Solving for x

Denzel

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The length of a rectangle is (2x -1) and width (x+3), given that the area is 294cm^2 , determine the value of x. I've attempted using
x = [−b ± √(b^2 − 4ac)]/(2a) , The answer is supposed to be x=11.
 
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MarkFL

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Hello, and welcome to FMH! :)

Are you given any information regarding the width of the rectangle?
 

JeffM

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Last edited:

Denzel

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The length of a rectangle is (2x -1) and width (x+3), given that the area is 294cm^2 , determine the value of x. I've attempted using
x=−b±√b2−4ac/2a, The answer is supposed to be x=11.
 

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MarkFL

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Okay, so the area is the product of the width and the height:

\(\displaystyle A(x)=(2x-1)(x+3)\)

Can you expand that to get this in the form required for part a) of the question?
 

Denzel

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Okay, so the area is the product of the width and the height:

\(\displaystyle A(x)=(2x-1)(x+3)\)

Can you expand that to get this in the form required for part a) of the question?
2x^2 +5x-3
 

MarkFL

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2x^2 +5x-3
Yes, so now equate this to 294, and then solve for \(x\):

\(\displaystyle 2x^2+5x-3=294\)

\(\displaystyle 2x^2+5x-297=0\)

Can you now factor?
 

Denzel

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Yes, so now equate this to 294, and then solve for \(x\):

\(\displaystyle 2x^2+5x-3=294\)

\(\displaystyle 2x^2+5x-297=0\)

Can you now factor?
Okay. I tried..I don't know if I'm going correct( if it's even the right way...) or how to continue.
 

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MarkFL

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Okay, if you're using the quadratic formula, then you want:

\(\displaystyle x=\frac{-5\pm\sqrt{2401}}{4}\)

Now, 2401 is a perfect square. We should be able to easily see that:

\(\displaystyle 50^2=2500\)

Hence:

\(\displaystyle 49^2=(50-1)^2=50^2-2\cdot50+1=2500-99=2401\)

And so, we would find (after discarding the negative root since it will lead to negative measures of length):

\(\displaystyle x=\frac{-5+49}{4}=\,?\)
 

Denzel

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Okay, if you're using the quadratic formula, then you want:

\(\displaystyle x=\frac{-5\pm\sqrt{2401}}{4}\)

Now, 2401 is a perfect square. We should be able to easily see that:

\(\displaystyle 50^2=2500\)

Hence:

\(\displaystyle 49^2=(50-1)^2=50^2-2\cdot50+1=2500-99=2401\)

And so, we would find (after discarding the negative root since it will lead to negative measures of length):

\(\displaystyle x=\frac{-5+49}{4}=\,?\)
x=11 THANK You. It always seems easy after I understand it... Then I ask, "Why did I not understand in the first place?"
 

MarkFL

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Yes, let's try factoring:

\(\displaystyle 2x^2+5x-297=0\)

We want two factors of 2(-297)=-594 whose sum is 5, or equivalently, we want two factors of 594 whose difference is 5. We find:

\(\displaystyle 594=2\cdot3^3\cdot11=22\cdot27\)

These are the factors we want, and since 22 is divisible by 2, we then find:

\(\displaystyle 2x^2+5x-297=(2x+27)(x-11)=0\)
 

Jomo

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Please note that you listed the quadratic formula incorrectly. Maybe that is why you were getting the wrong answer. Do you see why your formula is wrong?
 

Denzel

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Please note that you listed the quadratic formula incorrectly. Maybe that is why you were getting the wrong answer. Do you see why your formula is wrong?
Oh. I see the error, ( just fixed it ) but no, I had it as b^2 in my book, only made a typo on here, I just didn't understand. ^-^
 

lookagain

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… using
x=−b±√b^2−4ac/2a, ...
You edited that thread post, but that formula, written out in horizontal style, is wrong in at least three places:

1) There are grouping symbols missing around the radicand.
2) There are grouping symbols missing around the denominator.
3) There are grouping symbols missing around the numerator.

Look at this:

x = [−b ± √(b^2 − 4ac)]/(2a)
 

Otis

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-b±√b2−4ac/2a
Adding to lookagain's comments: Yes -- without the grouping symbols, someone unfamiliar with the formula might read your typing as:

\(\displaystyle -b \pm \sqrt{b^2} - \frac{4ac}{2a}\)

It takes time to learn how to type math expressions, using a keyboard, but the basic rule for radicals and ratios containing more than a single number/symbol is to enclose them in grouping symbols.

💡 In the forum guidelines, there's a link for Formatting Math as Text.



I'm more concerned about how you wrote the quadratic formula in your work. You need to divide everything by 4. You also wrote b instead of -b. That is, write:

\(\displaystyle \frac{-5 \pm \sqrt{2401}}{4} \quad \text{ instead of } \quad 5 \pm \frac{\sqrt{2401}}{4}\)

Cheers

😎
 
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