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This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.

I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help

This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.

I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help

Thanks, Diana

How did you get: 4(x^2+1)(x^2+2)+8?

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\geq 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 + \sqrt2\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.

This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.

I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help

Thanks, Diana

How did you get: 4(x^2+1)(x^2+2)+8?

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\geq 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 + \sqrt2\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.

First, I squared both sides of the equation. So I got something like (x^2+1)+2sqrt((x^2+1)(x^2+2))+(x^2+2)=4 Then I subbed relevant numbers into the discriminant formula which turns into 4(x^2+1)(x^2+2)+8. Now here, I'm thinking that 4(x^2+1)(x^2+2)+8 is definitely greater than zero, and if discriminant is greater than 0, then there are two roots, no?

P.S. I totally get your reasoning, it's just that I don't undersand why I can't reach that same answer using the discriminant rule, unless I'm not doing something right.

First, I squared both sides of the equation. So I got something like (x^2+1)+2sqrt((x^2+1)(x^2+2))+(x^2+2)=4 Then I subbed relevant numbers into the discriminant formula which turns into 4(x^2+1)(x^2+2)+8. Now here, I'm thinking that 4(x^2+1)(x^2+2)+8 is definitely greater than zero, and if discriminant is greater than 0, then there are two roots, no?

P.S. I totally get your reasoning, it's just that I don't undersand why I can't reach that same answer using the discriminant rule, unless I'm not doing something right.

Thanks heaps!

In the above reasoning there no quadratics shown. So, what is the quadratics you find the discriminant of? How did you get this quadratics? _________________

First, I squared both sides of the equation. So I got something like (x^2+1)+2sqrt((x^2+1)(x^2+2))+(x^2+2)=4 Then I subbed relevant numbers into the discriminant formula which turns into 4(x^2+1)(x^2+2)+8. Now here, I'm thinking that 4(x^2+1)(x^2+2)+8 is definitely greater than zero, and if discriminant is greater than 0, then there are two roots, no?

P.S. I totally get your reasoning, it's just that I don't undersand why I can't reach that same answer using the discriminant rule, unless I'm not doing something right.

Thanks heaps!

In the above reasoning there no quadratics shown. So, what is the quadratics you find the discriminant of? How did you get this quadratics?

bro bunuel,

I did the same way .. squared.. then squared again..

in the end I got, 16 * square(x)= - 7 and now there can never be square root of a negative number , hence no roots.. and hence zero.. hope this is correct. pls let me know _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.

I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help

Thanks, Diana

How did you get: 4(x^2+1)(x^2+2)+8?

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\geq 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 + \sqrt2\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.

Bunuel, Can you pls help me with the explanation:-X^2>=0, how did you get that ...Did you assume that since RHS is a positive term hence the LHS must be equal to 0...that implies x^2 must be greater than or equal to zero.

Also , can you pls post links to some the practice questions of above type...or the method used to determine the number of roots in the above cases.