Word Problem

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Sayuri

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Kiran drove from City A to City B, a distance of 270 mi. She increased her speed by 8 mi/h for the 374-mi trip from City B to City C. If the total trip took 10 h, what was her speed from City A to City B?

I got to 10r^2-564r+2160=0
and I'm not sure where to go from here
 
I agree with the equation.

Next, I'd use the quadratic formula. I don't think it will factor.
 
This is a prime candidate for the quadratic formula!

[MATH]10r^2-564r+2160=0[/MATH]​

If you ever have an equation in the form of [MATH]ax^2 + bx + c = 0[/MATH] and need to solve for [MATH]x[/MATH], which is the case here, the quadratic formula will get you where you need to go. It's kind of a scary-lookin' beast, but here it is:

[MATH]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/MATH]​

Note that [MATH]a \ne 0[/MATH], or else a division by zero will occur. Also note that if [MATH]b^2 < 4ac[/MATH], the figure inside the square root will be negative and you'll have to call for backup. Lastly, be mindful of that plus-or-minus [MATH]\pm[/MATH] sign, which indicates that you'll likely have two valid solutions for [MATH]x[/MATH]: one using the positive square root and one using the negative square root.

Plug in the numbers and turn the crank, and you're home free.


This is all just dandy if you happen to know the quadratic formula, but what if you don't? What if you don't have it memorized, but find yourself in need of it in the field? Being able to derive it manually is an important skill, and I'd like to invite everyone to give it a go (just to brush up on the process).
Start with the easy stuff:

[MATH]ax^2 + bx + c = 0[/MATH]
[MATH]ax^2 + bx = -c[/MATH]
[MATH]x^2 + \frac{b}{a}x = -\frac{c}{a}[/MATH]​

To consolidate the powers of [MATH]x[/MATH], we can complete the square and find what factor [MATH](x + ?)^2[/MATH] gives us the result we want. It'll give us an extra piece (which fills in a missing corner in the square), but it gives us the powers of [MATH]x[/MATH] that we want.

Let's use [MATH]x + \frac{b}{2a}[/MATH] as our factor. It expands into this:

[MATH]\left(x + \frac{b}{2a}\right)^2 \to x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}[/MATH]​

Since this introduces a [MATH]\frac{b^2}{4a^2}[/MATH] that wasn't there before, we need to remember to add it to the right-hand side in our equation:

[MATH]\left(x + \frac{b}{2a}\right)^2 = -\frac{c}{a} + \frac{b^2}{4a^2}[/MATH]​

Proceed with autopilot:

[MATH]\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{4ac}{4a^2}[/MATH]
[MATH]x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}} = \pm\left(\frac{1}{2a}\sqrt{b^2 - 4ac}\right) [/MATH]
[MATH]x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/MATH]​
 
Last edited:
@Sayuri: Of course, if you don't know the quadratic formula, you don't have to derive it in its entirety as Mr. Bland suggests.

The formula is derived by completing the square; and you can just do that directly with the numbers. (In that case, you'd probably leave it in fraction form, though there are other ways.)

That leaves open the question, what methods have you learned?
 
Sayuri said:
Kiran drove from City A to City B, a distance of 270 mi. She increased her speed by 8 mi/h for the 374-mi trip from City B to City C. If the total trip took 10 h, what was her speed from City A to City B?

I got to 10r^2-564r+2160=0
and I'm not sure where to go from here
Click to expand...

I would begin with (observing distances are in miles and time is in hours):

[MATH]270=rt\implies t=\frac{270}{r}[/MATH]
[MATH]374=(r+8)(10-t)=(r+8)\left(10-\frac{270}{r}\right)[/MATH]
Multiply by \(r\) knowing it must be positive:

[MATH]374r=(r+8)(10r-270)=10r^2-190r-2160[/MATH]
Arrange as:

[MATH]10r^2-564r-2160=0[/MATH]
It appears we disagree on the sign of the constant term.

Factor:

[MATH]2(r-60)(5r+18)=0[/MATH]
Discarding the negative root, we find:

[MATH]r=60[/MATH]
Let's check this solution. This solution implies:

[MATH]t=\frac{9}{2}[/MATH]
[MATH]60\cdot\frac{9}{2}=270[/MATH]
[MATH]68\cdot\frac{11}{2}=374[/MATH]
 
I had the right sign, but didn't notice it didn't match the OP. Then I copied a digit wrong and thought the discriminant implied it couldn't be factored.

Two wrongs don't always make a right.

But we still don't know why the OP was stuck -- not being able to factor it, or something else.
 
Dr.Peterson said:
I had the right sign, but didn't notice it didn't match the OP. Then I copied a digit wrong and thought the discriminant implied it couldn't be factored.

Two wrongs don't always make a right.

But we still don't know why the OP was stuck -- not being able to factor it, or something else.
Click to expand...
Yeah I just wasn't able to factor it, at least I couldn't figure it out. But I'll try the quadratic fromula!
 
Please divide the equation by 2 to get smaller numbers. Who knows, maybe you will see the factoring! If not, at least the numbers in the quadratic formula will be smaller.
 
Mr. Bland said:
This is a prime candidate for the quadratic formula!

[MATH]10r^2-564r+2160=0[/MATH]​

If you ever have an equation in the form of [MATH]ax^2 + bx + c = 0[/MATH] and need to solve for [MATH]x[/MATH], which is the case here, the quadratic formula will get you where you need to go. It's kind of a scary-lookin' beast, but here it is:

[MATH]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/MATH]​

Note that [MATH]a \ne 0[/MATH], or else a division by zero will occur. Also note that if [MATH]b^2 < 4ac[/MATH], the figure inside the square root will be negative and you'll have to call for backup. Lastly, be mindful of that plus-or-minus [MATH]\pm[/MATH] sign, which indicates that you'll likely have two valid solutions for [MATH]x[/MATH]: one using the positive square root and one using the negative square root.

Plug in the numbers and turn the crank, and you're home free.


This is all just dandy if you happen to know the quadratic formula, but what if you don't? What if you don't have it memorized, but find yourself in need of it in the field? Being able to derive it manually is an important skill, and I'd like to invite everyone to give it a go (just to brush up on the process).
Start with the easy stuff:

[MATH]ax^2 + bx + c = 0[/MATH]​
​
[MATH]ax^2 + bx = -c[/MATH]​
​
[MATH]x^2 + \frac{b}{a}x = -\frac{c}{a}[/MATH]​

To consolidate the powers of [MATH]x[/MATH], we can complete the square and find what factor [MATH](x + ?)^2[/MATH] gives us the result we want. It'll give us an extra piece (which fills in a missing corner in the square), but it gives us the powers of [MATH]x[/MATH] that we want.

Let's use [MATH]x + \frac{b}{2a}[/MATH] as our factor. It expands into this:

[MATH]\left(x + \frac{b}{2a}\right)^2 \to x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}[/MATH]​

Since this introduces a [MATH]\frac{b^2}{4a^2}[/MATH] that wasn't there before, we need to remember to add it to the prime numbers right-hand side in our equation:

[MATH]\left(x + \frac{b}{2a}\right)^2 = -\frac{c}{a} + \frac{b^2}{4a^2}[/MATH]​

Proceed with autopilot:

[MATH]\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{4ac}{4a^2}[/MATH]​
​
[MATH]x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}} = \pm\left(\frac{1}{2a}\sqrt{b^2 - 4ac}\right) [/MATH]​
​
[MATH]x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/MATH]​
Click to expand...

I just registered on this forum to say thank you because this formula helped me a lot :)
 
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