Question

What is `5x^2-28x+19=0`?

Answer 1

See explanation below!

Explanation:

Recall that a linear equation in one variable is of the form `ax+b=0`, where `a` and `b` are constants and `a≠0`.

For example: `" "` `3x+5=0`

A quadratic equation has an `x^2` (x-squared) term. ("Quadratum" is Latin for square.) The general quadratic equation in standard form looks like:

`ax^2+bx+c=0` `\ \ \` `\cdots` `\ \` `\cdots` where `a\ne 0`

If we want to find the `x` or `x's` that work, we might guess and substitute and hope we get lucky, or we might try one of these four methods:

• Guess and Check
• Solving by Square Roots (if b=0)
• Factoring
• Completing the Square
• The Quadratic Formula

We can solve graphically by equating the polynomial to `y` instead of to `0`, we get an equation whose graph is a parabola. The `x-\text{intercepts}` of the parabola (if any) correspond to the solutions of the original quadratic equation.

Answer 2

The solutions are `x=(14+-sqrt101)/5`.

Explanation:

One way to find the solutions to a quadratic is to use the quadratic formula:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Here's our quadratic:

`5x^2-28x+19=0`

The values are `a=5`, `b=-28`, and `c=19`. Plug in the values to the equation:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`color(white)x=(-(-28)+-sqrt((-28)^2-4(5)(19)))/(2(5))`

`color(white)x=(28+-sqrt((-28)^2-4(5)(19)))/10`

`color(white)x=(28+-sqrt(784-4(5)(19)))/10`

`color(white)x=(28+-sqrt(784-380))/10`

`color(white)x=(28+-sqrt(404))/10`

`color(white)x=(28+-sqrt(4*101))/10`

`color(white)x=(28+-sqrt(2^2*101))/10`

`color(white)x=(28+-2sqrt(101))/10`

`color(white)x=(14+-sqrt(101))/5`

This is as simplified as the answer gets. The two final solutions are:

`x=(14+sqrt101)/5`
and
`x=(14-sqrt101)/5`

Here's the graph of the function (with an altered scale):

graph{5x^2-28x+19[-3,8,-30,20]}