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.

$p^2+3p-10$`p`2+3`p`−10	$=$=	$0$0
$p^2+3p$`p`2+3`p`	$=$=	$10$10
$p^2+3p+\left(\frac{3}{2}\right)^2$`p`2+3`p`+(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
$x^2$`x`2	$=$=	$2$2
$x$`x`	$=$=	$\pm\sqrt{2}$±√2
AND
$x^2$`x`2	$=$=	$-5$−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.

$\left(2x+1\right)^2+2\left(2x+1\right)-3$(2`x`+1)2+2(2`x`+1)−3	$=$=	$0$0	substitute $j=2x+1$`j`=2`x`+1
$j^2+2j-3$`j`2+2`j`−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$2`x`+1
Then
$j$`j`	$=$=	$-3$−3	becomes
$2x+1$2`x`+1	$=$=	$-3$−3
$2x$2`x`	$=$=	$-4$−4
$x$`x`	$=$=	$-2$−2
And
$j$`j`	$=$=	$1$1	becomes
$2x+1$2`x`+1	$=$=	$1$1
$2x$2`x`	$=$=	$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

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

Variable substitution Method

Example 1

Example 2

Question 1

Question 2

Question 3

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