Solve by trial and improvement
So far we've learned a variety of methods to solve different types of equations. For example, we know how to solve linear equations, like $5\left(2x-3\right)=4\left(7x-8\right)$5(2x−3)=4(7x−8), and quadratic equations, like $3x^2+5x=3$3x2+5x=3. But we don't yet have the tools to solve an equation like $7x-1=3^x-1$7x−1=3x−1.
In this lesson we are going to learn how to solve equations through Trial and Improvement.
Let's look at a graph of the problem
Before we do our first trial, let's consider the graphs of $y=7x-1$y=7x−1 and $y=3^x-1$y=3x−1, as they are the two different sides of the equation $7x-1=3^x-1$7x−1=3x−1.
Remember that the solution to the equation $7*x-1=3^x-1$7*x−1=3x−1, is actually asking us for point where both curves have the same value, that is - where they cross over.
By considering the graphs of these two curves, we can find an approximate value for $x$x. We can see it is somewhere between $x=0$x=0 and $x=0.5$x=0.5. A good estimate is $x\approx0.1$x≈0.1.
Let's look at the numerical approach of trial and improvement
Now, let's consider another way to approach the problem of solving $7x-1=3^x-1$7x−1=3x−1.
$x=0$x=0
Let's substitute $x=0$x=0 into the left hand side ($LHS$LHS) of the equation, $7x-1$7x−1:
$LHS=7x-1$LHS=7x−1$=$=$7\times0-1$7×0−1$=$=$0-1=-1$0−1=−1
Similarly, substitute $x=0$x=0 into the right hand side ($RHS$RHS) of the equation, $3^x-1$3x−1:
$RHS=3^x-1$RHS=3x−1$=$=$3^0-1$30−1$=$=$1-1=0$1−1=0
If $x=0$x=0 was a solution, then the $LHS$LHS would equal the $RHS$RHS.
We can see that the $LHS$LHS and $RHS$RHS differ by $1$1.
We could consider this the first trial, and as this isn't the answer we need to improve our original guess of $x=0$x=0. So we try again with a different value for $x$x.
$x=0.5$x=0.5
I'm going to suggest we now try $x=0.5$x=0.5
So the left hand side ($LHS$LHS) of the equation, $7x-1$7x−1 becomes
$LHS=7x-1$LHS=7x−1$=$=$7\times0.5-1$7×0.5−1$=$=$3.5-1=2.5$3.5−1=2.5
And into the right hand side ($RHS$RHS) of the equation, $3^x-1$3x−1:
$RHS=3^x-1$RHS=3x−1$=$=$3^{0.5}-1$30.5−1$\approx$≈$1.732-1$1.732−1$\approx$≈$0.732$0.732
If $x=0.5$x=0.5 was a solution, then the $LHS$LHS would equal the $RHS$RHS. We can see that the $LHS$LHS and $RHS$RHS differ by $2.5-0.732=1.761$2.5−0.732=1.761. This is a much bigger difference than the previous attempt of substituting $x=0$x=0. So we have moved further away from the answer than $x=0$x=0.
We know this already, as from our graph above we could see that a better approximation was to be $0.1$0.1. (Which is much closer to $0$0 than $0.5$0.5.
At this point, we get to try the improvement part of trial and improvement. We make an educated decision to try a further point, that is closer to $0$0 than to $0.5$0.5
$x=0.1$x=0.1
So we can try $0.1$0.1. Substitute $x=0.1$x=0.1 into the left hand side ($LHS$LHS) of the equation, $7x-1$7x−1:
$LHS$LHS$=$=$7x-1=7\times0.1-1$7x−1=7×0.1−1$=$=$0.7-1=-0.3$0.7−1=−0.3
Substitute $x=0.1$x=0.1 into the right hand side ($RHS$RHS) of the equation, $3^x-1$3x−1:
$RHS$RHS$=$=$3^x-1=3^{0.1}-1$3x−1=30.1−1$\approx$≈$1.116-1$1.116−1$\approx$≈$0.116$0.116
If $x=0.1$x=0.1 was a solution, then the $LHS$LHS would equal the $RHS$RHS. We can see that the $LHS$LHS and $RHS$RHS differ by $0.116-\left(-0.3\right)=0.416$0.116−(−0.3)=0.416. This is better than the approximation with $x=0$x=0, and $x=0.5$x=0.5.
Our next step would be to try either side of $x=0.1$x=0.1, that is, $x=0.05$x=0.05 and $x=0.15$x=0.15.
