Saturday

September 20, 2014

September 20, 2014

Posted by **Randy** on Friday, July 13, 2012 at 7:10pm.

- Math -
**Mike**, Friday, July 13, 2012 at 10:35pmthe proof is simply to complete the square for a generalised quadratic equation. Like this:

ax2(xsquared) + bx + c = 0

Take 'a' outside:

a[x2 + bx/a + c/a] = 0

Divide through by 'a':

x2 + bx/a + c/a = 0

Complete the square:

(x + b/2a)2 - b2/4a2 + c/a = 0

Rearrange to find x:

(x + b/2a)2 = b2/4a2 - c/a

x + b/2a = (+/-)sqrt[b2/4a2 - c/a]

x = -b/2a (+/-) sqrt[b2/4a2 - c/a]

Finally, fiddle around so that (1/2a) can be taken out as a common factor:

x = -b/2a (+/-) sqrt[b2/4a2 - 4ac/4a2]

x = -b/2a (+/-) sqrt[(1/4a2)(b2 - 4ac)]

x = -b/2a (+/-) sqrt(1/4a2)sqrt(b2 - 4ac)

x = -b/2a (+/-) (1/2a)sqrt(b2 - 4ac)

x = [ -b (+/-) sqrt(b2 - 4ac) ] / 2a.

- Math -
**Reiny**, Friday, July 13, 2012 at 11:52pma bit easier to read ....

divide by a

x^2 + (b/a)x = -c/a

complete the square , add b^2/(2a^2) to both sides

x^2 + (b/a)x + b^2/(4a^2) = b^2/(4a^2) - c/a

write the left side as a square and add the two terms on the right

(x + b/(2a) )^2 = (b^2 - 4ac)/(4a^2)

take √ of both sides

x + b/(2a) = ± √(b^2 - 4ac)/(2a)

x = -b/(2a) ± √(b^2 - 4ac)/(2a)

=**(-b ± √(b^2 - 4ac)/(2a)**

**Answer this Question**

**Related Questions**

maths - If the sum of the squares of the roots of the quadratic equation ax^2 + ...

maths - If the ratio of the roots of the quadratic equation ax^2 + bx +c =0 is m...

algebra II - I'm quite clueless as to how to solve this. I know the quadratic ...

College Math : Linear Algebra - Suppose A is a square matrix satisfying the ...

Math - Write a quadratic equation having the given numbers as solutions. -5 and...

QUADRATIC EQUATION - r^2+8r=48 First bring the 48 to the other side of the ...

math - The solutions to a quadratic equation are 3/8 and 2/5. Write the ...

algebra II - How do you know if a quadratic equation will have one, two, or no ...

algebra II - How do you know if a quadratic equation will have one, two, or no ...

advanced math - Hi, I am trying to solve graphing quadratic equations. One of ...