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.

I need help with my math hw solving quadratic equations can someone help me understand it by helping guide me through I will surely appreciated the effort I'm here to learn :D

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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

The first question is \[x ^{2} -3x +2=0\]
You can either inspect the quadratic equation to see what 2 numbers multiply to be +2 and add up to be -3. Then you can write rewrite the quadratic in terms of two factors (x-2)(x-1) = 0 then by the zero product property if one of the factors is zero then the whole thing is zero so you can then solve the equations: x -2 =0 and x-1 = 0 and soon enough you'll find that x=2 and x=1 are the roots to the quadtratic.
Another approach is to use the quadratic formula which is a sure-fire way of finding the roots but the catch is that it must be in the standard form which is \[y = ax^2 + bx + c\] and then you would use the quadratic formula: \[x = {-b \pm \sqrt(b^2-4ac) \over 2a}\] the \[\sqrt(b^2-4ac)\] term is called the "discriminant" because if you get a negative number in that part of the equation then there are no real solutions and thus "discriminates" between real and imaginary solutions. Note that a quadratic equation has 2 roots...either both real or one real and one imaginary.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Sonic, I believe imaginary numbers come in pairs. So you have either two real roots or two imaginary roots.
Another method you can use to solve quadratic equations is called "completing the square" which requires the coefficient of the \[x ^2\] term to be one. The basic process involves you taking the b coefficient in \[x^2 +bx+c\] and then dividing by 2 it and then squaring it and then plugging it back into the equation which will give us:\[(x^2 + bx +({b \over 2})^2 ) - ({b \over 2})^2 + c = 0\]. You put that \[-({b \over 2})^2\] to maintain equality in the equation. It then follows that The first part in parenthesis is \[x^2 + bx + ({b \over 2})^2 = (x+ ({b \over 2}))^2\]. You can now have\[(x + ({b \over 2}))^2 -({b \over 2})^2 + c = 0\] Then you would find the roots by solving for x by moving all the terms other than the perfect square to the other side then taking the square root of both sides and finally solving for x - remember when you take the square root of something you have to put a plus minus there. I hope that was of help.

Not the answer you are looking for?

Search for more explanations.

Ask your own question