{64}\)
B. \(-\frac{1}{16}\)
C. \(-15\)
D. \(-16\)
E. \(-64\)

as \(x=3\) for \(x^2 + ax + 15 = 0\)
So \(9+3a+15=0\)
\(a=-8\)
put value of a in \(x^2 + ax - b = 0\)
So \(x^2 - 8x - b = 0\)
As roots are equal so sum of roots is \(x+x=\frac{-(-8)}{1}\)
\(2x=8\)
\(x=4\)
now put \(x=4\) in \(x^2 - 8x - b = 0\)
So \(16-8(4)-b=0\)
\(16-32-b=0\)
\(b=-16\)
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Re: D01-41  [#permalink]

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New post 03 Jul 2019, 12:19
VeritasKarishma wrote:
Yes, as shown by chetan2u above and in fact you can continue to use the same approach in the last steps also:

\(x^2 + ax + 15 = 0\)
If one root is 3, the other is 5 (to get the product 15). So sum of the roots is 8.
This means a = -8

The other equation becomes \(x^2 - 8x - b = 0\)
The sum of the roots is 8 and the roots are equal to they will be 4 and 4. Their product becomes 16. So b becomes -16.


This is the way I solved it as well but I was slightly confused, thinking it was +16 and not seeing that answer (only -16).
So after we get a = -8 and we put that into the 2nd equation, it states that "the two roots are equal"
Roots are what comes out when x=0, aka x-intercept/solution/zeroes
To get -8 in the middle term, we would need to factor it as (x-4)(x-4), so x=4 --> 4*4 = 16
This is the 3rd term, but since the equation has - b in it, we would need b = -16 in order to get 16 in the final equation: x^2 - 8x + 16 = 0
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Re: D01-41   [#permalink] 03 Jul 2019, 12:19

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