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

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.

"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

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

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?

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.

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:

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.

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.[/quote]

Very clear. Thank you so much Bunuel!

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