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Question 1 of 6
Identify the sum and product of roots of ax2+bx+c=0, given that:
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First, form a quadratic equation using the given roots
(x-α)(x-β) |
= |
0 |
Solving this will give out the roots above |
x2-βx-αx+αβ |
= |
0 |
Expand |
x2-(β+α)x+αβ |
= |
0 |
x2-(α+β)x+αβ |
= |
0 |
Since the equation formed previously comes from the roots or solution of ax2+bx+c=0, they are considered as identities
x2-(α+β)x+αβ |
≡ |
ax2+bx+c |
|
x2-(α+β)x+αβ |
≡ |
ax2a+bxa+ca |
Divide all terms on the right side by a |
|
x2-(α+β)x+αβ |
≡ |
x2+bax+ca |
Finally, combine equivalent coefficients
This is the product of roots
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Question 2 of 6
Find m if one of the roots in the equation below is -2
2x2-4x+m-1
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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 |
= |
−−42 |
|
α+β |
= |
2 |
Equation 1 |
αβ |
= |
ca |
|
αβ |
= |
m−12 |
Equation 2 |
Remember that one of the roots of the equation is -2. Label this as α
Substitute α to Equation 1 and solve for β
α+β |
= |
2 |
Equation 1 |
-2+β |
= |
2 |
Substitute α=-2 |
-2+β +2 |
= |
2 +2 |
Add 2 to both sides |
β |
= |
4 |
Finally, substitute α and β to Equation 2 and solve for m
αβ |
= |
m-12 |
Equation 2 |
|
(-2)(4) |
= |
m-12 |
Substitute values |
|
-8 |
= |
m-12 |
|
-8×2 |
= |
m-12×2 |
Multiply 2 to both sides |
|
-16 |
= |
m-1 |
-16 +1 |
= |
m-1 +1 |
Add 1 to both sides |
-15 |
= |
m |
m |
= |
-15 |
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Question 3 of 6
Find m if one of the roots in the equation below is -2
mx2-7x+m+6=0
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You can directly substitute the known root -2 as a value of x and solve for m
mx2-7x+m+6 |
= |
0 |
m(-2)2-7(-2)+m+6 |
= |
0 |
Substitute x=-2 |
4m+14+m+6 |
= |
0 |
Evaluate |
5m+20 |
= |
0 |
5m+20 -20 |
= |
0 -20 |
Subtract 20 from both sides |
5m÷5 |
= |
-20÷5 |
Divide both sides by 5 |
m |
= |
-4 |
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Question 4 of 6
For which values of m will the equation have real roots?
x2-mx+4=0
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Nature of the Roots |
Discriminant (Δ) |
Two real roots |
Δ>0 |
One real root |
Δ=0 |
No real roots |
Δ<0 |
First, compute for the discriminant
Δ |
= |
b2−4ac |
Discriminant Formula |
|
= |
(−m)2−4(1)(4) |
Substitute values |
|
= |
m2-16 |
A function that has real roots can have either one or two real roots, hence Δ≥0
Substitute the Δ computed previously, and then solve for k
Δ |
≥ |
0 |
m2-16 |
≥ |
0 |
(m-4)(m+4) |
≥ |
0 |
m=4 |
|
m=-4 |
To determine which region around m=4 and m=-4 would be included, plot these points and make a rough sketch of m2-16
Replace the x axis with m axis and draw an upward parabola since 1 is positive
Remember that Δ must be positive
Therefore, m≤-4 and m≥4
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Question 5 of 6
For which values of k will the roots of the equation be one double the other?
x2+2x-6k=0
Write fractions in the format “a/b”
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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 |
= |
−21 |
|
α+β |
= |
-2 |
Equation 1 |
αβ |
= |
ca |
|
αβ |
= |
−6k1 |
|
αβ |
= |
-6k |
Equation 2 |
Remember that one of the roots of the equation is double the other.
Using Equation 1, substitute 2α to β and solve for α
α+β |
= |
-2 |
Equation 1 |
α+2α |
= |
-2 |
Substitute β=2α |
3α |
= |
-2 |
3α÷3 |
= |
-2÷3 |
Divide both sides by 3 |
α |
= |
-23 |
Next,substitute 2α to β on Equation 2
α⋅β |
= |
-6k |
Equation 1 |
|
α⋅2α |
= |
-6k |
Substitute β=2α |
|
2α2÷-6 |
= |
-6k÷-6 |
Multiply both sides by -6 |
|
α2-3 |
= |
k |
|
k |
= |
α2-3 |
Finally, substitute α=-23 and solve for k
k |
= |
α2-3 |
|
k |
= |
(-23)2-3 |
Substitute α=-23 |
|
k |
= |
49-3 |
|
k |
= |
-427 |
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Question 6 of 6
Find p if the roots in the equation below are reciprocals
(p-2)x2+40x+2p-1=0
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First, list the coefficients of the quadratic equation individually
Slot the coefficients to the Product of Roots Formula to form a system of equations
Recall that the roots of the equation are reciprocal.
Substitute 1α to β and solve for p
αβ |
= |
2p-1p-2 |
|
α⋅(1α) |
= |
2p-1p-2 |
Substitute β=1α |
|
1 |
= |
2p-1p-2 |
|
1×(p-2) |
= |
2p-1p-2×(p-2) |
Multiply both sides by p-2 |
|
p-2 -p |
= |
2p-1 -p |
Subtract p from both sides |
-2 +1 |
= |
p-1 +1 |
Add 1 to both sides |
-1 |
= |
p |
p |
= |
-1 |