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.
2x^2 +5x-3Okay, 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?
Yes, so now equate this to 294, and then solve for \(x\):2x^2 +5x-3
Okay. I tried..I don't know if I'm going correct( if it's even the right way...) or how to continue.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?
x=11 THANK You. It always seems easy after I understand it... Then I ask, "Why did I not understand in the first place?"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):
\(\displaystyle x=\frac{-5+49}{4}=\,?\)
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. ^-^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?
You edited that thread post, but that formula, written out in horizontal style, is wrong in at least three places:… using
x=−b±√b^2−4ac/2a, ...
Adding to lookagain's comments: Yes -- without the grouping symbols, someone unfamiliar with the formula might read your typing as:-b±√b2−4ac/2a