Which of the following equation has 3 + 2*3^(1/2) as one of its roots

One of the ways is to approximate the value of \(3+2\sqrt{3}\) =3+2*1.7=6.4
Substitute 6.4 for x

Only option D. \(x^2-6x-3 = 0\) will work with the value. Answer D.

(X-(3+2sqrt(3))(x-(3-2sqrt(3))
=x^2-(3-2sqrt(3))x-(3+2sqrt(3))x-3
=x^2-6x-3
Answer D

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Alternate Solution

Given:

• One of the roots of a quadratic equation is \(3 + 2\sqrt{3}\)

To find:

• The original equation, with the given root \(3 + 2\sqrt{3}\)

Approach and Working:

• The value of \((3 + 2\sqrt{3})^2 = 9 + 12 + 12\sqrt{3} = 21 + 12\sqrt{3}\)

Replacing the value of \(x^2\) and \(x\) in the given options, we get:

• \(21 + 12\sqrt{3} + 18 + 12\sqrt{3} + 3\)

o Not equal to 0

• \(21 + 12\sqrt{3} – 18 – 12\sqrt{3} + 3\)

o Not equal to 0

• \(21 + 12\sqrt{3} + 18 + 12\sqrt{3} – 3\)

o Not equal to 0

• \(21 + 12\sqrt{3} – 18 – 12\sqrt{3} – 3\)

o Equal to 0

• \(21 + 12\sqrt{3} – 24 – 12\sqrt{3} – 3\)

o Not equal to 0

Hence, the correct answer is option D.

Answer: D

We are given that one of the root is 3+2√3, so another root is 3-2√3.
(We know that if one of the roots of a quadratic equation is a+√b, then as a conjugate root, the other one will be a–√b)
All the quadratic equation can be expressed as aX^2+bX+C=y
Let the two different roots be X1 and X2
X1=3+2√3
X2=3-2√3
X1 * X2 =-3
X1+X2 = 6

We also know that:
X1*X2 = c/a
X1+X2 = -b/a

A. x^2+6x+3=0
X1*X2 = 3
X1+X2 = -6
Not the answer we are looking for.
B. x^2−6x+3=0
X1*X2 = 3
X1+X2 = 6
Not the answer we are looking for.
C. x^2+6x−3=0
X1*X2 = -3
X1+X2 = -6
Not the answer we are looking for.
D. x^2−6x−3=0
X1*X2 = -3
X1+X2 = 6
Bingo, this is the answer choice we are looking for.
E. x^2−43√x−3=0

Could you please explain why the other root must necessarily. be \(3 – 2\sqrt{3}\) ?


I could theoretically have any other root and draw a parabola/quadratic that crosses both points on the x axis...

e.g.
y = (x - (\(3 + 2\sqrt{3}\))) * (x-(\(7 + \sqrt{3}\)))

...

In other words, I could tell you that one of the two roots in a quadratic is {insert any number}
How does that fact alone tell you anything about what the value of the second root is?
Wouldn't that entail some assumption about the shape of the parabola and where the point of reflection is?

Hey levloans,
We know that both the roots of a quadratic equation can be found out by applying \(\frac{-b +- \sqrt{b^2 -4ac}}{2a}\).

So, if one root is of the form \(\frac{-b + \sqrt{b^2 -4ac}}{2a}\) then other root will be of the form \(\frac{-b - \sqrt{b^2 -4ac}}{2a}\).

Notice the difference in sign.
Therefore, if one root is \(3- 2\sqrt{3}\) then another root will be \(3+ 2\sqrt{3}\) .
I hope this answers your query.
Regards,
Ashutosh

Thanks for your quick response, Ashutosh. I had considered this pattern, but still do not see why it is necessary that our roots are these.

Let me rephrase:

I understand that \(3 + 2\sqrt{3}\) & \(3- 2\sqrt{3}\) COULD be our roots.
But what says that they HAVE to?

i.e.

For example
Couldn't I have a quadratic with the roots (1) \(3+ 2\sqrt{3}\) and (2) 11 ?

This is a real quadratic with x-intercepts at \(3+ 2\sqrt{3}\) and 11...

in fact, just look at this graph for visual reference
https://www.wolframalpha.com/input/?i=y%3D(x-(3%2B2*sqrt(3)))*(x-11)


Just because I know the first root (1), it doesn't seem to me that I can make assumptions about what the second root (2) is, as that could be anything.
Or, I can know one of the x intercepts, but that doesn't tell me anything about what the other is, unless I know something else about the form of the quadratic.
... I can "peg" one of the x-intercepts (roots) and then move the other one to wherever I want and still have a valid parabola/quadratic.

I think your answer is in some way is affirming the consequent; or pulling in info from the output to make a judgment on the input, but the output is not given information, we can only start with the input of the question.

Please let me know me if I am missing something.

I think I have an easier solution than the ones posted here.

x = \(3+2\sqrt{3}\)

This can be rewritten as \(x - 3 = 2\sqrt{3}\)

Squaring both the sides, we get \(x^2+9-6x = 12\)

Subtracting both sides by 12, we have \(x^2-6x-3\), and hence, (D) is the correct answer choice

PLEASE HIT KUDOS IF YOU LIKE MY SOLUTION

I also did the above mentioned approach, however to reduce time on calculation, I did a bit of prethinking.

While eliminating answer choices, it is important to cancel out the roots. Hence, while solving I require one positive root and the other negative root.

This way, you'll be able to shortlist B & D. On plugging in value in D, you'll get the answer.

Hence, via pre thinking, you didn't even have to solve A & C.

Thanks

Which of the following equation has \(3+2\sqrt{3}\) as one of its roots?
If \(3+2\sqrt{3}\) is one root, other root is \(3-2\sqrt{3}\)
Sum of roots = 6
Product of roots = \((3+2\sqrt{3})(3-2\sqrt{3}) = 9 -12 =-3\)

Equation \(x^2 - 6x -3 =0\)

IMO D

If the roots of x^2 + ax + b = 0 are x1 and x2, then:
x1 + x2 = - b/a
x1*x2 = c/a

If x1 = 3+2√3, then x2 must be the conjugate of x1, that is 3-2√3

x1 = 3+2√3
x2 = 3-2√3

x1+x2= 6= -b/a. a=1, then b=-6.

x1*x2= 3^2 - (2√3)^2 = 9-12= -3. a=1 then c=-3.

Answer is obviously (D)
x^2 + ax + b = x^2 - 6x - 3

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From a while ago, but if anyone sees this question and has the same query:


https://study.com/academy/lesson/conjug ... ample.html


Edit:

Here is a better post about the conjugate root theorem (a bit beyond the GMAT, but may be helpful if this exact scenario were to pop up)

https://www.quora.com/If-one-root-of-qu ... other-root


The key, it appears, comes down to whether the coefficients are real number coefficients (which is always true on the GMAT - no imaginary numbers).

I think one of the reasons he chose not to answer is because the topic is a deep dive down a rabbit hole that is no where near the GMAT concepts.



Posted from my mobile device

Asked: Which of the following equation has \(3+2\sqrt{3}\) as one of its roots?

Other root will be \(3-2\sqrt{3}\)

Sum of roots = \(3+2\sqrt{3} + 3-2\sqrt{3} =6 \)
Product of roots = \((3+2\sqrt{3})*(3-2\sqrt{3}) = 3^2 - 2^2*3 = 9 - 12 = -3 \)

Equation : -
\(x^2 - 6x - 3 = 0\)

IMO D

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