# 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?

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#### 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

##### 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

##### 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

##### 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

##### 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:

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

#### 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)

#### 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