Question

Which equation is a quadratic equation?

Answer 1

the one with the x^2

Answer 2

y = −4(3x + 2)

y + 3x^3 = (x − 2)(3x + 8)

y − 3 = (2x^2 + 11)(x − 1) + 3x

y + 5x = −2x(4 − x) + 1

Answer 3

so it would be the third option?

Answer 4

I feel d go ahead and isolate y and expand

Answer 5

norp not the third

Answer 6

its the last one

Answer 7

(2x^2 + 11)(x − 1) + 3x
^ ^ beome 2x^3
if you distribute

Answer 8

coz it involves powers up to 2

Answer 9

okay thanks... i have another question...
What are the x-intercept(s) of the graph of y − 8 = x2 − 6x?

(−2, 0) and (4, 0)

(2, 0) and (−4, 0)

(2, 0) and (4, 0)

(−2, 0) and (−4, 0)

Answer 10

y − 8 = x2 − 6x
add 8 to both sides and plug in 0 into y
making y zero means that you will get the x axis as a result when you solve for x.

0= x2 − 6x+8 from here it is a standard quadratic equation. You can use your favorite method of finding the roots aka zeros aka x intercepts.

those methods include graphically, completing the square, quadratic formula or factoring. :)

Answer 11

as it turns out it can be factorable
0= x2 − 6x+8
is
0 = (x-4) (x-2)

I think you might now this but lets say you dont.
(x-4)* (x-2)=0
a * b =0

look at bottom equation. In order for a*b=0 to be true either a can be 0 or b can be 0
which means

x-4=0 or x-2=0

x=4 x=2


lets see

(4-4)*(2-4)=0
0 * (-2)=0
0=0 there fore true

same applies to x=2
(4-2) * (2-2)=0
(2) * (0)=0
0=0

there fore roots or x intercepts are
4 and 2

Answer 12

thank you so much! that helps a lot :)

Answer 13

yeah i can help