Hong Kong
Stage 4 - Stage 5

# Variable substitution Method

Lesson

Sometimes functions don't even look like quadratics, but with some clever substitutions we can make it look like a quadratic to enable us to solve them.

##### Example 1
Solve the equation $x^4+3x^2-10=0$x4+3x210=0.

Think: If we look at replacing every $x^2$x2 in the equation with a $p$p then we can rewrite the equation as a more familiar quadratic where $p$p is the variable.
Notice that $x^4=\left(x^2\right)^2$x4=(x2)2, so this will become $p^2$p2. We can then substitute $3x^2$3x2 with $3p$3p

Do: Our full substitution gives $x^4+3x^2-10=p^2+3p-10$x4+3x210=p2+3p10. From here we could solve in any number of ways! Let's solve for $p$p by completing the square.

 $p^2+3p-10$p2+3p−10 $=$= $0$0 $p^2+3p$p2+3p $=$= $10$10 $p^2+3p+\left(\frac{3}{2}\right)^2$p2+3p+(32​)2 $=$= $10+\left(\frac{3}{2}\right)^2$10+(32​)2 $\left(p+\frac{3}{2}\right)^2$(p+32​)2 $=$= $10+\frac{9}{4}$10+94​ $\left(p+\frac{3}{2}\right)^2$(p+32​)2 $=$= $\frac{49}{4}$494​ $p+\frac{3}{2}$p+32​ $=$= $\pm\frac{7}{2}$±72​ $p$p $=$= $\pm\frac{7}{2}-\frac{3}{2}$±72​−32​ $p$p $=$= $\frac{7}{2}-\frac{3}{2}$72​−32​  and $\frac{-7}{2}-\frac{3}{2}$−72​−32​ $p$p $=$= $\frac{4}{2}$42​  and $\frac{-10}{2}$−102​ $p$p $=$= $2$2   and  $-5$−5

BUT - remember that we made a substitution, and $p=x^2$p=x2. So we haven't finished yet, we still need to solve for $x$x. This is one of the most common mistakes, not finishing the question.

 $p$p $=$= $2$2 Then $x^2$x2 $=$= $2$2 $x$x $=$= $\pm\sqrt{2}$±√2 AND $x^2$x2 $=$= $-5$−5
Since there is no real number that gives $-5$5 when squared, $x^2=-5$x2=5 has no real solutions. So the real roots to this function are $x=\sqrt{2}$x=2 or $x=-\sqrt{2}$x=2.

##### Example 2
What are the roots of the quadratic equation $\left(2x+1\right)^2+2\left(2x+1\right)-3=0$(2x+1)2+2(2x+1)3=0

Think: We could expand it completely, collect like terms and then solve. This would involve a lot of extra algebraic manipulation. Or, we could make a clever substitution...

Do: Let's see what happens when we let $j=2x+1$j=2x+1.

 $\left(2x+1\right)^2+2\left(2x+1\right)-3$(2x+1)2+2(2x+1)−3 $=$= $0$0 substitute $j=2x+1$j=2x+1 $j^2+2j-3$j2+2j−3 $=$= $0$0 $\left(j+3\right)\left(j-1\right)$(j+3)(j−1) $=$= $0$0 So $j+3$j+3 $=$= $0$0 Where $j=-3$j=−3 $j-1$j−1 $=$= $0$0 Where $j=1$j=1 Remember that $j$j $=$= $2x+1$2x+1 Then $j$j $=$= $-3$−3 becomes $2x+1$2x+1 $=$= $-3$−3 $2x$2x $=$= $-4$−4 $x$x $=$= $-2$−2 And $j$j $=$= $1$1 becomes $2x+1$2x+1 $=$= $1$1 $2x$2x $=$= $0$0 $x$x $=$= $0$0
So the roots of the quadratic equation $\left(2x+1\right)^2+2\left(2x+1\right)-3=0$(2x+1)2+2(2x+1)3=0 are $x=-2$x=2 or $x=0$x=0.

Let's have a look at some other questions.

##### Question 1

Solve for $x$x: $x^4-20x^2+64=0$x420x2+64=0 .

Let $p$p be equal to $x^2$x2.

##### Question 2

Solve the following equation for $x$x:

$3\left(9x+10\right)^2+19\left(9x+10\right)+20=0$3(9x+10)2+19(9x+10)+20=0

You may let $p=9x+10$p=9x+10.

##### Question 3

Consider the equation

$\left(2^x\right)^2-9\times2^x+8=0$(2x)29×2x+8=0

1. The equation can be reduced to a quadratic equation by using a certain substitution.

By filling in the gaps, determine the correct substitution that would reduce the equation to a quadratic.

Let $m=\left(\editable{}\right)^{\editable{}}$m=()

2. Solve the equation for $x$x by using the substitution $m=2^x$m=2x.