I am finding the sum and product of roots of some quadratic equations. I picked two in particular and was trying to understand the reality of what I was doing but some how I got stucked.
Here are the two equations:
[math]1.~x^2+3x+7=0 \\ 2.~x^2+3x-10=0[/math]I used the formula [math]u+v = - \frac{b}{a}[/math] for sum of roots.
And [math]uv=\frac{c}{a}[/math] for product of roots.
Let's take a look at #2. [math]x^2+3x-10=0[/math]a=1, b=3, c =- 10.
The sum of the roots is [math]u+v = - \frac{b}{a} \\ \longrightarrow - \frac{3}{1}=-3[/math]The product of the roots is
[math]uv= \frac{c}{a} \\ \longrightarrow \frac{-10}{1} =-10[/math]Now one if decides to solve the equation
[math]x^2+3x-10=0[/math] using any of the method (factorization, completing the square method or quadratic formula), the solution would be [math]x=-5~~ \text{or }x=2[/math]The sum of the roots in reality is[math]-5+2=-3[/math]While the product of the roots is[math]-5(2)=-10[/math]From what I have done so far with the equation [math]x^2+3x-10=0[/math], it can be seen that if the formulae for finding sum and product of roots- [math]u+v = - \frac{b}{a}[/math] and [math]uv= \frac{c}{a}[/math] the same solution as [math]-3~~\text{and} -10[/math] would be obtained when the are actual roots of the equation is used to experiment it; that verifies that what was done is true in both cases.
Now let's test it in equation #1
[math]x^2+3x+7=0[/math]If the formula method is applied in finding the sum and product of roots of the equation, knowing fully well that a=1, b=3 and c=7, then the sum of the roots
[math]u+v = - \frac{b}{a}=-\frac{3}{1}= -3[/math] and the product of roots [math]uv=\frac{c}{a} = \frac{7}{1}=7[/math]When it comes to checking it with the actual roots of the equation.
This equation [math]x^2+3x+7=0[/math] can only be solved using either completing the square method or quadratic formula but not by factorization. The roots obtained when the equation is solved are imaginary roots which are [math]\frac{-3+\sqrt{-19} }{2}[/math] and [math]\frac{-3-\sqrt{-19} }{2}[/math]The sum of the roots is [math]\frac{-3+\sqrt{-19} }{2}+\frac{-3-\sqrt{-19}}{2}[/math] [math]=\frac{-3\cancel{+\sqrt{-19}}-3\cancel{-\sqrt{-19}} }{2}[/math] [math]=\frac{-3-3}{2} = \frac{-6}{2}=-3[/math]The product of the roots is
[math]\left(\frac{-3+\sqrt{-19} }{2}\right)×\left(\frac{-3-\sqrt{-19}}{2}\right)[/math] [math]=\frac{-3(-3-\sqrt{-19})+\sqrt{-19}(-3-\sqrt{-19})}{4}[/math] [math]=\frac{3\cancel{+3\sqrt{-19} }\cancel{ -3\sqrt{-19} }+19}{4}[/math] [math]=\frac{3+19}{4}=\frac{\cancel{22}^{11}}{\cancel{4}_2}=7\frac{1}{2}=7.5[/math]Where I am concerned now is why is the product of the roots different in the case of equation #1 when checking with product of the roots obtained by solving?
In case somebody don't understand my question, I am asking why is the product of the roots of the equation [math]x^2+3x+7=0[/math] obtained by using the formula [math]uv=\frac{c}{a} = \frac{7}{1}=7[/math] and that obtained by multiplying the actual imaginary roots different. As you can see the product obtained by using formula is 7 while the product obtained by multiplying the actual imaginary roots is 7.5. Why is it so? Why are they not the same as in equation #2 [math]x^2+3x-10=0[/math]? Or did I make mistake in my evaluation of the product of the imaginary roots?
Finally before I forget, I would want add this question- what is the signficance of finding the sum and product of roots of quadratic equations? Is it really worth finding? What do we stand to gain in doing so?Thank you.
