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BY L . D .
SOLVING QUADRATIC EQUATIONS USING THE QUADRATIC FORMULA
TABLE OF CONTENTS
• Slide 3: Formula• Slide 4: Solve 2x2 – 5x – 3 = 0• Slide 10: Solve 2x2 + 7x = 9• Slide 13: Solve x2 + x – 1 = 0
FORMULA
The Quadratic Formula:
We start with a problem structured like ax2 + bx + c = 0, then we use the a, b & c from that format in ournew formula:
x = -b ± √b2 – 4ac2a
PROBLEM 1
• Solve the equation
2x2 – 5x – 3 = 0
PROBLEM 1
• Solve the equation
2x2 – 5x – 3 = 0
The first thing we need to do is to see if this fits the first formula. It doessince it has the variable squared,the coefficient and the constant all equaling to zero.
Formula
ax2 + bx + c = 0
to
x = -b ± √b2 – 4ac 2a
PROBLEM 1
• Solve the equation
2x2 – 5x – 3 = 0
Next we find the a, b & c and put them into the second formula. To find them we just match up theplaces they are in formula one to the places we have in our problem. By doing this we discover thata = 2 b = -5 c = -3
Formula
ax2 + bx + c = 0
to
x = -b ± √b2 – 4ac 2a
PROBLEM 1
• Solve the equation
2x2 – 5x – 3 = 0
a = 2 b = -5 c = -3
Now that we have these, we canplace them in our second formula.
x = -(-5) ± √(-5)2 – 4(2)(-3) 2(2)
Formula
ax2 + bx + c = 0
to
x = -b ± √b2 – 4ac 2a
PROBLEM 1
Now we solve the equation we made!
x = -(-5) ± √(-5)2 – 4(2)(-3) 2(2)
x = 5± √25 + 24 4
x = 5 ± √49
4
x = 5 ± 7
4
x = 5 + 7 4
x = 12 4
x = 5 - 7 4
x = -2 4
or x = 3
or x = -1/2
PROBLEM 2
• Solve 2x2 + 7x = 9
PROBLEM 2
• Solve 2x2 + 7x = 9
So first we reorder it!
2x2 + 7x = 9 -9 -92x2 + 7x – 9 = 0
Formula
ax2 + bx + c = 0
to
x = -b ± √b2 – 4ac 2a
PROBLEM 2
2x2 + 7x – 9 = 0
Now we identify the a, b & c andplace it in formula two. Then we solve.x = -(7) ± √(7)2 – 4(2)(-9)
2(2)
x = -7 ± √49 – (- 72) 4
Formula
ax2 + bx + c = 0
to
x = -b ± √b2 – 4ac 2a
x = -7 ± √121 4
x = -7 ± 11 4
PROBLEM 2
x = -7 ± 11
4
x = -7 - 11
4
x = -18
4
x = -7 + 11
4
x = 4
4
or x = 1or x = -9/2
PROBLEM 3
• Solve x2 + x – 1 = 0
PROBLEM 3
• Solve x2 + x – 1 = 0
First we need to identify a, b & c
a = 1 b = 1 c = -1
Next we need to put them in our problem and solve.
x = -(1) ± √(1)2 – 4(1)(-1) 2(1)
Formula
ax2 + bx + c = 0
to
x = -b ± √b2 – 4ac 2a
PROBLEM 3
x = -(1) ± √(1)2 – 4(1)(-1) 2(1)
x = -1 ± √1+ 4 2
x = -1 ± √5 2
x = -1 + √5 2
x = -1 - √5 2
We can go no further since 5
can’t be squared.
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