B
given a quadratic equation in standard form
ax² + bx + c = 0
Then we can describe the nature of the roots using the discriminant
Δ = b² - 4ac
• if b² - 4ac > 0, then 2 real, distinct and irrational roots
• if b² - 4ac > 0 and a perfect square, then real and rational roots
• if b² - 4ac = 0, then real and equal roots
• if b² - 4ac < 0, then roots are not real
for x² + 9x + 14 = 0
with a = 1, b= 9 and c = 14, then
b² - 4ac = 9² - (4 × 1 × 14 ) = 81 - 56 = 25
Since b² - 4ac > 0 and a perfect square, then roots are real and rational
1. (f+g)(x)=7x^3-3x^3-8x^2+10x-2x-12=4x^3 - 8x^2 + 8x - 12
2. fg(x)=6-5(6x^2+x)=6-30x^2-5x
3. x^2+18x=44 | +36
x^2+18x+36=44+36
(x+6)^2=80
4. 2x+4y=20 |:2
x+2y=10
3x+2y=26
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3x-x=26-10
2x=16
x=16:2
x=8
x+2y=10
8+2y=10
2y=10-8
2y=2
y=2:2
y=1
5. 14x+y= -4
y= -4 -14x
y=3x^2-11x-4
-4 -14x = 3x^2-11x-4 |+4
-14x=3x^2-11x |+14x
0=3x^2+3x
3x(x+1)=0
x=0 or x=-1 is the solution
6. x^2-6x+8= x^2-2x-4x+8= x(x-2)-4(x-2)=(x-2)(x-4)
7. x^2-6x-27=0
x 1= (-6+V36+4*27)/2= (-6+V144)/2= (-6+12)/2=6/2=3
x2= (-6-12)/2= -18/2= -9
the point are ( 3,0) and (-9.0)
Answer:
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Step-by-step explanation:
We can use substitution.
We can substitute 2x + 3 for 'y' in the 2nd equation.
y = 3x + 5
2x + 3 = 3x + 5
Subtract 3 to both sides:
2x = 3x + 2
Subtract 3x to both sides:
-x = 2
Multiply -1 to both sides:
x = -2
Now we can plug this into any of the two equations to get our y-value:
y = 2x + 3
y = 2(-2) + 3
y = -4 + 3
y = -1
So our solution is (-2, -1)
Answer:
friction
lol you’re welcome
