The quadratic formula provides the solutions to a quadratic equation. Given `ax^2 + bx + c = 0`, then the solutions are given by:
`x=frac(-b±sqrt(b^2-4ac))(2a)`
Note that the squared term is calculated once with a positive value of the square root; and once with a negative value. Once the solutions have been found, substitute into the original equation to check there has been no arithmetical errors.
Note also that if `b^2 < 4ac` then there will be an attempt to find the square root of a negative number, and that therefore there are no real roots.
What, if any, are the solutions for `3x^2-48=0`?
Using the formula, remembering that `b=0` as there is no coefficient for `x`:
`x=frac(-(0)+-sqrt((0)^2 - 4(3)(-48)))(2(3)) `
`= +-frac(576)(8)`
` = +-4`
Substitute back to check: `3(4)^2 - 48=0`✔ and `3(-4)^2-48=0`✔
Answer: `x=+-4`
Determine the solutions for `2x^2+18x+28=0`.
Using the quadratic formula:
`x=frac(-b ± sqrt(b^2-4ac))(2a)`
Substituting:
`x=frac(-(18) ± sqrt((18)^2-4(2)(28)))(2(2))`
`x=frac(-18±sqrt(324 - 224))(4)`
`x=frac(-18±10)(2)`
`x=-7` and `x=-2`
Check:
`2(-7)^2+18(-7)+28=0` ✔ and
`2(-2)^2+18(-2)+28=0` ✔
Answer: `x=-2 and x=-7`