http://prntscr.com/l5i2b9
@Vocaloid
any ideas? the highest exponent is 3 and there's only 1 term
D
or C
close, binomial would be 2 terms (so ___ + ____) but it's only one term so monomial (C)
"no real solutions" so b^2 - 4ac must be negative, if you look at the equation a = 1 and b = 4 so try seeing what possible c value would make the discriminant negative
-5?
4^2 - 4(1)c < 0 16 - 4c < 0 16 < 4c c > 4 which is the only choice that works for c?
20
good so 20 = your sol'n
so you apply the distributive property A(B+C+D) = A*B + A*C + A*D following the same logic 3x(x^2 + 2x -4) = ?
D
yeah that's what I got too good
ooh boy anyway I don't really know how to explain this w/o giving it away but try applying the distributive property on x first to get x(6x-1) and then on 4 to get 4(6x-1)
add or subtract?
it's (x+4) so the +4 retains its sign making A the better option
|dw:1539393873804:dw|
the problem pretty much tells you that a = the cube root of 27x^3 = 3x and b = cube root of 8 = 2 so try to figure out what (a+b)(a^2 - ab + b^2) would be based on that
A?
yeah that's what i got too, well done
which one has a coefficient higher than the existing coefficient 4?
B
awesome
any ideas? try to imagine what vertical line would cut the parabola in two equal halves
either c or d
leaning to D
|dw:1539394263378:dw|
any ideas? the vertex (the pointy part of the parabola) should be on (-2,-9) and it should cross the x-axis on -5 and 1 to save time it's one of the first two choices
B
good
x-coordinate of the vertex is -b/(2a) so check your equation to see what the b and a values are, and plug 'em in
b is 6? or 4
ax^2 + bx + c so yes, b will be 6 since it's attached to the x
-6/2(4)
i think
a is 3, since the 3 is what's attached to the x^2
so -6/(2*3) = -1 = your solution
basically it's saying the length has to be 20 more than the width so if s is the width and s + 20 is the length, what's the area?
B?
yes
okay thank you voca !
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