Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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
hey Bunuel ! This Equation will have two complex roots +i(√7/4) and -i(√7/4). Since it is not mentioned in the problem that we have to consider only real roots, i marked C as the answer. Am i missing something ?
hey Bunuel ! This Equation will have two complex roots +i(√7/4) and -i(√7/4). Since it is not mentioned in the problem that we have to consider only real roots, i marked C as the answer. Am i missing something ?
Numbers on the GMAT are real by default (GMAT deals with only real numbers), so no need to consider complex roots.
_________________
close
56% (01:14) correct 
