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India
Class X

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 $$x4+3x210=0.
 
Think: If we look at replacing every $$x2 in the equation with a $$p then we can rewrite the equation as a more familiar quadratic where $$p is the variable. 
Notice that $$x4=(x2)2, so this will become $$p2. We can then substitute $$3x2 with $$3p
 
Do: Our full substitution gives $$x4+3x210=p2+3p10. From here we could solve in any number of ways! Let's solve for $$p by completing the square. 
 
$$p2+3p10 $$= $$0
$$p2+3p $$= $$10
$$p2+3p+(32)2 $$= $$10+(32)2
$$(p+32)2 $$= $$10+94
$$(p+32)2 $$= $$494
$$p+32 $$= $$±72
$$p $$= $$±7232
$$p $$= $$7232  and $$7232
$$p $$= $$42  and $$102
$$p $$= $$2   and  $$5
 
BUT - remember that we made a substitution, and $$p=x2. So we haven't finished yet, we still need to solve for $$x. This is one of the most common mistakes, not finishing the question. 
 
$$p $$= $$2
Then    
$$x2 $$= $$2
$$x $$= $$±2
AND    
$$x2 $$= $$5
Since there is no real number that gives $$5 when squared, $$x2=5 has no real solutions. So the real roots to this function are $$x=2 or $$x=2.
 
Example 2
What are the roots of the quadratic equation $$(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.
 
$$(2x+1)2+2(2x+1)3 $$= $$0 substitute $$j=2x+1
$$j2+2j3 $$= $$0  
$$(j+3)(j1) $$= $$0  
So      
$$j+3 $$= $$0 Where $$j=3
$$j1 $$= $$0 Where $$j=1
       
Remember that $$j $$= $$2x+1  
Then      
$$j $$= $$3 becomes
$$2x+1 $$= $$3  
$$2x $$= $$4  
$$x $$= $$2  
And      
$$j $$= $$1 becomes
$$2x+1 $$= $$1  
$$2x $$= $$0  
$$x $$= $$0  
So the roots of the quadratic equation $$(2x+1)2+2(2x+1)3=0 are $$x=2 or $$x=0.

 

Let's have a look at some other questions.

Question 1

Solve for $$x: $$x420x2+64=0 .

Let $$p be equal to $$x2.

Question 2

Solve the following equation for $$x:

$$3(9x+10)2+19(9x+10)+20=0

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

Question 3

Consider the equation

$$(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=()

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

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