Sum & Product of Roots 2
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Question 1 of 5
1. Question
Find the quadratic equation with roots:`x=3` and `x=-2`Hint
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Sum of Roots `alpha` and `beta`
$$\alpha+\beta=-\frac{\color{#9a00c7}{b}}{\color{#00880A}{a}}$$Product of Roots `alpha` and `beta`
$$\alpha\beta=\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$Standard Form with Sum & Product of Roots
`x^2+(alpha+beta)x+alphabeta=0`Method OneBracket the given roots and expand it until a quadratic equation is formed`x=3` `x=-2``(x-3)(x+2)` `=` `0` Solving this will give out the roots above `x^2+2x-3x-6` `=` `0` Expand `x^2-x-6` `=` `0` `x^2-x-6=0`Method TwoFind the sum and product of roots`x=3` `x=-2`Sum of Roots:`alpha+beta` `=` `3+(-2)` `=` `1` Product of Roots:`alphabeta` `=` `3(-2)` `=` `-6` Substitute values onto the Standard Form with Sum & Product of Roots`x^2+(alpha+beta)x+alphabeta` `=` `0` `x^2+(``1``)x+(``-6``)` `=` `0` `x^2+x-6` `=` `0` `x^2-x-6=0` -
Question 2 of 5
2. Question
Find `k` if one of the roots in the equation below is `4``x^2+2kx-8`- `k=` (-1)
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Sum of Roots `alpha` and `beta`
$$\alpha+\beta=-\frac{\color{#9a00c7}{b}}{\color{#00880A}{a}}$$Product of Roots `alpha` and `beta`
$$\alpha\beta=\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$First, list the coefficients of the quadratic equation individually`x^2+2kx-8``a=1` `b=2k` `c=-8`Slot the coefficients to the Sum and Product of Roots Formula to form a system of equationsSum of Roots:`alpha+beta` `=` $$-\frac{\color{#9a00c7}{b}}{\color{#00880A}{a}}$$ `alpha+beta` `=` $$-\frac{\color{#9a00c7}{-2k}}{\color{#00880A}{1}}$$ `alpha+beta` `=` `-2k` Equation `1` Product of Roots:`alphabeta` `=` $$\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$ `alphabeta` `=` $$\frac{\color{#007DDC}{-8}}{\color{#00880A}{1}}$$ `alphabeta` `=` `-8` Equation `2` Remember that one of the roots of the equation is `4`. Label this as `alpha`Substitute `alpha` to Equation `2` and solve for `beta``alphabeta` `=` `-8` Equation `2` `4beta` `=` `-8` Substitute `alpha=4` `4beta``divide4` `=` `-8``divide4` Divide both sides by `4` `beta` `=` `-2` Finally, substitute `alpha` and `beta` to Equation `1` and solve for `m``alpha+beta` `=` `-2k` Equation `1` `4+(-2)` `=` `-2k` Substitute values `2` `=` `-2k` `2``divide(-2)` `=` `-2k``divide(-2)` Divide both sides by `-2` `-1` `=` `k` `k` `=` `-1` `k=-1` -
Question 3 of 5
3. Question
For which values of `m` will the roots of the equation be equal?`x^2–mx+4=0`- `m=` (-4, 4) and (4, -4)
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Sum of Roots `alpha` and `beta`
$$\alpha+\beta=-\frac{\color{#9a00c7}{b}}{\color{#00880A}{a}}$$Product of Roots `alpha` and `beta`
$$\alpha\beta=\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$First, list the coefficients of the quadratic equation individually`x^2-mx+4=0``a=1` `b=-m` `c=4`Since the roots are equal, we can have `beta=alpha`Slot the coefficients to the Sum and Product of Roots Formula to form a system of equationsSum of Roots:`alpha+alpha` `=` $$-\frac{\color{#9a00c7}{b}}{\color{#00880A}{a}}$$ `alpha+alpha` `=` $$-\frac{\color{#9a00c7}{-m}}{\color{#00880A}{1}}$$ `2alpha` `=` `m` `2alpha``divide2` `=` `m``divide2` Divide both sides by `2` `alpha` `=` `m/2` Equation `1` Product of Roots:`alpha*alpha` `=` $$\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$ `alpha*alpha` `=` $$\frac{\color{#007DDC}{4}}{\color{#00880A}{1}}$$ `sqrt(alpha^2)` `=` `sqrt4` Get the square root of both sides `alpha` `=` `+-2` Substitute `alpha=+-2` to Equation `1` and solve for `m``alpha` `=` `m/2` `2` `=` `m/2` Substitute `alpha=2` `2``times2` `=` `m/2``times2` Multiply both sides by `2` `4` `=` `m` `m` `=` `4` `alpha` `=` `m/2` `-2` `=` `m/2` Substitute `alpha=-2` `2``times2` `=` `m/2``times2` Multiply both sides by `2` `-4` `=` `m` `m` `=` `-4` `m=4,-4` -
