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D01-41

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Intern
Joined: 06 Apr 2018
Posts: 43

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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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Joined: 06 Sep 2018
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Location: Pakistan
Concentration: Finance, Operations
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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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Joined: 05 Feb 2018
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03 Jul 2019, 12:19
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
Re: D01-41   [#permalink] 03 Jul 2019, 12:19

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