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Math Expert V
Joined: 02 Sep 2009
Posts: 58434

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Difficulty:   55% (hard)

Question Stats: 63% (01:28) correct 37% (01:27) wrong based on 169 sessions

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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. $$-64$$
B. $$-16$$
C. $$-15$$
D. $$-\frac{1}{16}$$
E. $$-\frac{1}{64}$$

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Math Expert V
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Official Solution:

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. $$-64$$
B. $$-16$$
C. $$-15$$
D. $$-\frac{1}{16}$$
E. $$-\frac{1}{64}$$

Since one of the roots of the equation $$x^2 + ax + 15 = 0$$ is 3, then substituting we'll get: $$3^3+3a+15=0$$. Solving for $$a$$ gives $$a=-8$$.

Substitute $$a=-8$$ in the first equation: $$x^2-8x-b=0$$.

Now, we know that it has equal roots thus its discriminant must equal to zero: $$d=(-8)^2+4b=0$$. Solving for $$b$$ gives $$b=-16$$.

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I thought the Discriminant was (b^2)-4ac.

This would yeild ((-8)^2)-4(1)(b) --> 64-4x.

So: 64-4x=0 -->64=4x -->x=16.

Can someone explain why this isn't the case? In Bunuel's solution we have addition, instead of subtraction in the discriminant.

Thanks
Math Expert V
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JackSparr0w wrote:
I thought the Discriminant was (b^2)-4ac.

This would yeild ((-8)^2)-4(1)(b) --> 64-4x.

So: 64-4x=0 -->64=4x -->x=16.

Can someone explain why this isn't the case? In Bunuel's solution we have addition, instead of subtraction in the discriminant.

Thanks

Yes, the discriminant of $$ax^2 + bx + c = 0$$ is $$b^2 - 4ac$$.

Thus the discriminant of $$x^2 - 8x - b = 0$$ is : $$(-8)^2 - 4(-b) = (-8)^2 + 4b$$.

Hope it's clear.
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Got it, thanks!
Intern  Joined: 21 May 2013
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Isn't it just more simple to acknowledge that x^2-8x-b=0 and then, knowing the roots are same, use reverse foil method find: 4+4=8, 4*4=16. ??

Knowing that Bunuel always uses the simplest method, please explain the importance of knowing the discriminant? I thought with reverse foil the discriminant was not important to remember.

Thanks !
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This question was duplicated from Q D01-41. I got this question twice from a quiz I was doing.
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Alternative method:

We know one of the roots of x^2 +ax + 15 = 0, so let x=3
a=-8 . . . . . by substituting x=3 in x^2 +ax + 15 = 0

Standard form of quadratic equation is ax^2+bx+c=0, let its root be p and q. so, (x-p)(x-q)=0
Solving (x-p)(x-q)=0, we get x^2 -(p+q)+pq=0

By comparing x^2 -(p+q)+pq=0 with x^2 +ax - b = 0 that is x^2 -8x - b=0 . . . . .as a=-8
we can deduce that p+q=8 and pq=-b
It is given that the roots are equal so both the roots p and q must be 4 as p+q=8
and pq=16

Substituting above values in pq=-b we get that -b=16 so b=-16

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We know one of the roots of x^2 +ax + 15 = 0 is 3.
Substitute x=3 in the equation.
We get x= -8 as the other root.
We know that sum of roots in equation ax^2 +bx + c = 0, is -b/a and product of the roots is c/a.
So, in equation x^2 +ax -b = 0, a becomes -8 => equation becomes, x^2 -8x - b = 0
(Let root 1= root 2 = r)
r + r = -8 => r = -4
r*r = -b => b = -16

Ans B

I was comfortable using the sum and product of roots concept and could solve faster that way. Hence, I thought I might add this as a solution.

Cheers! Re: M28-08   [#permalink] 18 Oct 2018, 23:23
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