# Quadratic Equations Using a Formula

• Jan 1st 2010, 09:53 AM
ExponentialP
Hi Happy New Year everyone hope you all have had a good one........ I'm doing Maths on New Years Day (Nerd)..... I'm having a bit of trouble with quadratic equations using a formula. I am using the formula to solve the equations but when I substitute the answers back in the equation the answers don't work....

This is the equation :
4²-3x-2=0

I use the formula :
x = -b ± √ b² - 4ac
_____________
2 a

I substitute the letters for the numbers in the formula so that it looks like

x = - - 3 ± √ -3² - 4 x 4 x -2
_____________________
2 x 4

x = 3 + 4.80
________
8

or

x = 3 - 4.80
_________
8

Can someone please tell me where I'm going wrong

I am stating roots correct to 2 dp
• Jan 1st 2010, 10:00 AM
Roam
Your a=4, b=-3 and c=-2. So, $\displaystyle \sqrt{b^2 -4ac} = \sqrt{41} = 6.4$. The answer has to be $\displaystyle \frac{3 \pm \sqrt{41}}{8}$.
• Jan 1st 2010, 10:02 AM
earboth
Quote:

Originally Posted by ExponentialP
Hi Happy New Year thank you
everyone hope you all have had a good one........ I'm doing Maths on New Years Day (Nerd)..... I'm having a bit of trouble with quadratic equations using a formula. I am using the formula to solve the equations but when I substitute the answers back in the equation the answers don't work....

This is the equation :
4²-3x-2=0

I use the formula :
x = -b ± √ b² - 4ac
_____________
2 a

I substitute the letters for the numbers in the formula so that it looks like

x = - - 3 ± √( -3)² - 4 x 4 x -2
_____________________
2 x 4

x = 3 + 4.80
________
8

or

x = 3 - 4.80
_________
8

Can someone please tell me where I'm going wrong

I am stating roots correct to 2 dp

See corrections.
• Jan 1st 2010, 10:03 AM
ExponentialP
Yes thats what I done, but my answers still don't add up
• Jan 1st 2010, 10:09 AM
earboth
Quote:

Originally Posted by ExponentialP
Yes thats what I done, but my answers still don't add up

If you calculate

$\displaystyle \sqrt{(-3)^2-4 \cdot 4 \cdot (-2)} = \sqrt{41}\approx 6.403$

but you still use $\displaystyle \sqrt{23} \approx 4.8$
• Jan 1st 2010, 10:11 AM
ExponentialP
Thank u earbooth am I doing something wrong at this point

x = 3 +6.40
________
8

or

x = 3 - 6.40
_________
8
• Jan 1st 2010, 10:26 AM
e^(i*pi)
Quote:

Originally Posted by ExponentialP
Thank u earbooth am I doing something wrong at this point

x = 3 +6.40
________
8

or

x = 3 - 6.40
_________
8

Both are equally valid solutions. Sub them back into the original equation if you want to be sure (Wink)
• Jan 2nd 2010, 04:27 AM
ExponentialP
I think I might be doing something wrong somewhere, I seem to be getting answers that don't make sense when I apply the formula to the equation, for instance take :

2x² - 7x = 3

The exercise I am doing is asking me to solve this equation using a formula... I'm not sure if this is where I am going wrong but I use

a= 2 b= -7 c= 0 ????

I then get

x = - (-7)± √ -7² - 4x2x0
____________________
2 x 2

The answer I get from this does not make sense, x = 7 ± √ -49
___________
4

Thanks

Happy New Year....... Again
• Jan 2nd 2010, 04:41 AM
earboth
Quote:

Originally Posted by ExponentialP
I think I might be doing something wrong somewhere, I seem to be getting answers that don't make sense when I apply the formula to the equation, for instance take :

2x² - 7x = 3

The exercise I am doing is asking me to solve this equation using a formula... I'm not sure if this is where I am going wrong but I use

a= 2 b= -7 c= 0 ????

I then get

x = - (-7)± √ -7² - 4x2x0
____________________
2 x 2

The answer I get from this does not make sense, x = 7 ± √ -49
___________
4

Thanks

Happy New Year....... Again

If you want to apply the quadratic formula the equation must be in the form

$\displaystyle ax^2+bx+c=0$

Therefore you have to transform the equation first:

$\displaystyle 2x^2 - 7x = 3~\implies~\boxed{2x^2-7x-3=0}$

So you have a = 2, b = -7, c = -3.
• Jan 2nd 2010, 04:45 AM
ExponentialP
Thank u !!!
• Jan 2nd 2010, 05:03 AM
ExponentialP
But I can't square root a negative number ??? What happens then ???
• Jan 2nd 2010, 05:30 AM
skeeter
Quote:

Originally Posted by ExponentialP
But I can't square root a negative number ??? What happens then ???

$\displaystyle 2x^2 - 7x - 3 = 0$

$\displaystyle b^2 - 4ac = (-7)^2 - 4(2)(-3) = 49+24 = 73$

$\displaystyle x = \frac{7 \pm \sqrt{73}}{4}$