Equation with one variable but two outcomes
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D
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Do you know how to solve a quadratic equation like 2x2 + 3x - 2 = 0?
Please show us what you have tried and exactly where you are stuck.
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Please share your work/thoughts about this assignment.
I don't know how to solve a quadratic equation.
That's where I'm stuck at, I don't know what to do with the 2x², I mean you can root it, but then i would have to root 3x + 2 = 0 too right?
Which for me right now makes no sense...
All I have done yet was to put in possible numbers for X and then calculate the equation with my guessed numbers.
Thank you for your response and help!
D
Deleted member 4993
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Quadratic equations are generally taught at intermediate level algebra.
What class/course asked you to solve this problem? Intermediate algebra should be pre-req for that class.
However, you can solve this problem by graphing. Graph the function
y = 2x2 + 3x - 2
It will be a parabola.
The x-intercepts of the graph are the solutions you are seeking.
I'm currently studying for myself only with a math book called "Mathematics for Engineers and Scientists" (original Title in GER; Mathematik für Ingenieure und Naturwissenschaftler) which says "A Learnbook for basic studies"
I'm from a small European country so there are no "courses" here, but I thank you for your response so I can look into parabola and alike, its also in the book but some pages further behind.
I guess my basic understanding of math isn't big enough yet, so I'll just leave it aside for now look at it later on.
Big thanks and have a nice weekend
HallsofIvy
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I would start by factoring the "2" out: \(\displaystyle 2(x^2+ (3/2)x)= 2\).
Now, "complete the square":
If you are trying to solve an equation like this you should know enough algebra to know that \(\displaystyle (x+ a)^2= (x+ a)(x+ a)= x(x+ a)+ a(x+ a)= x^2+ ax+ ax+ a^2= x^2+ 2ax+ a^2\).
Compare "\(\displaystyle x^2+ (3/2)x\)" to "\(\displaystyle x^2+ 2ax+ a^2\). It is clearly missing the "\(\displaystyle a^2\)". To "complete the square" we need to add it.
But what is "a"? Compare "(3/2)x" with "2ax". We must have 2a= 3/2 or a= 3/4. Then \(\displaystyle a^2= 9/16\). We have to add 9/16 to make this a "perfect square".
But we can't just add 9/16. That would change the value of the expression. We want to change the way it is written without changing its value. We can do that by both adding and subtracting 9/16:
\(\displaystyle 2(x^2+ (3/2)x+ 9/16- 9/16)= 2(x^2+ (3/2)x + 9/16)- 9/8= 2(x+ 3/4)^2- 9/8= 2\).
Add 9/8 to both sides: \(\displaystyle 2(x+ 3/4)^2= 2+ 9/8= 16/9+ 9/8= 25/8\).
Divide both sides by 2: \(\displaystyle (x+ 3/4)^2= 25/16\).
Take the square root of both sides: \(\displaystyle x+ 3/4= \pm 5/4\).
Finally, subtract 3/4from both sides: \(\displaystyle x= \pm 5/4- 3/4\)
Taking the "+" \(\displaystyle x= 5/4- 3/4= 2/4= 1/2\).
Taking the "-" \(\displaystyle x= -5/4- 3/4= -8/4= -2\).
The two solutions to the equation are 1/2 and -2.
Of course we have to check: If x= 1/2 then \(\displaystyle x^2= 1/4\) so that \(\displaystyle 2x^2+ 3x= 2(1/4)+ 3/2= 1/2+ 3/2= 2\). Yes, that checks.
If x= -2 then \(\displaystyle x^2= 4\) so that \(\displaystyle 2x^2+ 3x= 2(4)+ 3(-2)= 8- 6= 2\). Yes, that also checks.
The two values of x that satisfy \(\displaystyle 2x^2+ 3x= 2\) are \(\displaystyle x= \frac{1}{2}\) and \(\displaystyle x= -2\).
Now, "complete the square":
If you are trying to solve an equation like this you should know enough algebra to know that \(\displaystyle (x+ a)^2= (x+ a)(x+ a)= x(x+ a)+ a(x+ a)= x^2+ ax+ ax+ a^2= x^2+ 2ax+ a^2\).
Compare "\(\displaystyle x^2+ (3/2)x\)" to "\(\displaystyle x^2+ 2ax+ a^2\). It is clearly missing the "\(\displaystyle a^2\)". To "complete the square" we need to add it.
But what is "a"? Compare "(3/2)x" with "2ax". We must have 2a= 3/2 or a= 3/4. Then \(\displaystyle a^2= 9/16\). We have to add 9/16 to make this a "perfect square".
But we can't just add 9/16. That would change the value of the expression. We want to change the way it is written without changing its value. We can do that by both adding and subtracting 9/16:
\(\displaystyle 2(x^2+ (3/2)x+ 9/16- 9/16)= 2(x^2+ (3/2)x + 9/16)- 9/8= 2(x+ 3/4)^2- 9/8= 2\).
Add 9/8 to both sides: \(\displaystyle 2(x+ 3/4)^2= 2+ 9/8= 16/9+ 9/8= 25/8\).
Divide both sides by 2: \(\displaystyle (x+ 3/4)^2= 25/16\).
Take the square root of both sides: \(\displaystyle x+ 3/4= \pm 5/4\).
Finally, subtract 3/4from both sides: \(\displaystyle x= \pm 5/4- 3/4\)
Taking the "+" \(\displaystyle x= 5/4- 3/4= 2/4= 1/2\).
Taking the "-" \(\displaystyle x= -5/4- 3/4= -8/4= -2\).
The two solutions to the equation are 1/2 and -2.
Of course we have to check: If x= 1/2 then \(\displaystyle x^2= 1/4\) so that \(\displaystyle 2x^2+ 3x= 2(1/4)+ 3/2= 1/2+ 3/2= 2\). Yes, that checks.
If x= -2 then \(\displaystyle x^2= 4\) so that \(\displaystyle 2x^2+ 3x= 2(4)+ 3(-2)= 8- 6= 2\). Yes, that also checks.
The two values of x that satisfy \(\displaystyle 2x^2+ 3x= 2\) are \(\displaystyle x= \frac{1}{2}\) and \(\displaystyle x= -2\).
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