What is the quadratic formula and how do you use it?

[Kool]

For any equation of the form [ ax2 + bx + c = 0 ] the values of x ( the roots of the equation ) will always be [ ( -b ± ✓( b2 - 4ac ) ) / 2a ].

Theres a nice song to remember it by, you can probably find it. I always have to sing it to myself when using the quadratic formula.

The ± thing in there indicates that there are 2 solutions, one where the operator is adding and one where it is subtracting.

For Example: Solving x2 + 2x -15:

x = ( -2 ± ✓( 22 - 4 * 1 * -15 ) ) / 2 * 1

= ( -2 ± ✓( 4 -- 60 ) ) / 2

= ( -2 ± ✓( 64 ) ) / 2

= [ ( -2 + 8 ) / 2 = 3 ] OR [ ( -2 - 8 ) / 2 = -5 ]

Its basically just plugging the numbers in.

There are always 2 solutions because when square rooting, the solution can be either the positive root or the negative root. Because ( -x )2 = x2

If the value in the square root is negative, the solutions are complex numbers. If you get this ever, just say there are 'No Real Solutions'. This is the case when the graph of the equation you are solving does not go through the x-axis.

By calculating [ b2 - 4ac ] you can determine the nature of the solutions. If it is less than 0, the roots are not real numbers. If it is equal to 0, then there is one repeated root ( i.e. both roots are the same ). If it is greater than 0, there are 2 real roots.

In the case the roots are not real, can usually ignore them unless they are part of what you are learning. But usually they are a more advanced topic that is done much later. Look into 'Imaginary Numbers' and 'Complex Numbers' if you want to know more.

I hope this was what you were looking for

[Kool]

This is the equation: x = [-b +/- sqrt(b2 - 4ac)] / 2a

So if you have an equation like 3x2 - 15x + 4 = 0, a would be 3, b would be -15, and c would be 4. Then you just plug those numbers into the quadratic equation and it will tell you that x = 0.283 or 4.717. (The reason that there are 2 values for x is that this is a parabola, and it crosses the x axis in 2 places, so at those points, 3x2 - 15x + 4 = 0.)

Here's the proof for why it works:

ax2 + bx + c = 0 {Starting point}

ax2 + bx = -c {Subtract c from both sides}

(ax2 + bx)/a = -c / a {Divide both sides by a}

x2 + (b / a)x = -c / a {Simplify each term on the left}

x2 + (b / a)x + (b / 2a)2 = -c / a + (b / 2a)2 {Add (b / 2a)2 to both sides to complete the square}

[x + (b / 2a)]2 = -c / a + (b / 2a)2 {Factored left side}

[x + (b / 2a)]2 = -c / a + (b2 / 4a2 ) {Multiplied out the (b/2a)2 on the right side}

[x + (b / 2a)]2 = -c / a (4a / 4a) + (b2 / 4a2 ) {Multiplied -c/a by (4a / 4a)}

[x + (b / 2a)]2 = (-4ac / 4a2 ) + (b2 / 4a2 ) {Multiplied out previous step}

[x + (b / 2a)]2 = (-4ac + b2 ) / 4a2 {Put the two numerators over their common denominators on the right}

[x + (b / 2a)]2 = (b2 - 4ac) / 4a2 {Changed the order of b2 and -4ac}

x + (b / 2a) = +/- sqrt[(b2 - 4ac) / 4a2 ] {Took the square root of both sides}

x + (b / 2a) = +/- sqrt(b2 - 4ac) / 2a {Simplified the denominator on the right, since the square root of 4a2 = 2a}

x = -(b / 2a) +/- sqrt(b2 - 4ac) / 2a {Subtracted b/2a from both sides}

x = [-b +/- sqrt(b2 - 4ac)] / 2a {Put the two numerators over their common denominators on the right}

[Guest]

?? This looks like a response to some prior post. Where is the original post with a question?

[Guest]

Hello