Which of the following equations has 1 + √2 as one of its roots?

To solve this problem, we need to use the following two facts:

1) If a quadratic equation has integer coefficients only, and if one of the roots is a + √b (where a and b are integers), then a - √b is also a root of the equation.

2) If r and s are roots of a quadratic equation, then the equation is of the form x^2 – (r +s)x + rs = 0.

Since we know that 1 - √2 is a root of the quadratic equation, we can let:

r = 1 + √2

and

s = 1 - √2

Thus, r + s = (1 + √2) + (1 - √2) = 2 and rs = (1 + √2)(1 - √2) = 1 – 2 = -1.

Therefore, the quadratic equation must be x^2 – 2x – 1 = 0.

Answer: D
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x = 1 + √2
x - 1 = √2
Squaring both sides
(x - 1)^2 = (√2)^2
X^2 + 1 - 2x = 2
x^2 – 2x – 1 = 0

Option D

Note that
(B) is (x - 1)^2 so its roots are 1 and 1.
(C) is (x + 1)^2 so its roots are -1 and -1.

Now note that all 3 of the remaining options have x^2. When you put \(x = (1 + \sqrt{2})^2\) in them, you will get \((+2\sqrt{2})\) term. It should get canceled out by another term to get 0. So you should get \((-2\sqrt{2})\) term.
Option (D) has -2x which will give a term \(-2\sqrt{2}\).

So answer (D)
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If x1 = 1 + √2 then x2 = 1 - √2.

By the Viete theorem:
b = -(x1 + x2) = -(1 + √2 + 1 - √2) = -2
c = x1 * x2 = (1 + √2)(1 - √2) = 1 - 2 = -1
is the answer.

Hi All,

While this prompt might look a bit 'scary', you can answer it without doing a lot of complex math (but you need to pay attention to what each of the 5 equations implies (and whether you can actually get a sum of 0 in the end or not).

To start, we're told that (1 + √2) is a 'root' of one of those equations, which means that when you plug that value in for X and complete the calculation, you will get 0 as a result.

We know that √2 is greater than 1, so (1 + √2) will be GREATER than 2 (it's actually a little greater than 2.4, but you don't have to know that to answer this question).

So, when you plug that value into X^2 (which appears in all 5 answers), you get a value that is GREATER than 4. To get that "greater than 4" value down to 0, we have to subtract something.... Also keep in mind that...

(squaring a value greater than 2) > (doubling that same value)

So subtracting 2X from X^2 would NOT be enough to get us down to 0... we would ALSO need to subtract the 1....

With those ideas in mind, there's only one answer that matches....

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Check the links below:

Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm

Theory on Algebra: https://gmatclub.com/forum/algebra-101576.html
Algebra - Tips and hints: https://gmatclub.com/forum/algebra-tips- ... 75003.html

DS Algebra Questions to practice: https://gmatclub.com/forum/search.php?se ... &tag_id=29
PS Algebra Questions to practice: https://gmatclub.com/forum/search.php?se ... &tag_id=50

Hope it helps.
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A and C are automatically out as they are roots of 1 and -1 respectively

for rest of option substitute 1 - root(2)...which ever equation gets satisfied is the answer.

Only equation D satisfies

Ans: d

Another method to solve this question is :
x=\((-b±\sqrt{(b^2-4ac)})/2a\) , using equation for finding roots of quadratic equation
where b is coefficient of x, a is coef of x^2 and c is constant.
Substituting values for option A gives
x=\(-1±\sqrt{2}\)
Since we need root as \(1+\sqrt{2}\), b must be negative, with other coef same as A
which is option D
Answer D.

Hi Bunuel

Could you please put your input here ?
I can not get the concept or people are talking about something I already miss
BR.

This problem involves irrational roots of a quadratic equation:

For a quadratic equation: ax^2+bx+c = 0 where, the coefficients a,b and c are real.

If we have one root as m +√n than it will also have a conjugate root m -√n where m, n are rational and n is not a perfect square.

According to the question, one root is 1+√2 than the other root will be its conjugate i.e. 1-√2.

