Question

A quadratic equation in the variable y whose roots are α and β is
pls answer fast I'll mark you as the brainliest​

Answer 1

Answer:

Yes how r u where r u when r u what r u do

Answer 2

Answer:

Everybody knows the quadratic formula. There are a couple of variants I call the Shakespeare Quadratic Formula ( 2b or −2b ). I do a zillion math problems on Quora, and this little shortcut often saves an annoying simplification step. When we put a coefficient on our coefficient, we can get a formula without a fraction:

x2−2bx+c has zeros x=b±b2−c−−−−−√

A less pretty but often useful variation is ax2−2bx+c has zeros

x=1a(b±b2−ac−−−−−−√)

These are trivial to show from the regular quadratic formula so I won’t bother.

I remembered one more thing. It’s not really a trick, but it’s an interesting fact.

Let’s say we didn’t know what we were doing and try to solve

2x2−2x−1=0

2x2=2x+1

x=1+12x

x=1+12(1+12x)

x=1+12+1x

x=1+12+1x

x=1+12+11+12x

It doesn’t seem like we’re getting anywhere because we have x on both sides. If we continue we’ll get

x=1+12+11+12+11+…

If we stop at various stages we get

x1=1

x2=1+12=32=1.5

x3=1+12+11=43≈1.33

x4=1+12+11+12=118=1.375

x5=1+12+11+12+11=1511≈1.3636

We could go on like this. From the Shakespeare Quadratic Formula 2x2−2x−1=0 has roots

x=12(1±3–√)

It’s a quite pleasant fact that the positive root is x≈1.3660 and as we take more and more terms of our continued fraction, the values approach this root.