Here are the two equations:
[math]1.~x^2+3x+7=0 \\ 2.~x^2+3x-10=0[/math]I used the formula [math]u+v = - \frac{b}{a}[/math] for sum of roots.
And [math]uv=\frac{c}{a}[/math] for product of roots.
Let's take a look at #2. [math]x^2+3x-10=0[/math]a=1, b=3, c =- 10.
The sum of the roots is [math]u+v = - \frac{b}{a} \\ \longrightarrow - \frac{3}{1}=-3[/math]The product of the roots is
[math]uv= \frac{c}{a} \\ \longrightarrow \frac{-10}{1} =-10[/math]Now one if decides to solve the equation
[math]x^2+3x-10=0[/math] using any of the method (factorization, completing the square method or quadratic formula), the solution would be [math]x=-5~~ \text{or }x=2[/math]The sum of the roots in reality is[math]-5+2=-3[/math]While the product of the roots is[math]-5(2)=-10[/math]From what I have done so far with the equation [math]x^2+3x-10=0[/math], it can be seen that if the formulae for finding sum and product of roots- [math]u+v = - \frac{b}{a}[/math] and [math]uv= \frac{c}{a}[/math] the same solution as [math]-3~~\text{and} -10[/math] would be obtained when the are actual roots of the equation is used to experiment it; that verifies that what was done is true in both cases.
Now let's test it in equation #1
[math]x^2+3x+7=0[/math]If the formula method is applied in finding the sum and product of roots of the equation, knowing fully well that a=1, b=3 and c=7, then the sum of the roots
[math]u+v = - \frac{b}{a}=-\frac{3}{1}= -3[/math] and the product of roots [math]uv=\frac{c}{a} = \frac{7}{1}=7[/math]When it comes to checking it with the actual roots of the equation.
This equation [math]x^2+3x+7=0[/math] can only be solved using either completing the square method or quadratic formula but not by factorization. The roots obtained when the equation is solved are imaginary roots which are [math]\frac{-3+\sqrt{-19} }{2}[/math] and [math]\frac{-3-\sqrt{-19} }{2}[/math]The sum of the roots is [math]\frac{-3+\sqrt{-19} }{2}+\frac{-3-\sqrt{-19}}{2}[/math] [math]=\frac{-3\cancel{+\sqrt{-19}}-3\cancel{-\sqrt{-19}} }{2}[/math] [math]=\frac{-3-3}{2} = \frac{-6}{2}=-3[/math]The product of the roots is
[math]\left(\frac{-3+\sqrt{-19} }{2}\right)×\left(\frac{-3-\sqrt{-19}}{2}\right)[/math] [math]=\frac{-3(-3-\sqrt{-19})+\sqrt{-19}(-3-\sqrt{-19})}{4}[/math] [math]=\frac{3\cancel{+3\sqrt{-19} }\cancel{ -3\sqrt{-19} }+19}{4}[/math] [math]=\frac{3+19}{4}=\frac{\cancel{22}^{11}}{\cancel{4}_2}=7\frac{1}{2}=7.5[/math]Where I am concerned now is why is the product of the roots different in the case of equation #1 when checking with product of the roots obtained by solving?
In case somebody don't understand my question, I am asking why is the product of the roots of the equation [math]x^2+3x+7=0[/math] obtained by using the formula [math]uv=\frac{c}{a} = \frac{7}{1}=7[/math] and that obtained by multiplying the actual imaginary roots different. As you can see the product obtained by using formula is 7 while the product obtained by multiplying the actual imaginary roots is 7.5. Why is it so? Why are they not the same as in equation #2 [math]x^2+3x-10=0[/math]? Or did I make mistake in my evaluation of the product of the imaginary roots?
Finally before I forget, I would want add this question- what is the signficance of finding the sum and product of roots of quadratic equations? Is it really worth finding? What do we stand to gain in doing so?Thank you.
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