Sum & Product of Roots 1
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Question 1 of 6
1. Question
Identify the sum and product of roots of `ax^2+bx+c=0`, given that:`x=alpha``x=beta`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`
$$\alpha\beta=\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$First, form a quadratic equation using the given roots`x=alpha` `x=beta``(x-alpha)(x-beta)` `=` `0` Solving this will give out the roots above `x^2-betax-alphax+alphabeta` `=` `0` Expand `x^2-(beta+alpha)x+alphabeta` `=` `0` `x^2-(alpha+beta)x+alphabeta` `=` `0` Since the equation formed previously comes from the roots or solution of `ax^2+bx+c=0`, they are considered as identities`x^2-(alpha+beta)x+alphabeta` `≡` `ax^2+bx+c` `x^2-(alpha+beta)x+alphabeta` `≡` `(ax^2)/a+(bx)/a+c/a` Divide all terms on the right side by `a` `x^2``-(alpha+beta)``x+``alphabeta` `≡` $$x^2+\color{#00880A}{\frac{b}{a}}x+\color{#9a00c7}{\frac{c}{a}}$$ Finally, combine equivalent coefficientsCoefficients of `x`:`-(alpha+beta)` `=` `b/a` `alpha+beta` `=` `(-b)/a` This is the sum of rootsConstants:`alphabeta` `=` `c/a` This is the product of roots`alpha+beta=(-b)/a``alphabeta=c/a` -
Question 2 of 6
2. Question
Find `m` if one of the roots in the equation below is `-2``2x^2-4x+m-1`- `m=` (-15)
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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^2-4x+m-1``a=2` `b=-4` `c=m-1`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` `=` $$-\frac{\color{#9a00c7}{-4}}{\color{#00880A}{2}}$$ `alpha+beta` `=` `2` Equation `1` Product of Roots:`alphabeta` `=` $$\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$ `alphabeta` `=` $$\frac{\color{#007DDC}{m-1}}{\color{#00880A}{2}}$$ Equation `2` Remember that one of the roots of the equation is `-2`. Label this as `alpha`Substitute `alpha` to Equation `1` and solve for `beta``alpha+beta` `=` `2` Equation `1` `-2+beta` `=` `2` Substitute `alpha=-2` `-2+beta` `+2` `=` `2` `+2` Add `2` to both sides `beta` `=` `4` Finally, substitute `alpha` and `beta` to Equation `2` and solve for `m``alphabeta` `=` `(m-1)/2` Equation `2` `(-2)(4)` `=` `(m-1)/2` Substitute values `-8` `=` `(m-1)/2` `-8``times2` `=` `(m-1)/2``times2` Multiply `2` to both sides `-16` `=` `m-1` `-16` `+1` `=` `m-1` `+1` Add `1` to both sides `-15` `=` `m` `m` `=` `-15` `m=-15` -
Question 3 of 6
3. Question
Find `m` if one of the roots in the equation below is `-2``mx^2-7x+m+6=0`- `m=` (-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`
$$\alpha\beta=\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$You can directly substitute the known root `-2` as a value of `x` and solve for `m``mx^2-7x+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``divide5` `=` `-20``divide5` Divide both sides by `5` `m` `=` `-4` `m=-4` -
Question 4 of 6
4. Question
For which values of `m` will the equation have real roots?`x^2-mx+4=0`Hint
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Nature of the Roots Discriminant (`Delta`) Two real roots `Delta``>``0` One real root `Delta=0` No real roots `Delta``<``0` Discriminant Formula
$$\Delta={\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$$First, compute for the discriminant`x^2-mx+4=0``a=1` `b=-m` `c=4``Delta` `=` $${\color{#9a00c7}{b}}^2-4\color{#00880A}{a}\color{#007DDC}{c}$$ Discriminant Formula `=` $${\color{#9a00c7}{(-m)}}^2-4\color{#00880A}{(1)}\color{#007DDC}{(4)}$$ Substitute values `=` `m^2-16` A function that has real roots can have either one or two real roots, hence `Delta``≥``0`Substitute the `Delta` computed previously, and then solve for `k``Delta` `≥` `0` `m^2-16` `≥` `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 `m^2-16`Replace the `x` axis with `m` axis and draw an upward parabola since `1` is positive
Remember that `Delta` must be positiveTherefore, `m``≤``-4` and `m``≥``4``m``≤``-4` and `m``≥``4` -
Question 5 of 6
5. Question
For which values of `k` will the roots of the equation be one double the other?`x^2+2x-6k=0`Write fractions in the format “a/b”- `k=` (-4/27)
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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+2x-6k=0``a=1` `b=2` `c=-6k`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` `=` $$-\frac{\color{#9a00c7}{2}}{\color{#00880A}{1}}$$ `alpha+beta` `=` `-2` Equation `1` Product of Roots:`alphabeta` `=` $$\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$ `alphabeta` `=` $$\frac{\color{#007DDC}{-6k}}{\color{#00880A}{1}}$$ `alphabeta` `=` `-6k` Equation `2` Remember that one of the roots of the equation is double the other.Using Equation `1`, substitute `2alpha` to `beta` and solve for `alpha``alpha+beta` `=` `-2` Equation `1` `alpha+2alpha` `=` `-2` Substitute `beta=2alpha` `3alpha` `=` `-2` `3alpha``divide3` `=` `-2``divide3` Divide both sides by `3` `alpha` `=` `-2/3` Next,substitute `2alpha` to `beta` on Equation `2``alpha*beta` `=` `-6k` Equation `1` `alpha*2alpha` `=` `-6k` Substitute `beta=2alpha` `2alpha^2``divide-6` `=` `-6k``divide-6` Multiply both sides by `-6` `(alpha^2)/(-3)` `=` `k` `k` `=` `(alpha^2)/(-3)` Finally, substitute `alpha=-2/3` and solve for `k``k` `=` `(alpha^2)/(-3)` `k` `=` `(((-2)/3)^2)/(-3)` Substitute `alpha=-2/3` `k` `=` `(4/9)/(-3)` `k` `=` `-4/27` `k=-4/27` -
Question 6 of 6
6. Question
Find `p` if the roots in the equation below are reciprocals`(p-2)x^2+40x+2p-1=0`- `p=` (-1)
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`
$$\alpha\beta=\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$First, list the coefficients of the quadratic equation individually`(p-2)x^2+40x+2p-1=0``a=p-2` `b=40` `c=2p-1`Slot the coefficients to the Product of Roots Formula to form a system of equations`alphabeta` `=` $$\frac{\color{#007DDC}{c}}{\color{#00880A}{a}}$$ `alphabeta` `=` $$\frac{\color{#007DDC}{2p-1}}{\color{#00880A}{p-2}}$$ Recall that the roots of the equation are reciprocal.Substitute `1/alpha` to `beta` and solve for `p``alphabeta` `=` `(2p-1)/(p-2)` `alpha*(1/alpha)` `=` `(2p-1)/(p-2)` Substitute `beta=1/alpha` `1` `=` `(2p-1)/(p-2)` `1``times(p-2)` `=` `(2p-1)/(p-2)``times(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` `p=-1`
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