Solving for x

Denzel

New member
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

Super Moderator
Staff member
Hello, and welcome to FMH! Are you given any information regarding the width of the rectangle?

Last edited:

Denzel

New member
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

Super Moderator
Staff member
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

New member
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

MarkFL

Super Moderator
Staff member
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

New member
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

MarkFL

Super Moderator
Staff member
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

Denzel

New member
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

MarkFL

Super Moderator
Staff member
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:

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

• Denzel

Jomo

Elite Member
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

New member
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

Senior Member
… 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)

• topsquark and Denzel

Otis

Senior Member
-b±√b2−4ac/2a

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

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 • Denzel