Question

How do you solve `\frac { 1} { 4} x ^ { 2} + 4x - 3= 0`?

Answer 1

There are two answers:
`x_1 = 2(-4+\sqrt(19))`
`x_2 = 2(-4-\sqrt(19))`

Explanation:

Use delta formula:
The delta formula says:
`ax^2+bx+c=0 => x= \frac{-b+-\sqrt(\Delta)}{2a}`
where `\Delta = b^2-4ac`

and just put the values in the formula:
`\Delta = 4^2-4\frac{1}{4}\times-3 = 16 + 3 = 19`
and `x` will be:

`x = \frac{-4+-\sqrt(19)}{2\times\frac{1}{4}} = 2(-4+-\sqrt(19))`

There are two answers:
`x_1 = 2(-4+\sqrt(19))`
`x_2 = 2(-4-\sqrt(19))`

Answer 2

`x=-8+2sqrt(19)color(white)("XX")orcolor(white)("XX")x=-8-2sqrt(19)`

Explanation:

Given
`color(white)("XXX")1/4x^2+4x-3=0`

This becomes easier if we multiply both sides by `4` to get rid of the fraction:
`color(white)("XXX")x^2+16x-12=0`

We can then solve this using the quadratic equation or by completing the square method.

Completing the square

`x^2+16x-12=0`

`color(white)("XXX")rarr x^2+16x=12`

Noting that if `x^2+16x` are the first 2 terms of a squared binomial, then the third term must be `(16/2)^2=8^2 (=64)`
`color(white)("XXX")rarr x^2+16x+8^2=12+64`

`color(white)("XXX")rarr (x+8)^2=76`

`color(white)("XXX")rarr x+8=+-sqrt(76)=+-2sqrt(19)`

`color(white)("XXX")rarr x=-8+-2sqrt(19)`

Quadratic formula

`ax^2+bx+c=0 rarr x=(-b+-sqrt(b^2+4ac))/(2a)`

In this case
`1x^2+16x-12=0 rarr x=(-16+-sqrt(16^2-4 * 1 * (-12)))/(2 * 1)`

`color(white)("XXXXXXXXXXXXXXX")=(-16+sqrt(304))/2`

`color(white)("XXXXXXXXXXXXXXX")=(-16+4sqrt(19))/2`

`color(white)("XXXXXXXXXXXXXXX")=-8+-2sqrt(19)`