Question

1. consider the equation $81x^4 - 225x^2 + 144 = 0$.

3a what substitution can we make to reduce this equation to a monic quadratic in the variable $p$?
$p = 9x^2$
3 great effort!
3b use the substitution $9x^2 = p$ and solve the resulting equation for $p$.
enter your next step here

Explanation:

Step1: Substitute \( p = 9x^2 \)

Given \( 81x^4 - 225x^2 + 144 = 0 \), note that \( 81x^4=(9x^2)^2 = p^2 \) and \( 225x^2 = 225\times\frac{p}{9}=25p \) (since \( p = 9x^2\Rightarrow x^2=\frac{p}{9} \), but actually we can rewrite \( 81x^4=(9x^2)^2=p^2 \) and \( 225x^2 = 225\times\frac{p}{9}=25p \) is wrong, correct way: \( 81x^4=(9x^2)^2 = p^2 \), \( 225x^2=225\times\frac{p}{9}? \) No, wait, \( p = 9x^2\Rightarrow x^2=\frac{p}{9} \), so \( 225x^2=225\times\frac{p}{9} = 25p \)? Wait no, original equation is \( 81x^4-225x^2 + 144 = 0 \). Since \( 81x^4=(9x^2)^2=p^2 \) and \( 225x^2 = 225\times\frac{p}{9} \)? No, \( 9x^2=p\Rightarrow x^2=\frac{p}{9} \), so \( 225x^2=225\times\frac{p}{9}=25p \). So substitute \( p = 9x^2 \) into the equation: \( p^2-25p + 144 = 0 \)

Step2: Solve the quadratic equation \( p^2 - 25p + 144 = 0 \)

We can factor the quadratic. We need two numbers that multiply to \( 144 \) and add up to \( - 25 \) (wait, the quadratic is \( p^2-25p + 144 \), so we need two numbers \( a \) and \( b \) such that \( a + b=-25 \) and \( ab = 144 \). Wait, \( 144 = 16\times9 \), and \( 16 + 9=25 \). So \( p^2-16p-9p + 144 = 0 \)

Step3: Factor by grouping

\( p(p - 16)-9(p - 16)=0 \)

Step4: Factor out \( (p - 16) \)

\( (p - 16)(p - 9)=0 \)

Step5: Solve for \( p \)

Set each factor equal to zero: \( p - 16 = 0\Rightarrow p = 16 \); \( p - 9 = 0\Rightarrow p = 9 \)

Answer:

\( p = 9 \) or \( p = 16 \)