Show that the equation 2^x= 2 − x^2 has at least two real roots??

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Show that the equation 2^x= 2 − x^2 has at least two real roots??

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Ankur, This is a quadratic equation, \[x^2+2x-2=0\]To determine the nature of the solutions to a quadratic, we use something called the 'discriminant', defined as\[\Delta = b^2-4ac\]where a,b, and c are the coefficients of the quadratic,\[ax^2+bx+c=0\]A quadratic has two real solutions if\[\Delta >0\] one real solution when\[\Delta = 0\]and no real solutions when\[\Delta <0\]In your case\[a=1,b=2,c=-2\]so\[\Delta = 2^2-4(1)(-2)=12>0\]which means your quadratic will have two real solutions.
The function \[2^x-2+x^2\] is continuous, negative at 0 and positive at both -2 and 2. Thus there are roots in those two intervals.
I hope I have copied the equation correctly. The equation is \[2^{x}=2-x ^{2}\] I set x to =2 I get the following\[2^{2}=2-2^{2}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

4=2-4 Duh I don't think so if I let x=0 then I have 1=2-1 hey 1=1 that is a solution! What about x=1, then 2 =2-2=0 No 1 is not a solution. If x=3 then 8=2-9 8=-7 No What is the solution and is the term \[2^{x}\] makes question this problem unless is something do with logs.
Only that 0^2 = 0 and not 1...
That was a \[2^{0}\]
I assume you were discussing when I let x = 0
Yes indeed and you mixed up something. If you insert x=0 you get \[2^0 = 2 - 0^2 ⇔ 1 = 2 - 0\] which is false.
Ah yes nowhereman, so even 0 is not a solution!
Sorry, Ankur, I read your equation incorrectly.

Not the answer you are looking for?

Search for more explanations.

Ask your own question