Quadratic Equations
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 $$x4+3x2−10=0.
Think: If we look at replacing every $$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 $$x4+3x2−10=p2+3p−10. From here we could solve in any number of ways! Let's solve for $$p by completing the square.
| $$p2+3p−10 | $$= | $$0 |
| $$p2+3p | $$= | $$10 |
| $$p2+3p+(32)2 | $$= | $$10+(32)2 |
| $$(p+32)2 | $$= | $$10+94 |
| $$(p+32)2 | $$= | $$494 |
| $$p+32 | $$= | $$±72 |
| $$p | $$= | $$±72−32 |
| $$p | $$= | $$72−32 and $$−72−32 |
| $$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+2j−3 | $$= | $$0 | |
| $$(j+3)(j−1) | $$= | $$0 | |
| So | |||
| $$j+3 | $$= | $$0 | Where $$j=−3 |
| $$j−1 | $$= | $$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: $$x4−20x2+64=0 .
Let $$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
$$(2x)2−9×2x+8=0
The equation can be reduced to a quadratic equation by using a certain substitution.
By filling in the gaps, determine the correct substitution that would reduce the equation to a quadratic.
Let $$m=()
Solve the equation for $$x by using the substitution $$m=2x.