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Question 1 of 5
Find the quadratic equation with roots:
x=3 and x=-2
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Method One
Bracket the given roots and expand it until a quadratic equation is formed
(x-3)(x+2) |
= |
0 |
Solving this will give out the roots above |
x2+2x-3x-6 |
= |
0 |
Expand |
x2-x-6 |
= |
0 |
Method Two
Find the sum and product of roots
Substitute values onto the Standard Form with Sum & Product of Roots
x2+(α+β)x+αβ |
= |
0 |
x2+(1)x+(-6) |
= |
0 |
x2+x-6 |
= |
0 |
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Question 2 of 5
Find k if one of the roots in the equation below is 4
x2+2kx-8
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First, list the coefficients of the quadratic equation individually
Slot the coefficients to the Sum and Product of Roots Formula to form a system of equations
α+β |
= |
−ba |
|
α+β |
= |
−−2k1 |
|
α+β |
= |
-2k |
Equation 1 |
αβ |
= |
ca |
|
αβ |
= |
−81 |
|
αβ |
= |
-8 |
Equation 2 |
Remember that one of the roots of the equation is 4. Label this as α
Substitute α to Equation 2 and solve for β
αβ |
= |
-8 |
Equation 2 |
4β |
= |
-8 |
Substitute α=4 |
4β÷4 |
= |
-8÷4 |
Divide both sides by 4 |
β |
= |
-2 |
Finally, substitute α and β to Equation 1 and solve for m
α+β |
= |
-2k |
Equation 1 |
|
4+(-2) |
= |
-2k |
Substitute values |
|
2 |
= |
-2k |
|
2÷(-2) |
= |
-2k÷(-2) |
Divide both sides by -2 |
|
-1 |
= |
k |
k |
= |
-1 |
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Question 3 of 5
For which values of m will the roots of the equation be equal?
x2–mx+4=0
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First, list the coefficients of the quadratic equation individually
Since the roots are equal, we can have β=α
Slot the coefficients to the Sum and Product of Roots Formula to form a system of equations
α+α |
= |
−ba |
|
α+α |
= |
−−m1 |
|
2α |
= |
m |
2α÷2 |
= |
m÷2 |
Divide both sides by 2 |
|
α |
= |
m2 |
Equation 1 |
α⋅α |
= |
ca |
|
α⋅α |
= |
41 |
|
√α2 |
= |
√4 |
Get the square root of both sides |
α |
= |
±2 |
Substitute α=±2 to Equation 1 and solve for m
α |
= |
m2 |
|
2 |
= |
m2 |
Substitute α=2 |
|
2×2 |
= |
m2×2 |
Multiply both sides by 2 |
|
4 |
= |
m |
m |
= |
4 |
α |
= |
m2 |
|
-2 |
= |
m2 |
Substitute α=-2 |
|
2×2 |
= |
m2×2 |
Multiply both sides by 2 |
|
-4 |
= |
m |
m |
= |
-4 |
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Question 4 of 5
For which values of k will the roots of the equation be one twice the other?
kx2-9x+2k=0
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First, list the coefficients of the quadratic equation individually
Slot the coefficients to the Sum and Product of Roots Formula to form a system of equations
α+β |
= |
−ba |
|
α+β |
= |
−−9k |
|
α+β |
= |
9k |
Equation 1 |
αβ |
= |
ca |
|
αβ |
= |
2kk |
|
αβ |
= |
2 |
Equation 2 |
Remember that one of the roots of the equation is twice the other.
Using Equation 2, substitute 2α to β and solve for α
α⋅β |
= |
2 |
Equation 2 |
α⋅2α |
= |
2 |
Substitute β=2α |
2α2 |
= |
2 |
2α2÷2 |
= |
2÷2 |
Divide both sides by 2 |
√α2 |
= |
√1 |
Get the square root of both sides |
α |
= |
±1 |
Next, substitute 2α to β on Equation 1
α+β |
= |
9k |
Equation 1 |
|
α+2α |
= |
9k |
Substitute β=2α |
|
3α×k |
= |
9k×k |
Multiply both sides by k |
|
3αk÷3α |
= |
9÷3α |
Divide both sides by 3α |
|
k |
= |
93α |
Finally, substitute the values of α=±1 and solve for k
k |
= |
93α |
|
k |
= |
93(1) |
Substitute α=1 |
|
k |
= |
3 |
k |
= |
93α |
|
k |
= |
93(-1) |
Substitute α=-1 |
|
k |
= |
-3 |
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Question 5 of 5
If the roots of 2x2-3x-2=0 are α and β, find:
6α+6β
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First, list the coefficients of the quadratic equation individually
Slot the coefficients to the Sum and Product of Roots Formula
Manipulate the given expression until you can substitute the sum and product of roots.
|
|
6α+6β |
|
|
= |
6α+6βαβ |
|
|
= |
6(α+β)αβ |
|
|
= |
6(32)-1 |
|
|
= |
-6(32) |
|
|
= |
-9 |