Roots of the parabola

[rachelmaddie]

I want to make sure my work is correct.

Find the roots of the parabola given by the following equation: 2x^2 + 5x - 9 = 2x

The formula to solve a quadratic equation of the form ax^2 + bx + c = 0
is equal to x = -b(+ / -) (b^2 - 4ac)/2a

2x^2 + 5x - 2x - 9 = 0
2x^2 + 3x - 9 = 0
where a = 2, b = 3, c = -9

Substitute in the formula
x = -3(+ / -) (3^2 - 4(2)(-9)/2(2)
x = -3(+ / -)sqrt81/4
x = -3(+ / -)9/4
x = -3(+)9/4 = 1.5
x = -3(-)9/4 = -3


Solution: The roots are x = -3 or x = 1.5

 

[topsquark]

Well, the way you wrote the quadratic equation needs some work, but as -3 and 3/2 both solve the original equation it looks like you did good. (You can always do that check by hand.)

Just a note: Since this is a Mathematics problem I'd use 3/2 instead of 1.5.

And
x = ( -b +/- sqrt( b^2 - 4ac ) )/ (2a)

-Dan

 

[rachelmaddie]

Well, the way you wrote the quadratic equation needs some work, but as -3 and 3/2 both solve the original equation it looks like you did good. (You can always do that check by hand.)

Just a note: Since this is a Mathematics problem I'd use 3/2 instead of 1.5.

And
x = ( -b +/- sqrt( b^2 - 4ac ) )/ (2a)

-Dan

Where does the quadratic equation need work?

 

[tkhunny]

x = -3(+ / -) (3^2 - 4(2)(-9)/2(2)

Let's see...
1) It needs parentheses to separate numerator and denominator properly.
x = [-3(+ / -) (3^2 - 4(2)(-9)]/[2(2)]
2) Your square root seems to be missing.
x = [-3(+ / -) sqrt((3^2 - 4(2)(-9))]/[2(2)]
Be MUCH more careful. Write something that might mean what you intend.

 

[rachelmaddie]

x = -3(+ / -) (3^2 - 4(2)(-9)/2(2)

Let's see...
1) It needs parentheses to separate numerator and denominator properly.
x = [-3(+ / -) (3^2 - 4(2)(-9)]/[2(2)]
2) Your square root seems to be missing.
x = [-3(+ / -) sqrt((3^2 - 4(2)(-9))]/[2(2)]
Be MUCH more careful. Write something that might mean what you intend.

Find the roots of the parabola given by the following equation: 2x^2 + 5x - 9 = 2x

The formula to solve a quadratic equation of the form ax^2 + bx + c = 0
is equal to x = (-b + / - sqrt(b^2 - 4ac ) ) (2a)

2x^2 + 5x - 2x - 9 = 0
2x^2 + 3x - 9 = 0
where a = 2, b = 3, c = -9

Substitute in the formula
x = [-3(+ / -) sqrt((3^2 - 4(2)(-9))]/[2(2)]
x = -3(+ / -)sqrt81/4
x = -3(+ / -)9/4
x = -3(+)9/4 = 1.5
x = -3(-)9/4 = -3


Solution: The roots are x = -3 or x = 3/2

Is this better?

 

[tkhunny]

Substitute in the formula
x = [-3(+ / -) sqrt((3^2 - 4(2)(-9))]/[2(2)]
x = -3(+ / -)sqrt81/4

Started out better. Then not so good.

x = [-3(+ / -)sqrt(81)]/4 -- Why did so many of the grouping symbols just vanish?

 

[amathproblemthatneedsolve]

How I'd do it:

[MATH]x=\frac{-3(+/-)\sqrt{3^2-4(2)(-9)}}{2(2)}[/MATH]
[MATH]x=\frac{-3(+/-)\sqrt{81}}{4}[/MATH]
[MATH]x=\frac{-3(+/-)9}{4}[/MATH]
[MATH]x=\frac{-3(+)9}{4}=3/2[/MATH]
[MATH]x=\frac{-3(-)9}{4}=-3[/MATH]
Solution: The roots are x = -3 or x = 3/2

 

[Otis]

… $-3$$($$+$$)$$9$
[OR]
$-3$$($$-$$)$$9$ …

Hi amptns. Those grouping symbols are unnecessary. (I don't know whether Rachel's interested in learning LaTeX.)

TIP: The code for symbol ± is \pm

Using text, I'd type:

x = [-b ± √(b^2 - 4ac)] / [2a]

x = [-3 ± √(3^2 - (4)(2)(-9))] / [2(2)]

?

 

[rachelmaddie]

Started out better. Then not so good.

x = [-3(+ / -)sqrt(81)]/4 -- Why did so many of the grouping symbols just vanish?

Can you help me with the grouping symbols for this whole portion?
Substitute in the formula
x = 3(+ / -) (3^2 - 4(2)(-9)/2(2)
x = -3(+ / -)sqrt81/4
x = -3(+ / -)9/4
x = -3(+)9/4 = 1.5
x = -3(-)9/4 = -3

 

[Dr.Peterson]

Can you help me with the grouping symbols for this whole portion?
Substitute in the formula
x = 3(+ / -) (3^2 - 4(2)(-9)/2(2)
x = -3(+ / -)sqrt81/4
x = -3(+ / -)9/4
x = -3(+)9/4 = 1.5
x = -3(-)9/4 = -3

You never put grouping symbols around an operation. You wrote (+/-) presumably as an easy substitute for the proper symbol, [MATH]\pm[/MATH]. The (+) and (-) are just bad practice, perhaps thought of as putting a circle around the operator to emphasize it. Don't do any of this in writing, and do it in typing only if you have no alternative.

But you must put parentheses around the numerator and denominator of a fraction (division) in which either contains more than one number or variable. I often use brackets for the sake of contrast.

So, assuming I couldn't write the proper plus-or-minus symbol, and didn't want to use LaTeX formatting, I'd type your work here as:

x = [-3 +/- sqrt(3^2 - 4(2)(-9))]/[2(2)]​

x = [-3 +/- sqrt(81)]/4​

x = [-3 +/- 9]/4​

x = [-3 + 9]/4 = 1.5​

or [-3 - 9]/4 = -3​

I made a couple corrections other than parentheses/brackets there!