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How many roots does the equation \(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) have?

A. 0 B. 1 C. 2 D. 3 E. 4

We know that \(x^2 \ge 0\) so the least value of the left hand side (LHS) of the equation is for \(x=0\):

\(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4\). Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real \(x\) can satisfy given equation.

How many roots does the equation \(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) have?

A. 0 B. 1 C. 2 D. 3 E. 4

We know that \(x^2 \ge 0\) so the least value of the left hand side (LHS) of the equation is for \(x=0\):

\(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4\). Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real \(x\) can satisfy given equation.

Answer: A

Awkward question. I was trying to square both site and ultimately got stuck somewhere. Nice approach to this ques.
_________________

easy approach. but for ppl who have been trying to square. What can be done is to take rt(x2+2) to the right side of the equation. Square it and solve for roots. x2 turns out to be negative. Hence no solution

How many roots does the equation \(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) have?

A. 0 B. 1 C. 2 D. 3 E. 4

We know that \(x^2 \ge 0\) so the least value of the left hand side (LHS) of the equation is for \(x=0\):

\(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4\). Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real \(x\) can satisfy given equation.

Answer: A

Bunel,

Pls could you tell how to approach a Q asking to find no. of roots?

We know that x2≥0x2≥0 so the least value of the left hand side (LHS) of the equation is for x=0x=0:

x2+1−−−−−√+x2+2−−−−−√=1+2√≈2.4x2+1+x2+2=1+2≈2.4. Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real xx can satisfy given equation.

The correct answer is A

My Doubt is that since the question is asking for roots, dont we have to find the discriminant (b^2-4ac) and then determine the no. of roots?

How many roots does the equation \(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) have?

A. 0 B. 1 C. 2 D. 3 E. 4

We know that \(x^2 \ge 0\) so the least value of the left hand side (LHS) of the equation is for \(x=0\):

\(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4\). Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real \(x\) can satisfy given equation.

Answer: A

Hi, Would you please explain how \(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4\) is derived.

How many roots does the equation \(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) have?

A. 0 B. 1 C. 2 D. 3 E. 4

We know that \(x^2 \ge 0\) so the least value of the left hand side (LHS) of the equation is for \(x=0\):

\(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4\). Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real \(x\) can satisfy given equation.

Answer: A

Hi, Would you please explain how \(\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4\) is derived.

Thanks.

By substituting x=0 into the equation.
_________________

The Official Solution already posted is the kind of logic that is given a handy reward on the GMAT, but for those of you who started to simplify, I wanted to show an alternative route to getting this correct.

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

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

\(x^2+1+2(\sqrt{x^2+1})(\sqrt{x^2+2})+x^2+2=4\)

\(2x^2+2(\sqrt{x^4+3x^2+2})=1\)

\(2(x^2+\sqrt{x^4+3x^2+2})=1\)

\(x^2+\sqrt{x^4+3x^2+2}=\frac{1}{2}\)

\(\sqrt{x^4+3x^2+2}=\frac{1}{2}-x^2\)

\(x^4+3x^2+2=(\frac{1}{2}-x^2)^2\)

\(x^4+3x^2+2=\frac{1}{4}-x^2+x^4\)

\(4x^2+\frac{7}{4}=0 => 16x^2+7=0\)

This is a quadratic equation of the form \(ax^2+bx+c=0\), where \(a=16\), \(b=0\), and \(c=7\)

The discriminant of a quadratic equation is given by the equation \(D=b^2-4ac\)

When \(D>0\), there are exactly two real solutions When \(D=0\), there is exactly one real solutions When \(D<0\), there are no real solutions

In this case \(D=0^2-4(16)(7)<0\), so there are no real solutions.

Let \(\sqrt{x^2+1} = A\) and let \(\sqrt{x^2+2} = B\) \(=>\) \(\sqrt{A} + \sqrt{B} = 2\) Rearrange and square both sides \(=>\) \(A = (2-\sqrt{B})(2-\sqrt{B})\) \(A = 4-4\sqrt{B}+B\) Substitute original expressions back in \(=>\) \(x^2+1 = 4-4\sqrt{x^2+2}+x^2+2\) \(4\sqrt{x^2+2}=5\) Square both sides, rearrange \(=>\) \(16x^2=-7\) which is not true for any real x, so 0 solutions