Download Solving Quadratic Equations

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ax² + bx + c = 0
• x² + 8x + 16 = f(x)
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To make the chart, you simply take any number and
plug it in for x in the equation and the value you get is
the y value.
So -2 plugged into “x” in the equation and solved
would give you 4.
get a graph, simply plug the equation into your calculator.
• To graph by hand all you have to do is graph the ordered
pairs that you just found by plugging in the x to find y.
x
y
-2
4
-1
9
8
16
2
36
3
49
Either by graphing by hand
or graphing on your
calculator, your graph
should look like this.
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Quadratic equations CAN have two solutions or just one.
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The graph of the parabola (the “U” shape) that you get,
should be touching the line in one or two places, if it isn’t,
then the equation has “no solution.”
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When graphing by hand, you can usually see the graph
without graphing all of the coordinates that you found.
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This method is fairly simple, as long as the numbers aren’t
large, which makes it more complex to try to graph!
Y= (ax²) + (bx) + (c)
Quadratic
Term
Linear
term
Constant
For solving quadratic equations by factoring you take
your problem and try to factor it.
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x² + 8x + 16
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(x + 4) (x + 4)=0
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(x+4)=0 (x+4)=0
- 4 -4
-4 -4
x=-4
x= -4
• To do this, you find out what
multiplies to 16 and adds to 8
so we would have (4 x 4)=16
and (4+4)=8.
• Your next step would be to
equal each to zero.
So your answer to this equation would be x = - 4
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Make sure that you always set both equations equal to zero!
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Sometimes problems aren’t able to be factored, keep your eye
out for those!
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If you have a problem where there is a number in front of the x²
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You multiply the number in the quadratic term to
the number in the constant.
2x² + 8x + 8
x² +8x + 16
(2x + 4) (2x + 4) = 0
2
2
x+2=0
x+2=0
- 2 -2
-2 -2
X= -2
x= -2
Now you bring back the 2 to put in front of the x.
Since that is divisible by 2 then you divide by 2.
Set each equal to zero and solve from there.
Your answer would be x = -2
1.
2.
3.
4.
5.
6.
The quadratic and linear term need to be on the left side, and the constant on
the right!
Find the perfect square trinomial for the left side.
Add that number to both sides of the equation.
Factor the left side.
Find the square root of both halves of the equation.
Write the solutions
x + 8x - 9
+9 Add 9 to move it to the other side.
x +8x = 9
Divide 8 by 2then square that number to get 16, add the 16 to both
x + 8x + 16 = 9 +16 sides!
(x+4) (x + 4) = 25
(x+4)2 = 25
√(𝑥 + 4)2 = √25
x+4=5
-4
X=1
Since your terms are alike, you can square them!
Now find the square root of both of them.
x+4 = -5One of your answers is always a negative, the other a positive.
-4
x = -9 Solve out the two equations, to find your TWO answers!
Fun tip: The Quadratic Formula will work for ALL quadratic equations!!
−𝑏 ± 𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
4x² + 8x - 16
−8 ± 82 − 4(4)(−16)
𝑥=
2(4)
−8 ± 64 + 256
𝑥=
8
−8 ± 320
𝑥=
8
𝑥=
−8 ± 6 5
8
Plug your equation into the
quadratic formula.
Work the multiplying out first.
For this problem, we have to factor
what’s under the radical.
Sometimes your answer is left with the
division and the radical in this
method!
The discriminant is -b² -4ac
Value of discriminate
Nature of solutions
Negative
2 imaginary solutions
Zero
1 real solution
Positive(perfect square)
2 reals - rational
Positive (not perfect square)
2 reals - irrational