there are 4 ways to solve quadratics:
1. Complete the square
2. the quadratic formula
3. factorise
4. graphs (yes you can use graphs to use quadratic)

You're best choices will be the formula (you find it online) and graphs (but you'll have to use excel to ensure that you get an accurate answer) - input some numbers in the formula and plot an XY scatter graph in excel and take interception of the shape with X line ... it should work very well

Reply 8

username1560589

Original post by RAlexO

You'll rarely get a precise answer solving them graphically, If you include that then you have to include numerical methods as well.
Also, they're unlikely to be allowed to use Excel in their exam.

Reply 9

RAlexO

Original post by morgan8002

You'll rarely get a precise answer solving them graphically, If you include that then you have to include numerical methods as well.
Also, they're unlikely to be allowed to use Excel in their exam.

I did not know he needed it for an exam... I thought it is some sort of homework or essay :smile:

Reply 10

Zacken

Original post by Andy98

Let's use a general formula:

ax^2 + bx + c = 0

Step 1 - take a out of the terms involving x:

a(x^2+ \frac{bx}{a})+c=0

Step 2 - look at the

x^2+ \frac{bx}{a}

divide through by x and divide the second term by 2 (and square the bracket):

a(x+ \frac{b}{2a})^2 + c =0

Step 3 - Take c to the other side, divide through by a, square root the equation and solve for x.

\displaystyle a\left(x^2 + \frac{bx}{a}\right) + c = 0 \iff a\bigg(\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\bigg) + c

Reply 11

Andy98

Original post by Zacken

\displaystyle a\left(x^2 + \frac{bx}{a}\right) + c = 0 \iff a\bigg(\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\bigg) + c

Oh yeah, oops :colondollar:

Reply 12

Zacken

Original post by Andy98

Oh yeah, oops :colondollar:

Which is, by the way, precisely how you derive the quadratic formula. :wink:

Reply 13

Andy98

Original post by Zacken

Which is, by the way, precisely how you derive the quadratic formula. :wink:

It is?

Reply 14

Zacken

Original post by Andy98

It is?

\displaystyle a\left(x^2 + \frac{bx}{a}\right) + c = 0 \iff a\bigg(\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\bigg) + c = 0

So, isolating 'x':

\displaystyle a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} = -c \iff a\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a} - c = \frac{b^2 - 4ac}{4a}

\displaystyle \iff \left(x + \frac{b}{2a}\right) = \frac{b^2 - 4ac}{4a^2}

Square rooting and isolating x:

\displaystyle x = \frac{-b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Reply 15

Andy98

Original post by Zacken

\displaystyle a\left(x^2 + \frac{bx}{a}\right) + c = 0 \iff a\bigg(\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\bigg) + c = 0

So, isolating 'x':

\displaystyle a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} = -c \iff a\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a} - c = \frac{b^2 - 4ac}{4a}

\displaystyle \iff \left(x + \frac{b}{2a}\right) = \frac{b^2 - 4ac}{4a^2}

Square rooting and isolating x:

\displaystyle x = \frac{-b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Wow! I never realised that. Nice!

Reply 16

Zacken

Original post by Andy98

Wow! I never realised that. Nice!

Learn something new everyday. :biggrin:

Reply 17

Andy98

Original post by Zacken

Learn something new everyday. :biggrin:

True

Reply 18

vectorpi

Original post by Andy98

Let's use a general formula:

ax^2 + bx + c = 0

Step 1 - take a out of the terms involving x:

a(x^2+ \frac{bx}{a})+c=0

Step 2 - look at the

x^2+ \frac{bx}{a}

divide through by x and divide the second term by 2 (and square the bracket):

a(x+ \frac{b}{2a})^2 + c =0

Step 3 - Take c to the other side, divide through by a, square root the equation and solve for x.

You have to account for the extra term when the bracket is squared by subtracting (b^2)/4a^2, so in the case of 5x^2 + 9x + 3 = 0 you get:

5[x^2 + 1.8x + 0.6] = 0
X^2 + 1.8x + 0.6 = 0
(X + 0.9)^2 + 0.6 - 0.81 = 0
(X + 0.9)^2 - 0.21 = 0
(X + 0.9)^2 = 0.21
X + 0.9 = +/- sqrt(0.21)
∴ x = -0.9 +/- sqrt(0.21)