1. ## Quadratic Equations Using a Formula

Hi Happy New Year everyone hope you all have had a good one........ I'm doing Maths on New Years Day ..... 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

2. Your a=4, b=-3 and c=-2. So, $\sqrt{b^2 -4ac} = \sqrt{41} = 6.4$. The answer has to be $\frac{3 \pm \sqrt{41}}{8}$.

3. 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 ..... 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.

4. Yes thats what I done, but my answers still don't add up

5. Originally Posted by ExponentialP
Yes thats what I done, but my answers still don't add up
If you calculate

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

but you still use $\sqrt{23} \approx 4.8$

6. Thank u earbooth am I doing something wrong at this point

x = 3 +6.40
________
8

or

x = 3 - 6.40
_________
8

7. 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

8. 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

9. 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

$ax^2+bx+c=0$

Therefore you have to transform the equation first:

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

So you have a = 2, b = -7, c = -3.

10. Thank u !!!

11. But I can't square root a negative number ??? What happens then ???

12. Originally Posted by ExponentialP
But I can't square root a negative number ??? What happens then ???
$2x^2 - 7x - 3 = 0$

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

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