Topics
Factorization and Quadratics
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CanadaON
Grade 12

Solve quadratics using the variable substitution method

Lesson
Practice
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+3x2−10=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+3x2−10=p2+3p−10. 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: $x^4-20x^2+64=0$x4−20x2+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)2−9×2x+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.

Outcomes

12F.C.3.2

Factor polynomial expressions in one variable, of degree no higher than four, by selecting and applying strategies

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