Question 4 of 5
4. Question
For which values of `k` will the roots of the equation be one twice the other?`kx^2-9x+2k=0`- `k=` (-3, 3) and (3, -3)
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Sum of Roots `alpha` and `beta`
$$\alpha+\beta=-\frac{\color{#9a00c7}{b}}{\color{#00880A}{a}}$$Product of Roots `alpha` and `beta`
$$\alpha\beta=\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$First, list the coefficients of the quadratic equation individually`kx^2-9x+2k``a=k` `b=-9` `c=2k`Slot the coefficients to the Sum and Product of Roots Formula to form a system of equationsSum of Roots:`alpha+beta` `=` $$-\frac{\color{#9a00c7}{b}}{\color{#00880A}{a}}$$ `alpha+beta` `=` $$-\frac{\color{#9a00c7}{-9}}{\color{#00880A}{k}}$$ `alpha+beta` `=` `9/k` Equation `1` Product of Roots:`alphabeta` `=` $$\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$ `alphabeta` `=` $$\frac{\color{#007DDC}{2k}}{\color{#00880A}{k}}$$ `alphabeta` `=` `2` Equation `2` Remember that one of the roots of the equation is twice the other.Using Equation `2`, substitute `2alpha` to `beta` and solve for `alpha``alpha*beta` `=` `2` Equation `2` `alpha*2alpha` `=` `2` Substitute `beta=2alpha` `2alpha^2` `=` `2` `2alpha^2``divide2` `=` `2``divide2` Divide both sides by `2` `sqrt(alpha^2)` `=` `sqrt1` Get the square root of both sides `alpha` `=` `+-1` Next, substitute `2alpha` to `beta` on Equation `1``alpha+beta` `=` `9/k` Equation `1` `alpha+2alpha` `=` `9/k` Substitute `beta=2alpha` `3alpha``timesk` `=` `9/k``timesk` Multiply both sides by `k` `3alphak``divide3alpha` `=` `9``divide3alpha` Divide both sides by `3alpha` `k` `=` `9/(3alpha)` Finally, substitute the values of `alpha=+-1` and solve for `k``k` `=` `9/(3alpha)` `k` `=` `9/(3(1))` Substitute `alpha=1` `k` `=` `3` `k` `=` `9/(3alpha)` `k` `=` `9/(3(-1))` Substitute `alpha=-1` `k` `=` `-3` `k=3,-3` -
Question 5 of 5
5. Question
If the roots of `2x^2-3x-2=0` are `alpha` and `beta`, find:`6/alpha+6/beta`- (-9)
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Sum of Roots `alpha` and `beta`
$$\alpha+\beta=-\frac{\color{#9a00c7}{b}}{\color{#00880A}{a}}$$Product of Roots `alpha` and `beta`
$$\alpha\beta=\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$First, list the coefficients of the quadratic equation individually`2x^2-3x-2``a=2` `b=-3` `c=-2`Slot the coefficients to the Sum and Product of Roots FormulaSum of Roots:`alpha+beta` `=` $$-\frac{\color{#9a00c7}{b}}{\color{#00880A}{a}}$$ `=` $$-\frac{\color{#9a00c7}{-3}}{\color{#00880A}{2}}$$ `=` `3/2` Product of Roots:`alphabeta` `=` $$\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$ `=` $$\frac{\color{#007DDC}{-2}}{\color{#00880A}{2}}$$ `=` `-1` Manipulate the given expression until you can substitute the sum and product of roots.`6/alpha+6/beta` `=` `(6alpha+6beta)/(alphabeta)` `=` `(6(alpha+beta))/(alphabeta)` `=` `(6(3/2))/(-1)` `=` `-6(3/2)` `=` `-9` `6/alpha+6/beta=-9`
Quizzes
- Sum & Product of Roots 1
- Sum & Product of Roots 2
- Sum & Product of Roots 3
- Sum & Product of Roots 4
- Solving Equations by Factoring 1
- Solving Equations Using the Quadratic Formula
- Completing the Square 1
- Completing the Square 2
- Intro to Quadratic Functions (Parabolas) 1
- Intro to Quadratic Functions (Parabolas) 2
- Intro to Quadratic Functions (Parabolas) 3
- Graph Quadratic Functions in Standard Form 1
- Graph Quadratic Functions in Standard Form 2
- Graph Quadratic Functions by Completing the Square
- Graph Quadratic Functions in Vertex Form
- Write a Quadratic Equation from the Graph
- Write a Quadratic Equation Given the Vertex and Another Point
- Quadratic Inequalities 1
- Quadratic Inequalities 2
- Quadratics Word Problems 1
- Quadratics Word Problems 2
- Quadratic Identities
- Graphing Quadratics Using the Discriminant
- Positive and Negative Definite
- Applications of the Discriminant 1
- Applications of the Discriminant 2
- Solving Reducible Equations