In the equation x^2 + bx + 12 = 0, x is a variable and b is a constant

Question

In the equation x^2 + bx + 12 = 0, x is a variable and b is a constant. What is the value of b ?

(1) x - 3 is a factor of x^2 + bx + 12.
(2) 4 is a root of the equation x^2 + bx + 12 = 0.

ydmuley · Accepted Answer

In the equation x^2 + bx + 12 = 0, x is a variable and b is a constant. What is the value of b ?

(1)\(x - 3\) is a factor of \(x^2 + bx + 12\)

as \((x-3)\) is one of the factor, this means that \(x = 3\) is one of the roots of the equation \(x^2 + bx + 12\)

In order to get 12 = we only have one value after 3 and that is 4 as \(4 * 3 = 12\)

Hence, \(bx = 4x + 3x = 7x\), value of \(b = 7\)

Hence, (1) ===== is SUFFICIENT

(2) 4 is a root of the equation \(x^2 + bx + 12 = 0\)

As 4 is one of the roots, the other root will be 3

Hence, \(bx = 4x + 3x = 7x\), value of \(b = 7\)

Hence, (2) ===== is SUFFICIENT

Hence, Answer is D

Did you like the answer? 1 Kudos Please :good

Bunuel · Answer

SOLUTION

In the equation x^2 + bx + 12 = 0, x is a variable and b is a constant. What is the value of b ?

(1) x-3 is a factor of x^2 + bx + 12:

The above implies that x = 3 is a root of x^2 + bx + 12 = 0 (if x - 3 is a factor of x^2 + bx + 12 then x^2 + bx + 12 = (x - 3)(x - k) = 0, for some k, so x = 3 is on of the roots of the equation). Substituting x = 3 into the equation yields 3^2 + 3b + 12 = 0, which gives b = -7. Sufficient.

(2) 4 is a root of x^2 + bx + 12 = 0.

The same here, substitute x = 4 to get 4^2 + 4b + 12 = 0, which gives b = -7. Sufficient.

Answer: D.

energy2pe · Answer

I'm a bit confused as to the term 'root' and specifically the phrase '4 is a root of x^2' here. How should this be interpreted?

I thought of it as the factor of 12 (i.e. 4x = 12).

Thanks ahead of time.

BrentGMATPrepNow · Answer

"root" is the same as "solution"

We can say "x = 3 is a  solution  to the equation 2x = 6" or we can say "3 is a  root  of the equation 2x = 6"

Cheers,
Brent

Pritishd · Answer

Sharing an alternate way to solve using the roots of the equation.

Let p and q be the roots of the equation, then the sum and product of the roots will be:

p + q   =   \frac{-b}{a}   =   -b  (---->  1 )

p * q   =   \frac{c}{a}   =   12  (---->  2 )

Statement-1:  If  (x-3)  is a factor then one of the roots of the equation will be  3 . Let  p   =   3  and substitute in (---->  2 )
 3 * q   =   12  therefore  q   =   4 
Substituting   p = 3  and  q = 4  in (---->  1 ) we get  b = -7   ( Sufficient )

Statement-2:  We are directly given that  4  is one of the roots. Let's assume   q = 4   and substitute in (---->  2 )
 p * 4   =   12  therefore  p = 3 
Substituting  p = 3  and  q = 4   in (---->  1 ) we get  b = -7   ( Sufficient )

Ans.  D

HoudaSR · Answer

Bunuel   TargetTestPrep   chetan2u  Can you please tell me why when I set x= 3 or x=4 and I work backwards I don't land at the equation x^2-7x+12=0
When I set for example x=3
x-3=0
(x-3)^2=0
I land at x^2-6x+9=0 and not x^2-7x+12=0

On the official guide for a similar question where the root of the quadratic was 1+sqrt2, the solution method was to work backwards to obtain the equation x^2-2x-1=0 (question 197 in the OG 2022), however this doesn't work here. Can you please explain why?

Bunuel · Answer

Want to elaborate your question? Not clear what you are doing there? Where does the red parts come from?

We got that b = -7, hence the given equation is x^2 - 7x + 12 = 0 and its solutions are x = 3 and x = 4.

