Solving equations by quadratic formula and quadratic method
Quiz yourself by thinking what should be in
each of the black spaces below before clicking
on it to display the answer.
Help!
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| What is the quadratic formula? How many solutions will you get? | x = [ -b ± sqrt(b^2 - 4ac) ] / 2a
2 solutions
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| What is the first step when solving a quadratic equation? | Write equation in standard form
ax^2+bx+c=0
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| Is 3x^2+10=25 in standard form? If not, make corrections. | -No this equation is not in standard form and does not equal 0
-subtract 25 from both sides
3x^2+10-25=0 is correct form
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| Identify values for a,b, & c in this equation? 3x^2 + 10x - 25 = 0 | a=3
b=10
c=-25
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| Identify values for a,b, & c in this equation? x^2 - 4x + 9 = 0 | a=1
b=-4
c=9
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| Solve for x using the quadratic formula for this equation: x^2 + 2x - 8 = 0 | Identify values for a,b, & c. Plug a=1, b=2,c=-8 into the quadratic formula and simplify.
x = [ -2 ± sqrt(2^2 - 4(1)(-8) ] / 2(1)
x = [-2 ± sqrt (4+32) ] / 2
x = [-2 ± sqrt (36) ] / 2 [Look for a perfect square factor]
x = -2 ± 6 / 2
x = -4, x = 2
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| What part of the quadratic formula is called the discriminant and what is this used for? | b^2 - 4ac
to determine the type of solutions produced by the equation
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| When using the discriminant, what type of solutions will a positive, negative and an answer of 0 produce? | positive = two real solutions
0 = one real solution
negative = two complex but not real solutions
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| How would you solve for x with a triangle that has a known hypotenuse of 36 and one side that is 8 feet longer than the last side? | Use the equation x^2 + (x+8)^2 = 36^2
Simplify, combine like terms, and set equation = 0
2x^2+16x-1232=0 (Factoring out 2 optional)
Plug in values into quadratic formula & simply
Only use positive answer for solution for distance (x=about 21)
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| In a equation like (p+2)^2 = 9(p+2) - 20, you can substitute (p+2) with x to make the equation less intimidating. Try to solve. | x^2 = 9x - 20
Write in standard form, factor or use quadratic equation
(x-4)(x-5)=0 set each value equal to 0 and solve
x=4, x=5
plug x values into x=p+2 & solve for p since we used substitution
p=2, p=3. Plug p values back into original equation
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Created by:
marynowak11
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