Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

How can you tell when a quadratic equation has no real solutions? A. when the radicand is negative B. when b in the quadratic formula is greater than the radicand C. when the radicand equals zero D. when the radicand is not a perfect square I believe the answer is A? is that right?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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

A
its A using the discriminant \[b^2 - 4ac\] < 0 no real solutions the radical will be negative
whats radicand anyway??

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

i go with campbell
the square root symbol
b^2-4ac
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
Thanks!! :) I also have this one... For which value of x does the graph of y = 2x2 − 7x + 6 cross the x-axis? A. −3/2 B. −2/3 C. 2 D. 3 I think its C.
\[b^2-4ac\] is the discriminant
You are correct.
you can substitute each value to find y = 0 or factorise (2x - 3)(x - 2) = 0 x = 3/2 and 2
Thanks so much, here's another one i'm stuck on... What are the approximate solutions of 4x2 + 3 = −12x to the nearest hundredth? A. x ≈ −3.23 and x ≈ 0.23 B. x ≈ −2.72 and x ≈ −0.28 C. x ≈ 0.28 and x ≈ 2.72 D. x ≈ −0.23 and x ≈ 3.23 I think its C, but I'm not sure??
put it in standard form 4x^2 + 12x + 3 = 0 use the general quadratic formula \[x = (-b \pm \sqrt{b^2 - 4ac})/2a\] in your question a = 4, b = 12 and c = 3
C is correct.
thanks so much!! :)
No. It's B
Both roots are negative.
B? Really? are you sure?
Okay, i'll trust you! :)
\[x=\frac{-12\pm \sqrt{96}}{8}\]
I have one more question that's really confusing me, would you mind sticking around for one more?
\[x=\frac{-12+9.80}{8} or x=\frac{-12-9.80}{8}\]
ok
*note* don't ask me why the numbers are spreed out like that, I just copied and pasted it onto here, that how it looks on my paper. Which part of the quadratic formula tells you whether the quadratic equation can be solved by factoring? −b b2 − 4ac 2a Use the part of the quadratic formula that you chose above and find its value given the following quadratic equation: 2x2 + 7x + 3 = 0
b^2-4ac
is that the answer?
Well which part would you choose? -b? 2a????
\[b^2-4ac=7^2-4(2)(3)=49-24=25\]
I don't know... This is really confusing me for some reason...
so the answer is b^2-4ac?
I'm still confused... :/
or is the answer this: b² - 4ac = (7)² - 4(2)(3) = 49 - 24 = 25.
The discriminant, which is b^2-4ac, is the part of the quadratic formula which tells you about the roots.
Do you understand that the question has two parts? READ THE QUESTION!!!
PART 1: Which part of the quadratic formula tells you whether the quadratic equation can be solved by factoring?
The answer to that is :b^2-4ac
PART 2: Use the part of the quadratic formula that you chose above and find its value given the following quadratic equation: 2x2 + 7x + 3 = 0
The answer to that is 25
ohhh okay, I understand now
Thank you!!
yw
I just wasn't looking at the problem correctly, sorry for the confusion and thanks again for breaking it down for me

Not the answer you are looking for?

Search for more explanations.

Ask your own question