Now, if we know the (sum of the roots) and (product of the roots) we can create a quadratic equation as:

x^2 - (sum of roots) x + product of the roots = 0

Sum of roots = (1+√2) + (1-√2) = 2
Product of roots = (1+√2) * (1-√2) = 1 - (√2)^2 = -1

Quadratic Equation = x^2 -2x+1 = 0 i.e. option D.

I just plugged the approximate value of 1+ √2 as 1+ approx 1.4= approx 2.4 and tested on answer choices.

D) satisfies as 5. something - 4.something - 1 is approx 0.

Hi VeritasKarishma

In the highlighted part, is the squared (^2) mistakenly put?

Could you explain the quoted part?
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EMPOWERgmatRichC
Thanks for the extra-ordinary explanation.
But how do someone convinced that those equations are perfectly fine ( i mean- they are not randomly written)?
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Hi Asad,

To start, NOTHING about an Official GMAT question is ever 'random'; every aspect of the question (including what appears in the 5 answer choices) was specifically chosen for a reason - and sometimes the reason is to present a pattern that you can take advantage of.

From your question, I assume that you're asking about how we know for sure that the each of those equations actually does have a solution (or multiple solutions). Since we're looking for just one thing - the equation that has (1 + √2) as a root - it doesn't really matter what the other 4 equations are (since they won't have that root, none of them will be the answer to the question that was asked).

GMAT assassins aren't born, they're made,
Rich

Yes, it should be
When you put \(x = (1 + \sqrt{2})\) in them, you will get \((+2\sqrt{2})\) term.
Basically, it means \(x^2 = ( 1 + \sqrt{2})^2 = 1 + 2 + 2*\sqrt{2}\)

Recall that
\((a + b)^2 = a^2 + b^2 + 2ab\)
\((a - b)^2 = a^2 + b^2 - 2ab\)

B) x^2 – 2x + 1 = 0
Look at the left side. It is (x - 1)^2 because when you expand it, you get x^2 - 2x + 1.
\((x - 1)^2 = (x - 1)*(x - 1) = 0\)
So x = 1, 1

C) x^2 + 2x + 1 = 0
Similarly, (x +1)^2 = x^2 + 2x + 1
So here (x + 1)(x + 1) = 0
x = -1, -1
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can someone explain this solution and why it's allowed?

They used 1 root to find the ORIGNINAL equation that has both roots????? how does that even make sense when we typically use both roots?

working backwards:
x = 1+ sqrt 2
x-1 = sqrt 2
(x-1)^2 = 2
FOIL
x^2 -2x +1 =2
x^2 -2x -1 = 0

how does this give us the actual original equation with 2 roots based on 1 root? Is this some special property, is it consistent? Can it be proved>?

Hi Lazybum,

Normally, having just one root would NOT be enough information to create the original equation. For example, if we had the root X = 3, then that does not give us much to work with and there could be lots of different possible equations. For example, we could quickly create the following two equations by making the "other" root -1 or +1, respective:

X^2 - 4X + 3 = 0
X^2 - 2X - 3 = 0

In THIS question though, the answer choices are 5 equations, so we know that ONE of them MUST have a root of (1 + root2). Since the root includes a square-root in it, we can 'work backwards' as the prompt describes and get the correct answer. However, that approach really only works here because of how the prompt/answers are designed.

GMAT assassins aren't born, they're made,
Rich

Video solution from Quant Reasoning:
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Solution:

​option (b) and option(c) are expanded versions of (x-1)^2 and (x+1)^2 respectively.

Their roots would be 1,1 and -1,-1 respectively and not 1 + √2 .You can eliminate them.

Now plug in 1 + √2 in the remaining options. If you observe closely, you do not need to solve.

To have a 0 of the RHS, you need to eliminate the + √2 that would be present from the x^2 terms.

This is possible in option (d) because it has a -2x . -2x would generate a -2√2 term when -2 is multiplied with 1 + √2.

Thus option(d).

Hope you are clear
Devmitra Sen(GMAT Quant Expert)
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