5. Equations Reducible to Quadratic Form (Part 2)

This lesson will discuss equations that can be reduced (or converted) to quadratic equations, using simple substitutions. I’ll focus on examples where the variable occurs as a power.

Example 1 Solve for x: 9^x – 4.3^x + 3 = 0

Solution To convert an equation to a quadratic equation, we first pick a term involving the variable.

Next, we’ll check if the square (or square root) of this term is also around somewhere.

Sometimes, the square (or the square root) will be there right in front of you. In other cases, you’ll have to do small manipulations to convert the other term to the required form.

If we find a term y and its square y², we’re done – the equation can then be converted to the form ay² + by + c = 0.

Here, the terms involving x are 9^x and 3^x. Observe that:

9^x = (3²)^x = 3^2x = (3^x)²

We got what we wanted. Now, we’ll simply substitute 3^x = y, so that the given equation becomes y² – 4y + 3 = 0.

This can be easily solved using any of the methods discussed here. I’ll use the factorization method to solve.

We’ll get:

(y – 1)(y – 3) = 0

This gives

y = 1 and y = 3

But things don’t end here. We were supposed to find the value of x. So, we’ll substitute y back to 3^x, giving us the following equations:

3^x = 1

3^x = 3

Solving the first equation gives us x = 0 and the second one gives x = 1. And these were the solutions we were looking for.

Try substituting these values in the original equation (i.e 9^x – 4.3^x + 3 = 0) to verify that the solutions are correct.

 

Example 2 Solve for y: 2^{3 + y} + 2^{3 – y} = 34

Solution Here, the terms involving the variable are 2^{3 + y} and 2^{3 – y}.

Using the laws of exponents, these can be respectively expressed as 2³.2^y and 2³.2^-y. Now, if we substitute 2^y as x, these become 8x and 8/x. So, the original equation changes to:

8x + 8/x = 34

⇒ 4x + 4/x = 17

We do not see x and x­² yet. But if we rearrange the terms, we’ll get:

4x² – 17x + 4 = 0

Nice! A quadratic equation. This can be solved easily. Let’s try to factorize the LHS.

4x² – 16x – x + 4 = 0

⇒ 4x(x – 4) – (x – 4) = 0

⇒ (x – 4)(4x – 1) = 0

⇒ x = 4 or x = 1/4

Since we need the value of y, we’ll need to substitute x back to 2­^y. We’ll get:

2^y = 4 ⇒ y = 2

2^y = 1/4 ⇒ y = -2

Once again, try substituting these into the original equation to verify that the solutions are correct.

That’s all for this lesson. In the next lesson, I’ll cover a few more ‘types’ of equations that can be converted to quadratic equations.