$x=0.05$x=0.05
For $x=0.05$x=0.05,
$LHS$LHS$=$=$7x-1=7\times0.05-1$7x−1=7×0.05−1$=$=$0.35-1=-0.65$0.35−1=−0.65
$RHS$RHS$=$=$3^x-1=3^{0.05}-1$3x−1=30.05−1$\approx$≈$1.056-1$1.056−1$\approx$≈$0.056$0.056
If $x=0.05$x=0.05 was a solution, then the $LHS$LHS would equal the $RHS$RHS. We can see that the $LHS$LHS and $RHS$RHS differ by $0.056-\left(-0.65\right)=0.706$0.056−(−0.65)=0.706. This is worst than the approximation with $x=0.1$x=0.1. So we have moved further away from the answer.
$x=0.15$x=0.15
For $x=0.15$x=0.15,
$LHS$LHS$=$=$7x-1=7\times0.15-1$7x−1=7×0.15−1$=$=$1.05-1=0.05$1.05−1=0.05
$RHS$RHS$=$=$3^x-1=3^{0.05}-1$3x−1=30.05−1$\approx$≈$1.179-1$1.179−1$\approx$≈$0.179$0.179
If $x=0.15$x=0.15 was a solution, then the $LHS$LHS would equal the $RHS$RHS. We can see that the $LHS$LHS and $RHS$RHS differ by $0.179-0.05=0.129$0.179−0.05=0.129. This is worst than the approximation with $x=0.1$x=0.1, and the best approximation so far.
Applet
This applet will step through the results we have just found
The actual answer to this (solved using technology) is for $x=0.1727$x=0.1727. From our trial and improvement we can see that we were getting close to that.
The differences we are calculating are actually the differences in distance between the curves. We want these differences to be as close to $0$0 as possible as that will mean we are approaching the point of intersection. This close up of the graph above shows the differences we just worked out.
For ever and ever
This process can be repeated ad infinitum, but we will pause it here. With the use of computer programs, this type of process is relatively easy. For questions in this subtopic, we will only repeat one or two iterations, and choose the best approximation to solve the problem.
Worked Examples
QUESTION 1
We wish to solve the equation $x^2+5x=10$x2+5x=10.
Start with an initial approximation of $x=1$x=1. Evaluate $x^2+5x$x2+5x when $x=1$x=1.
Is the initial approximation of $x=1$x=1 too low or too high?
Too low.
AToo high.
BNow test $x=2$x=2. Evaluate $x^2+5x$x2+5x when $x=2$x=2.
Is the approximation of $x=2$x=2 too low or too high?
Too low.
AToo high.
BNow complete the table of values to test other approximations. Leave all answers as exact values.
$x$x $x^2+5x$x2+5x $1.5$1.5 $\editable{}$ $1.6$1.6 $\editable{}$ $1.7$1.7 $\editable{}$ $1.8$1.8 $\editable{}$ $1.9$1.9 $\editable{}$ Which of these values is the solution closest to?
$2.55$2.55
A$1.15$1.15
B$1.55$1.55
C$1.75$1.75
DEvaluate $x^2+5x$x2+5x when $x=1.55$x=1.55
QUESTION 2
The equation $x^3+3x^2=270$x3+3x2=270 has a solution between $x=5$x=5 and $x=6$x=6. We wish to solve for $x$x, correct to 1 decimal place.
Evaluate $x^3+3x^2$x3+3x2 when $x=5.3$x=5.3
Does $x=5.3$x=5.3 result in a value that is too low or too high?
Too low
AToo high
BEvaluate $x^3+3x^2$x3+3x2 when $x=5.7$x=5.7
Does $x=5.7$x=5.7 result in a value that is too low or too high?
Too high.
AToo low.
BEvaluate $x^3+3x^2$x3+3x2 when $x=5.5$x=5.5
Does $x=5.5$x=5.5 result in a value that is too low or too high?
Too low
AToo high
BHence which of the following is the next best approximation to the solution of the equation, correct to 1 decimal place?
$x=5.4$x=5.4
A$x=5.6$x=5.6
B
QUESTION 3
The populations of two competing species of animals is given by $258t+5$258t+5 and $\frac{400}{t-5}$400t−5, where each expression represents the respective populations $t$t years from now. Tracy is looking to determine approximately when their populations will be equal.
She first equates the two functions and forms a quadratic equation.
Write down the quadratic equation in the form $at^2+bt+c=0$at2+bt+c=0, where $a>0$a>0.
She tests various $t$t values as presented in the table. Complete the table, rounding all values to two decimal places if necessary.
$t$t $5$5 $6$6 $7$7 $8$8 $9$9 $258t^2-1285t-425$258t2−1285t−425 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Using the table, in which interval of time will the two populations be equal?
A value of $t$t that is slightly less than $5$5.
AA value of $t$t that is slightly less than $6$6.
B$t=5$t=5
CA value of $t$t that is slightly greater than $5$5.
DDetermine the value of $258t^2-1285t-425$258t2−1285t−425 for $t=5.5$t=5.5
Will the populations of the two competing species be equal closer to $5$5 years from now or closer to $5.5$5.5 years from now?
$5.5$5.5 years from now
A$5$5 years from now
B