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Mathematical Formulae

Rearranging Equations II (Quadratic Equations)

So far we have looked at only formulae in which the formula can be evaluated by substituting a value ONCE and then doing many operations on that value, but not all exzpressions are of that type. For some more than one substitution is necessary.

Equation Type

For an expression of the ONE term type we can always use the methods given so far.

Given the expression    y=(6+e^((3-2*b)/7))^3, we can rearrange to make b the subject of the formula because once we have a value for b, we need only do 6 operations to obtain y. These are in order, multiply by -2, add 3, divide by 7, take the exponent, add 6 then finally take the cube root. If we take y, cube it, subtract 6, take the natural logarithm, multiply by 7, subtract 3 and divide by -2, b can be obtained.

In contrast a simple expression like y = x2 + 8x cannot because, as written, we have to substitute x TWICE to evaluate y, once to evaluate x2 and once to evaluate 8x.

Click on the number of substitutions of x needed to evaluate the expression:

y = 6log (x + 1)      1 2 3

y = 2(7x3 - 20)      1 2 3

y = x + 1/x      1 2 3

y = x3 - 5x2 + 19x      1 2 3

y = xex      1 2 3

There is a nice way to convert quadratic expressions to a form which requires only one substitution.

Rearranging Quadratic Equations

Recall the form of a "perfect square": y = (x + b)2 = x2 + 2bx + b2

Notice that the first expression requires only one substitution whereas the right expression needs two. If we can write our expression in the bracketed form we can rearrange it.

Let's look at the expression    y = x2 + 6x - 5  

We could write it as y = x2 + 2*3x + 32 - 32 - 5

in which case we can also write it as y = (x + 3)2 - 14

Now it only requires one substitution of x.  It can be rearranged by taking y, adding 14, taking the square root then subtracting 3.


Lets check.  If x = 0, using the original expression, y = -5.  Now put the value of y into our new expression:


That was easy with simple numbers.  Now lets consider the more general form of a quadratic:

y = ax2 + bx + c

By rearranging, we can make one side equal zero, then use the quadraticformula to solve for x.

In this case subtract y from both sides to obtain 0 on the left side.

0 = -x2 + 2x + 29 - y

Now we can use the quadratic formula.Remember that for 0 = ax2 + bx + c, the solutions are:x=(-b +/- sqrt(b^2-4ac))/(2*a) .

In our example, a = -1, b = 2 and c = (29-y).  Substituting these values into the quadratic formula we obtain:

x=(-2 +/- sqrt(2^2-4(-1)(29-y)))/(2*(-1))

This expression can be simplified by multiplying and collecting like terms.

x=(-2 +/- sqrt(4+4(29-y)))/(2*(-1)) x=(-2 +/- sqrt(4(30-y)))/(2*(-1))

Recognise that 4 has a known square root, so that can be evaluated

x=(-2 +/- sqrt(4)*sqrt(30-y))/(2*(-1)) x=(-2 +/- 2*sqrt(30-y))/(2*(-1))

And -2 is common to numerator and denominator so may be cancelled.

x= 1 +/- sqrt(30-y)




More Quadratic Exercises

<< Rearranging Equations I (Simple Equations) | Formulae Index | Rearranging Equations III (Harder Examples)>>


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