3. The Quadratic Formula
At the end of the last section (Completing the Square), we derived a general formula for solving quadratic equations. Here is that general formula:
For any quadratic equation `ax^2+ bx + c = 0`, the solutions for x can be found by using the quadratic formula:
The expression under the square root, `b^2− 4ac`, can tell us how many roots we'll get. (There's no magic here - just a consideration of what the square root of `b^2− 4ac` is.)
If ` b^2− 4ac = 0`, then we'll have one root only, `x = −b/(2a)`.
If ` b^2− 4ac > 0`, then we'll have two roots, one involving the "+" sign and the other involving the "−" sign in the formula.
If ` b^2− 4ac < 0`, then we'll have no real roots, since you cannot find the square root of a negative number.
The expression `b^2 − 4ac` is called the discriminant and in some books you will see it written with a Greek upper case Delta, like this `Delta = b^2 − 4ac`.
Solve `2x^2- 7x - 5 = 0` using the quadratic formula.
By inspection, we can see that: `a = 2`, `b = -7` and `c = -5`.
Substituting these into the quadratic formula, we get :
`=(7-sqrt89)/4 or (7+sqrt89)/4`
`=-0.6085 or 4.108`
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Solve `2x^2= 4x + 3`
Firstly, in order to identify a, b and c , we must re-arrange the expression in the proper form ie. all the terms to the left, leaving zero on the right:
`2x^2- 4x - 3 = 0`
Only then, we can see that:
`a = 2`, `b = -4` and `c = -3`
Substituting these into the quadratic formula, we get:
`=(4-sqrt40)/4 or (4+sqrt40)/4`
`x=-0.581 or 2.581`
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Solve `6r^2= 6r + 1` using the quadratic formula.
Rearranging so it's in quadratic form:
Using the Quadratic Equation for the variable `r`:
`=(6-2sqrt15)/12 or (6+2sqrt15)/12`
`=(3-sqrt15)/6 or (3+sqrt15)/6`
`r=-0.145 or 1.145`