Hi all, Many thanks to those who have offered valued help to my other posts.

I think i'm starting to get the hang of this math lark

Ok, down to the question, i have several questions that i have done my best to answer, I wondered if anyone would mind just looking over them for mistakes or wrong use of solutions please?

(i) Solve by Factorization of quadratic formula:

x(2) + 6x + 8 = 0

By Factorization: (x + 4) (x + 2) = 0

X = -4 X = -2

By Quadratic Equation: -b +- sqrt b(2) - 4ac
2a 2a

- 6 +- sqrt 6(2) - (4x1x8)
2x1 2x1

= -3 +- 2
2
X = -4 X = -2.

.................................................. ................................................

(ii) 2x(2) + 11x + 14 = 0

- 11 +- sqrt 11(2) - (4 x 2 x 14)
2x2 2x2

= -2.75 +- 3
4

x = -3.5 x = -2

.................................................. ................................................

(iii) 6x(2) + 7x - 4 = 0

Using Quadratic Equation ( to 3 d.p)

- 7 +- sqrt 7(2) - (4 x 6 (-4)
2x6 2x6

-0.583 +- 12.041
12

x = 0.420 x = - 1.587

.................................................. .................................................

(iv) -7 + 11x - 3x(2) = 0

Using Quadratic Equation ( to 3 d.p)

*** do i transpose this so that x(2) is at the front***

3x(2) - 11x + 7 = 0 ???? (x = 0.550 x =-4.220)

Continueing with -7 + 11x - 3x(2) = 0

- 11 +- sqrt 11(2) - (4 x (-7) x 3
-14 -14

x = -0.236 x = 1.810

Phew, that the lot, I'm sorry to ask stupid questions, just that this is all new to me, and confidence is low

2. Originally Posted by Parton-Bill
Hi all, Many thanks to those who have offered valued help to my other posts.

I think i'm starting to get the hang of this math lark

Ok, down to the question, i have several questions that i have done my best to answer, I wondered if anyone would mind just looking over them for mistakes or wrong use of solutions please?

(i) Solve by Factorization of quadratic formula:

x(2) + 6x + 8 = 0

By Factorization: (x + 4) (x + 2) = 0

X = -4 X = -2 ........ OK

By Quadratic Equation: -b +- sqrt b(2) - 4ac
2a 2a

- 6 +- sqrt 6(2) - (4x1x8)
2x1 2x1

= -3 +- 2
2
X = -4 X = -2. ........ OK

.................................................. ................................................

(ii) 2x(2) + 11x + 14 = 0

- 11 +- sqrt 11(2) - (4 x 2 x 14)
2x2 2x2

= -2.75 +- 3
4

x = -3.5 x = -2 ........ OK

.................................................. ................................................

(iii) 6x(2) + 7x - 4 = 0

Using Quadratic Equation ( to 3 d.p)

- 7 +- sqrt 7(2) - (4 x 6 (-4)
2x6 2x6

-0.583 +- 12.041
12

x = 0.420 x = - 1.587 ........ OK

.................................................. .................................................

(iv) -7 + 11x - 3x(2) = 0

Using Quadratic Equation ( to 3 d.p)

*** do i transpose this so that x(2) is at the front*** ........ Yes

3x(2) - 11x + 7 = 0 ???? (x = 0.550 x =-4.220) ........ Where did you get these results from?

...
$\displaystyle 3x^2 - 11x + 7 = 0~\implies~x=\dfrac{11}{2 \cdot 3} \pm\dfrac{\sqrt{ 121 - 4 \cdot 3 \cdot 7}}{2 \cdot 3}$

Therefore: $\displaystyle x=\dfrac{11}{6} \pm\dfrac{\sqrt{ 37}}{6} ~\implies~x\approx 2.847~\vee~x\approx 0.820$

3. ## Danke :)

Thanks very much for that, wasn't sure on the last one. will remember that.

Thanks Again

4. Originally Posted by earboth
$\displaystyle 3x^2 - 11x + 7 = 0~\implies~x=\dfrac{11}{2 \cdot 3} \pm\dfrac{\sqrt{ 121 - 4 \cdot 3 \cdot 7}}{2 \cdot 3}$

Therefore: $\displaystyle x=\dfrac{11}{6} \pm\dfrac{\sqrt{ 37}}{6} ~\implies~x\approx 2.847~\vee~x\approx 0.820$
The figures i calculated allowed for 7 to be -7, despite the tranposition, this would remain negative wouldn't it?