Question

Part 1: Create your own quadratic equation that cannot be solved by factoring, but can be solved using the quadratic formula. Identify the values of a, b, and c, and find the solutions using the quadratic formula. Show all work to receive credit.

Part 2: Using complete sentences, explain how you know that the equation from Part 1 cannot be solved by factoring, but can be solved by using the quadratic formula.

Answer 1

Make something in ax^2+bx+c where b^2-4ac is not a neat square root.

Answer 2

how do i know when its a perfect square ?

Answer 3

Take several sequential numbers, square them, and then make sure it is between two of those.

Answer 4

\[2x^2-3x+\frac{ 1 }{ 2 }=0\]

Answer 5

3x^2 + 5x + 6 is that good ?

Answer 6

That doesn't really help unless he knows why. Your polynomial has imaginary roots, Gaby.

Answer 7

(x − 5) × (x − 5)

(x − 5) × (x − 5) = x2 + -5x + -5x + 25 = x2 + -10x + 25 and x2 + -10x + 25
is a perfect square trinomial

Answer 8

numbers that i can use ? NO FRACTIONS i havent larned that lol

Answer 9

Whenever you multiply a binomial by itself twice, the resulting trinomial is called a perfect square trinomial

Answer 10

OK, just pick a, b, and c so that b^2-4ac is >0 but not a perfect square number.

Answer 11

1 2 3 ?

Answer 12

Not in any order, no.

Answer 13

lol good night @Gabylovesyou
thanks for taking over, @Numb3r1

Answer 14

No problem, @nincompoop

Answer 15

goodnight .-. and @Numb3r1 how about x^2 + x - 4 ?

Answer 16

That works.

Answer 17

now to solve it ..

Answer 18

That works because when you take the quadratic, you have:\[\frac{(b \pm \sqrt{ b^{2}-4ac})}{2a}=\frac{(1 \pm \sqrt{ 1^{2}-(4\times1\times-4)}}{2\times1}=\frac{1 \pm \sqrt{ 17}}{2}\]

Answer 19

The root 17 would require pretty elaborate factorings which humans don't normally do.

Answer 20

............I do it.......

Answer 21

x = 1 +- sqrt(1^2 - 4(1)(-4)) all over 2(1)
x = 1 +- sqrt(1 + 16) all over 2
x = 1 +- sqrt(17) all over 2

Answer 22

is that correct @Numb3r1

Answer 23

there is usually -b in the starting of the formula not just 'b'

Answer 24

As nubeer pointed out the formula is
\[x= \frac{(-b \pm \sqrt{ b^{2}-4ac})}{2a} \]
notice the "minus b" in front

that means your answer should be
\[ x= \frac{-1 ± \sqrt{17}}{2}\]
for the problem
\[ x^2 + x - 4 =0 \]

we can test to see if the answer makes the equation true. Do we get 0 ?
one of the x answers is -0.5 + sqrt(17)/2= 1.561552813

(1.561552813)^2 + 1.561552813 -4 = a very small number
close enough to 0

(we won't get *exactly* zero if we use decimals because the sqrt(17) 's digits go on forever.
if we keep x in the form
\[ x= \frac{-1 + \sqrt{17}}{2}\]
and use algebra, we would get exactly 0.