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5. Equations in Quadratic Form

In this section, we'll come across equations that are in fact quadratic, but they may not look like it at first glance.

We'll use either of the following methods to solve such equations:

  • Factoring
  • Quadratic formula: `x=(-b+-sqrt(b^2-4ac))/(2a)`

Example 1

Solve: x4 − 20x2 + 64 = 0

Answer

Here, if we let u = x2, we can rewrite the equation so it looks like an ordinary quadratic equation:

u2 − 20u + 64 = 0

We now factor to give:

(u − 16)(u − 4) = 0

So the solutions for u are 16 or 4.

So `x^2= 16` or `x^2= 4`.

These give us:

x = −4 or 4 x = −2 or 2

So the complete set of solutions is: `x = −4, −2, 2, 4`.

Is it correct?

The sketch shows:

`y = x^4 - 20x^2 + 64`, showing intersections with the x-axis

We can see from where the graph cuts the x-axis that the solutions are correct.

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Example 2

Solve: `4x+3sqrtx=1`

Answer

Here, if we write `u=sqrt(x)` we have:

`4x+3sqrtx=1`

`4u^2+3u-1=1`

`(4u-1)(u+1)=0`

So `u = 1/4`; or `u= -1`.

DANGER! Always think carefully about your answer. You can often get answers which are not true solutions.

`sqrt(x)=1/4` means `x= 1/16`

Check by substitution: `4(1/16)+3(1/4)=1`. OK.

But `sqrt(x)=-1` is not possible (`sqrt(x)` is always `≥ 0`).

We conclude there is only one root: `x=1/16`

To give a better idea what our solution looks like, let's have a look at the graph of `y = 4x + 3sqrt(x) - 1`.

The intersection with the x-axis will tell us the solution for the original equation.

This is an interesting curve since it starts at `(0,-1)` (we cannot have negative `x`-values and the curve does not continue down the `y`-axis). There is one intersection with the x-axis, at `x = 1/16 = 0.0625`.

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Exercises

1. Solve: ` 4x^4+ 15x^2= 4`

Answer

`4x^4+15x^2=4`, so

`4x^4+15x^2-4=0`

Let `u=x^2` to make things easier. Then we have:

`4u^2+15u-4=0`

`4u^2+16u-u-4=0`

`4u(u+4)-1(u+4)=0`

`(4u-1)(u+4)=0`

So `u=1/4` or `u=-4`.

But `u=x^2`, so

`x^2=1/4` or `x^2=-4`

`x=+-1/2` are the only real solutions.

Once again, let's have a look at the graph of the function, to better understand the situation. This time it's `y = 4x^4 + 15x^2 - 4`.

The intersection with the x-axis will tell us the solution for the original equation.

Graph of `y = 4x^4 + 15x^2 - 4`

The two intersections with the x-axis are at `x = -0.5` and `x=0.5`.

2. Solve `10x^-2+3x^-1-1=0`

Answer

Let `u=x^-1`. Then we have:

`10u^2+3u-1=0`

` (5u-1)(2u+1)=0`

So `u=1/5 or -1/2`

This gives us `x^-1=1/5` so `x = 5`

and `x^-1=-1/2` so `x=-2`.

Always CHECK your answers!!

Here's the graph of the function, to help clarify the situation. This time we sketch `y = 10x^-2 + 3x^-1 -1`.

Graph of `y = 10x^-2 + 3x^-1 -1`

The two intersections with the x-axis are at `x = -2` and `x=5`.

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