Question

To open these doors, you must match the number and type of solutions for the following two fuctions in standard form.
f(x) = x2 + 6x – 16
g(x) = x2 +6x + 1
Match the following descriptions of the solutions to each of the functions above.
Hint: they each have their own match.
Two real irrationals solutions.
Two real rationals solutions.

Answer 1

Have you heard of the discriminant?

Answer 2

no :

Answer 3

Okay, how about the quadratic formula?
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 4

the discriminant is the thing under the square root sign, and it's sometimes referred to by the capital Greek letter \(\Delta\)

\[\Delta = b^2-4ac\]

If \(\Delta = 0\) you have a "perfect square" and the solution is just \[x=\frac{-b}{2a}\]This is also the formula for the x-coordinate of the vertex of a parabola written in the form \(y=ax^2+bx+c\)

If \(\Delta > 0\), you'll have two real solutions, assuming that the square root simplifies to something that is rational (read: can be represented by the ratio of two integers). \(\sqrt{4} = 2\) is rational. \(\sqrt{3}\approx 1.7320508\) is not rational.

If \(\Delta < 0 \), you'll have two complex solutions, in the form of a conjugate pair, which can be written as \(a\pm bi\) where \(i = \sqrt{-1}\) Again, they will be rational if the square root can be simplified to something that doesn't require a radical sign, or irrational otherwise.

Answer 5

Sorry, I misspoke a bit:

If \(\Delta > 0\), you'll have two real solutions. They'll be rational, assuming that the square root simplifies to something that doesn't need a radical sign. Otherwise, they'll be irrational.

Answer 6

can you write out the steps, sorry im a visual learner

Answer 7

Write out your first equation, in order of descending exponents, and with a 0 on the right side of the equals sign:

\[x^2 + 6x – 16 = 0\]\[ax^2+bx+c = 0\]
Can you tell me the values of \(a,b,c\) to make those equivalent?

Answer 8

a x2
b 6
c 16

Answer 9

-16*

Answer 10

\(a=1\), because \(ax^2 = x^2\) If we divide both sides by \(x^2\), we're left with \(a=1\). Agreed?

Answer 11

\(b=6, c=-16\) are correct

So let's calculate \(\Delta\) for that equation:
\[\Delta = b^2-4ac = (6)^2-4(1)(-16)=\]

Answer 12

Can you evaluate that for me, please?

Answer 13

that looks like another langue to me

Answer 14

12-4-64

Answer 15

No, \[(6)^2 = 6*6,\text{ not } 6+6\]

\[\Delta=(6)^2-4(1)(-16) = 6*6 - 4*1*(-16) = 36 - 4*(-16)\]\[\qquad = 36 - (-64) = 36 + 64 = 100\]
Subtracting a negative number is the same as adding a positive number

Answer 16

what would be the final outcome og that

Answer 17

Well, so \(\Delta = 100\)
Which of the 3 cases that I outlined earlier is that?

Answer 18

:,( not sure

Answer 19

Okay, here are the three cases again:
\[\Delta = 0\]Perfect square, only 1 solution (actually, it's 2 copies of the same solution, but that doesn't matter for your purposes)

We don't have that case, agreed?

Answer 20

agreed

Answer 21

Okay, here's another case:

\[\Delta < 0\]
That means we'll have the square root of a negative number involved, which means complex roots rather than real ones.
We don't have this case, either, right?

Answer 22

no sir

Answer 23

That leaves us with \[\Delta > 0\]Two real roots, but we don't know (yet) if they are rational or irrational. Rational is just a fancy term for "can be written as a fraction". All the counting numbers are rational. All the fractions you can make by dividing any counting number by any other counting number are rational. The negative values are, as well.

Numbers where you can't write them as a fraction, and thus can't write them as a decimal which either terminates (like 1.25) or has a fixed pattern (like 1/3 = 0.3333333...) are irrational. Any square root that you can't simplify to something without a square root sign
(\(\sqrt{5}\) would be an example) is irrational. \(\pi\) is irrational (happy Pi day today, by the way!)

Answer 24

If the square root of \(\Delta\) doesn't need a square root sign to be written, then we have a pair of real, rational roots. Can you simplify \[\sqrt{100}\]so that it does not need a square root sign?

Answer 25

no right

Answer 26

No? Are you sure? Can't you think of a number \(x\) such that \(x*x = 100\)?

Answer 27

50 ?

Answer 28

50*50 = 100? You understand that "*" means multiply, not add, right?

Sorry for the delay, got an important phone call...

Answer 29

No, 10*10 = 100...so \[\sqrt{100} = \sqrt{10*10} = 10\]

Because that is rational, the roots of the first equation are real and rational.

Answer 30

Doing the other one:

\[x^2 +6x + 1=0\]\[ax^2+bx+c=0\]\[a=1,\,b=6,\,c=1\]
\[\Delta = b^2-4ac = (6)^2-4(1)(1) = 36-4 = 32\]\[\Delta > 0\]so we have two real roots, but \[\sqrt{32} = \sqrt{2*2*2*2*2} = \sqrt{2*2}*\sqrt{2*2}*\sqrt{2} = 2*2*\sqrt{2} = 4\sqrt{2}\]which is irrational, so we have two real, irrational roots