HoudaSR · Answer

On a similar question to this one in the official guide, they worked backwards from the root of the equation to construct the quadratic equation.
the root given was 1+sqrt2

They started by setting x=1+sqrt2
subtracted 1 from both sides x-1=sqrt2
Squared both sides (x-1)^2=2
Expanded the left side x^2-2x+1=2
Subtracted 2 from both sides to obtain the final quadratic equation: x^2-2x-1=0

I have tried using the same method in this question to construct the quadratic equation by setting x=3 
Subtracted 3 from both sides x-3=0
Squared both sides (x-3)^2=0
And I obtained the following equation x^2-6x+9=0, however that is different from x^2+bx+12=0

I wanted to understand why this backwards solving method worked in that example on the OG and not in this one.

See picture below:

Bunuel · Answer

In  this question , you are given a specific root (1+√2) and asked to find which quadratic equation from the five options has that root. The official solution performs some manipulations to show that x^2 – 2x – 1 = 0 has that root.

It's important to note two things. First, you can use this manipulation to get a quadratic for any number. For example:

x = 7;
x - 1 = 6;
(x - 1)^2 = 6^2; 
x^2 - 2x + 1 = 36;
x^2 - 2x - 35 = 0.

Second, it does not mean that this equation is the only quadratic that has that root. There are infinitely many quadratics that have (1+√2) as a root. For example, one of the roots of x^2 - 2√2x + 1 = 0 is also (1+√2). Generally, for any k in (x-(1+√2))(x-k)=0, you get a quadratic that has (1+√2) and k as roots. If the options in that question were different and x^2 - 2√2x + 1 = 0 were an option instead of x^2 – 2x – 1 = 0, the manipulations in the solution would not work. Thus, that solution is specific to that option and cannot be generalized. It's worth mentioning that the solutions from the official guides may not always be the most reliable, as their quality is not always on par with the quality of the questions themselves.

Finally, using that approach, you can get a quadratic that has 3 as the root (x^2-6x+9=0), as you did. However, that won't be the only quadratic with that root, and it won't necessarily match x^2 + bx + 12 = 0.

I hope this clarifies your doubt.

HoudaSR · Answer

In  this question , you are given a specific root (1+√2) and asked to find which quadratic equation from the five options has that root. The official solution performs some manipulations to show that x^2 – 2x – 1 = 0 has that root.

It's important to note two things. First, you can use this manipulation to get a quadratic for any number. For example:

x = 7;
x - 1 = 6;
(x - 1)^2 = 6^2; 
x^2 - 2x + 1 = 36;
x^2 - 2x - 35 = 0.

Second, it does not mean that this equation is the only quadratic that has that root. There are infinitely many quadratics that have (1+√2) as a root. For example, one of the roots of x^2 - 2√2x + 1 = 0 is also (1+√2). Generally, for any k in (x-(1+√2))(x-k)=0, you get a quadratic that has (1+√2) and k as roots. If the options in that question were different and x^2 - 2√2x + 1 = 0 were an option instead of x^2 – 2x – 1 = 0, the manipulations in the solution would not work. Thus, that solution is specific to that option and cannot be generalized. It's worth mentioning that the solutions from the official guides may not always be the most reliable, as their quality is not always on par with the quality of the questions themselves.

Finally, using that approach, you can get a quadratic that has 3 as the root (x^2-6x+9=0), as you did. However, that won't be the only quadratic with that root, and it won't necessarily match x^2 + bx + 12 = 0.

I hope this clarifies your doubt.

Very clear. Thank you so much  Bunuel !

demawtf · Answer

In the equation x^2 + bx + 12 = 0, x is a variable and b is a constant. What is the value of b ?

(1) x - 3 is a factor of x^2 + bx + 12

so it means x-3 is a component of the quadratic equation --> we have "+bx" and "+12" this means that the other way to see the quadratic would be (x-b1)(x-b2), one of the terms is x-3 so (x-3)(x-b2) --> which is the term that -3*b2=12? --> -4 so the sum is (-4) + (-3) = -7 --> SUFFICIENT

(2) 4 is a root of the equation x^2 + bx + 12 = 0.
same here, 4 is directly the value of one of the components of "bx" --> SUFFICIENT

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

In the equation x^2 + bx + 12 = 0, x is a variable and b is a constant

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

In the equation x^2 + bx + 12 = 0, x is a variable and b is a constant

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards