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# Which of the following equation has 3+2[m][square_root] 3 [/square_roo

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Joined: 04 Jan 2015
Posts: 2457
Which of the following equation has 3+2[m][square_root] 3 [/square_roo  [#permalink]

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30 Jun 2018, 10:10
00:00

Difficulty:

85% (hard)

Question Stats:

55% (01:58) correct 45% (01:50) wrong based on 122 sessions

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Question-2

Which of the following equation has $$3+2\sqrt{3}$$ as one of its roots?

Options

A. $$x^2+6x+3 = 0$$
B. $$x^2-6x+3 = 0$$
C. $$x^2+6x-3 = 0$$
D. $$x^2-6x-3 = 0$$
E. $$x^2-4 \sqrt{3}x-3 = 0$$

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Which of the following equation has 3+2[m][square_root] 3 [/square_roo  [#permalink]

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Updated on: 01 Jul 2018, 05:25
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.

Originally posted by Hero8888 on 30 Jun 2018, 14:04.
Last edited by Hero8888 on 01 Jul 2018, 05:25, edited 2 times in total.
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Re: Which of the following equation has 3+2[m][square_root] 3 [/square_roo  [#permalink]

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30 Jun 2018, 15:14
1
(X-(3+2sqrt(3))(x-(3-2sqrt(3))
=x^2-(3-2sqrt(3))x-(3+2sqrt(3))x-3
=x^2-6x-3

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Re: Which of the following equation has 3+2[m][square_root] 3 [/square_roo  [#permalink]

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30 Jun 2018, 18:54
3
1
Here's my solution: If one root is 3+2$$\sqrt{3}$$, then other root is 3-2$$\sqrt{3}$$.

Sum of roots = 6
Product of roots = 9-(4*3) = -3

Hence equation is: $$x^{2}$$ - (sum of roots)*x + (product of roots) = $$x^{2}$$- 6x -3

Hence D
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Re: Which of the following equation has 3+2[m][square_root] 3 [/square_roo  [#permalink]

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04 Jul 2018, 02:45

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:
We know that if one of the roots of a quadratic equation is $$a + \sqrt{b}$$, then as a conjugate root, the other one will be $$a – \sqrt{b}$$

Therefore, the roots of the given equation will be:
• $$3 + 2\sqrt{3}$$
• $$3 – 2\sqrt{3}$$

So, the equation can be:
• $$[x – (3 + 2\sqrt{3})] [x – (3 – 2\sqrt{3})] = 0$$
Or, $$(x – 3 – 2\sqrt{3}) (x – 3 + 2\sqrt{3}) = 0$$
Or, $$x^2 – 3x + 2\sqrt{3}x – 3x + 9 – 6\sqrt{3} – 2\sqrt{3}x + 6\sqrt{3} – 12 = 0$$
Or, $$x^2 – 6x – 3 = 0$$

Hence, the correct answer is option D.

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Which of the following equation has 3+2[m][square_root] 3 [/square_roo  [#permalink]

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04 Jul 2018, 02:57

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.

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Joined: 17 Jun 2018
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Re: Which of the following equation has 3+2[m][square_root] 3 [/square_roo  [#permalink]

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04 Jul 2018, 19:25
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
Re: Which of the following equation has 3+2[m][square_root] 3 [/square_roo &nbs [#permalink] 04 Jul 2018, 19:25
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# Which of the following equation has 3+2[m][square_root] 3 [/square_roo

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