Question

How do you solve the quadratic equation `5x^2 + 3x = 1` ?

Answer 1

`x_(1,2) = (-5 +- sqrt(13))/10`

Explanation:

The first thing to do here is rearrange your equation to quadratic form by getting all the terms on one side of the equation.

To do that, subtract `1` from both sides of the equation

`5x^2 + 3x - 1 = color(red)(cancel(color(black)(1))) - color(red)(cancel(color(black)(1)))`

`5x^2 + 3x - 1 = 0`

Now, the quadratic formula allows you to calculate the two solutions of a general form quadratic equation

`color(blue)(ax^2 + bx + c = 0)`

by using the equation

`color(blue)(|bar(ul(color(white)(a/a)x_(1,2) = (- b +- sqrt(b^2 - 4 * a * c))/(2 * a) color(white)(a/a)|)))`

In your case, you have

`{(a = color(white)(-)5), (b = color(white)(-)3), (c = -1) :}`

Plug these values into the quadratic formula to get

`x_(1,2) = (-5 +- sqrt(3^2 - 4 * 1 * (-1)))/(2 * 5)`

`x_(1,2) = (-5 +- sqrt(13))/10 implies {(x_1 = (-5 - sqrt(13))/10), (x_2 = (-5 + sqrt(13))/10) :}`

You can thus say that your original equation has two solutions

`x = (-5 - sqrt(13))/10" "` and `" "x = (-5 + sqrt(13))/10`

graph{5x^2 + 3x - 1 [-3, 3, -3, 3]}

Answer 2

`(-3 +- sqrt29)/10`

Explanation:

Bring the quadratic equation to standard form:
`y = 5x^2 + 3x - 1 = 0`
Use the new quadratic formula in graphic form:
`D = d^2 = b^2 - 4ac = 9 + 20 = 29` --> `d = +- sqrt29`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = -3/10 +- sqrt29/10 = (-3 +- sqrt29)/10`

NOTE. Using the improved quadratic formula gets simpler expressions and easier numeric computation. In addition, it shows students the graphic representation and interpretation of the axis of symmetry and the 2 x-intercepts.