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555-605 Level|   Algebra|                        
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AndrewN
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Which of the following equations has 1 + √2 as one of its roots? Updated on: 12 Mar 2022, 16:58
Originally posted by AndrewN on 24 Oct 2021, 15:22.
Last edited by AndrewN on 12 Mar 2022, 16:58, edited 1 time in total.
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Bunuel
Which of the following equations has 1 + √2 as one of its roots?

A) x^2 + 2x – 1 = 0
B) x^2 – 2x + 1 = 0
C) x^2 + 2x + 1 = 0
D) x^2 – 2x – 1 = 0
E) x^2 – x – 1= 0
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Hello, everyone. I solved this one is a slightly different manner by completing the square. Remember, to complete the square, manipulate the standard form of a quadratic equation (when it is set to 0) to isolate the "C" term.

\(Ax^2+Bx+C=0\)

\(Ax^2+Bx=-C\)

To complete the square, halve the "B" coefficient and square the result, then add this value to both sides. Take (A), for example. Just remember to first push the "C" term to the right-hand side. Also, the point of completing the square is to be able to put the derived quadratic back into binomial form and then take the square root. (See step #4 below.)

\(x^2+2x=1\)

\(x^2+2x+(\frac{2}{2})^2=1+(\frac{2}{2})^2\)

\(x^2+2x+1=2\)

\((x+1)^2=2\)

\(\sqrt{(x+1)^2}=+/-\sqrt{2}\)

\(x+1=+/-\sqrt{2}\)

\(x=-1+/-\sqrt{2}\)

Of course, this does NOT match our answer, but we are on the right track.

There is no need to complete the square with (B) or (C) (although you could), since those are both perfect squares and will produce roots of 1 and -1, respectively.

\(x^2-2x+1=(x-1)^2\) (root will be 1)

\(x^2+2x+1=(x+1)^2\) (root will be -1)

Picking up with (D), we can complete the square again.

\(x^2-2x=1\)

\(x^2-2x+(\frac{-2}{2})^2=1+(\frac{-2}{2})^2\)

\(x^2-2x+1=2\)

\((x-1)^2=2\)

\(\sqrt{(x-1)^2}=+/-\sqrt{2}\)

\(x-1=+/-\sqrt{2}\)

\(x=1+/-\sqrt{2}\)

This matches our root, 1 + √2, so the answer must be (D). Sometimes completing the square can be faster than dragging out the quadratic formula, particularly when the "B" coefficient is even. (Note that if you were to complete the square with (E), you would derive a fraction, and who wants to work with those when it is not necessary?)

I hope the above method may prove useful to others. Good luck with your studies.

- Andrew
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Which of the following equations has 1 + √2 as one of its roots? 12 Mar 2022, 16:43
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avigutman
Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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avigutman or any expert:
For some reason, the OG solution seems more intuitive to me. However, I was curious as to what would happen if instead I shifted the root 2 over, setting the equation equal to 1... rather than how the OG shifts the 1 over and sets the equation equal to root 2. When I do x-root2=1 and then square both sides, I get to a final equation of x^2-2xroot2+2. Why did I get a different equation doing this?
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Which of the following equations has 1 + √2 as one of its roots? 12 Mar 2022, 17:13
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woohoo921
I was curious as to what would happen if instead I shifted the root 2 over, setting the equation equal to 1... rather than how the OG shifts the 1 over and sets the equation equal to root 2. When I do x-root2=1 and then square both sides, I get to a final equation of x^2-2xroot2+2. Why did I get a different equation doing this?
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I don't have the OG solution in front of me, but I suspect that your use of the word "shift", woohoo921, might be causing you to make a careless error (I've seen it happen a lot). Instead, I suggest that you articulate to yourself which arithmetic operation you're performing on both sides of the equation, e.g. "I'm adding 1 to both sides of the equation, and subtracting root(2) from both sides of the equation."
If that doesn't help, please reply with the original equation, which arithmetic operations you performed on it, and the final equation.
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Which of the following equations has 1 + √2 as one of its roots? 12 Mar 2022, 17:31
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avigutman
woohoo921
I was curious as to what would happen if instead I shifted the root 2 over, setting the equation equal to 1... rather than how the OG shifts the 1 over and sets the equation equal to root 2. When I do x-root2=1 and then square both sides, I get to a final equation of x^2-2xroot2+2. Why did I get a different equation doing this?

I don't have the OG solution in front of me, but I suspect that your use of the word "shift", woohoo921, might be causing you to make a careless error (I've seen it happen a lot). Instead, I suggest that you articulate to yourself which arithmetic operation you're performing on both sides of the equation, e.g. "I'm adding 1 to both sides of the equation, and subtracting root(2) from both sides of the equation."
If that doesn't help, please reply with the original equation, which arithmetic operations you performed on it, and the final equation.
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avigutman
Thank you for your response!
The Official Guide says you can solve this problem by working backwards to construct a quadratic equation with 1 +root2 as a root that does not involve radicals. The OG then proceeds to set x=1+root2 and subtracts 1 from both sides, squares both sides, and then subtracts 2 from both sides to arrive at x^2-2x+1. My question was if I subtract root2 from the right-hand side of the equation instead of root 1 --> x-root2=1 and then solve through, I get x^2-2xroot2+2. Where did I go wrong? Should I have arrived at the same exact equation as the answer choice?
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Which of the following equations has 1 + √2 as one of its roots? 12 Mar 2022, 17:37
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woohoo921

The Official Guide says you can solve this problem by working backwards to construct a quadratic equation with 1 +root2 as a root that does not involve radicals. The OG then proceeds to set x=1+root2 and subtracts 1 from both sides, squares both sides, and then subtracts 2 from both sides to arrive at x^2-2x+1. My question was if I subtract root2 from the right-hand side of the equation instead of root 1 --> x-root2=1 and then solve through, I get x^2-2xroot2+2. Where did I go wrong? Should I have arrived at the same exact equation as the answer choice?
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So, you took x-root2=1 and at that point you decided to square both sides of the equation, right, woohoo921?
But, then you only typed the left hand side. What about the right hand side?
If the OG's final equation had zero on the right hand side, and your final equation has 1 on the right hand side, then it's the same equation, yes?
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Which of the following equations has 1 + √2 as one of its roots? 12 Mar 2022, 18:48
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woohoo921

The Official Guide says you can solve this problem by working backwards to construct a quadratic equation with 1 +root2 as a root that does not involve radicals. The OG then proceeds to set x=1+root2 and subtracts 1 from both sides, squares both sides, and then subtracts 2 from both sides to arrive at x^2-2x+1. My question was if I subtract root2 from the right-hand side of the equation instead of root 1 --> x-root2=1 and then solve through, I get x^2-2xroot2+2. Where did I go wrong? Should I have arrived at the same exact equation as the answer choice?

So, you took x-root2=1 and at that point you decided to square both sides of the equation, right, woohoo921?
But, then you only typed the left hand side. What about the right hand side?
If the OG's final equation had zero on the right hand side, and your final equation has 1 on the right hand side, then it's the same equation, yes?
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I did forget about the 1, but I still have a root 2 in the answer approaching the problem this way: x^2-xroot2+1

In other words, what equation would you get solving for: x-root2=1

Thank you for all of your help!!!
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Which of the following equations has 1 + √2 as one of its roots? 12 Mar 2022, 20:24
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woohoo921

What equation would you get solving for: x-root2=1
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Oh, I see what you mean, woohoo921. So, yeah, you're getting a different equation (which is also correct, but it's not in the answer choices). This can happen because [1+root(2)] is just one root out of two, and we don't know what the other root is. There's an infinite number of equations for which [1+root(2)] is a root, depending on the other root. Whoever wrote the OG solution got "lucky" that his or her specific manipulations happened to lead to an equation that appears within the answer choices.
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Which of the following equations has 1 + √2 as one of its roots? 13 Mar 2022, 07:29
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KarishmaB
Bunuel
Which of the following equations has 1 + √2 as one of its roots?

A) x^2 + 2x – 1 = 0
B) x^2 – 2x + 1 = 0
C) x^2 + 2x + 1 = 0
D) x^2 – 2x – 1 = 0
E) x^2 – x – 1= 0

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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KarishmaB
In your view, once you eliminate choices B and C, would it be a bad idea to plug in 1 +root(2) into x for the remaining answer choices and solving algebraically to see which one gives you a solution of zero... if you are really stuck? Thank you!
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Which of the following equations has 1 + √2 as one of its roots? 14 Mar 2022, 01:53
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woohoo921
KarishmaB
Bunuel
Which of the following equations has 1 + √2 as one of its roots?

A) x^2 + 2x – 1 = 0
B) x^2 – 2x + 1 = 0
C) x^2 + 2x + 1 = 0
D) x^2 – 2x – 1 = 0
E) x^2 – x – 1= 0

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)
KarishmaB
In your view, once you eliminate choices B and C, would it be a bad idea to plug in 1 +root(2) into x for the remaining answer choices and solving algebraically to see which one gives you a solution of zero... if you are really stuck? Thank you!
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Certainly not! Whatever helps you get the answer quickly. I try to give you an efficient method - the one that comes to my mind at that time. I could be using another method if facing the same question another time.
Solve it in whichever way you can and then try to look for a better method (while practicing only! :)).
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Which of the following equations has 1 + √2 as one of its roots? 06 Aug 2022, 12:16
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KarishmaB

I thought the conjugate root theorem can be applied only when one of the roots contains an imaginary number. However, neither [1 + (2)^1/2] or [1 - (2)^1/2] contains an imaginary number.

Please help explain.
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Which of the following equations has 1 + √2 as one of its roots? 07 Aug 2022, 06:17
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Vegita
KarishmaB

I thought the conjugate root theorem can be applied only when one of the roots contains an imaginary number. However, neither [1 + (2)^1/2] or [1 - (2)^1/2] contains an imaginary number.

Please help explain.
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It works for irrational roots too (assuming all co-efficients are rational numbers)

Think about it: \(ax^2 + bx + c = 0\)
a, b and c are rational numbers

Roots are \(\frac{-b+-\sqrt{b^2 - 4ac}}{2a}\)

That is, the roots are \(\frac{-b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a}\) and \(\frac{-b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}\) (split into rational and irrational parts). The only difference between the two is the sign before the irrational part.

If the co-efficients were irrational too, then we couldn't split them cleanly into the rational and the irrational parts as we did above.
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Which of the following equations has 1 + √2 as one of its roots? 25 Aug 2023, 09:14
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My approach:
(1+root2)^2=3+2root2
substitute X with this value in each answer choice
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Which of the following equations has 1 + √2 as one of its roots? 14 May 2024, 06:11
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­Annoying but I'd honestly just plug in 1+\(\sqrt{2}\) for x in the answer choices:

­
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Which of the following equations has 1 + √2 as one of its roots? 11 Apr 2025, 07:26
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