How do u solve this quadratic (1 Viewer)

Drdusk · Mar 29, 2020

Cujo10 said:
View attachment 28151

$= (3^x)^2 - 4(3^x) + 3$

$\text{Let u = }\hspace{1mm}3^x\Rightarrow u^2 - 4u + 3 = 0$

$\text{Which is a standard quadratic that you can solve for 'u' and then in the end substitute}\hspace{2mm}3^x\hspace{2mm}\text{where there's a 'u'}$

Arrowshaft · Mar 30, 2020

Drdusk said:
$= (3^x)^2 - 4(3^x) + 3$

$\text{Let u = }\hspace{1mm}3^x\Rightarrow u^2 - 4u + 3 = 0$

$\text{Which is a standard quadratic that you can solve for 'u' and then in the end substitute}\hspace{2mm}3^x\hspace{2mm}\text{where there's a 'u'}$

Ahem $u$ thank you very much, na jk

CM_Tutor · Apr 2, 2020

Cujo10 said:
$3^{2x} - 4(3^x) + 3 = 0$

This is an equation that is reducible to a quadratic, in this case by the substitution

$u = 3^x$

to produce the quadratic

$u^2 - 4u + 3 = 0$

Watch out for these being concealed by using index laws, such as by posing the above question as

$9^x - 3(3^x - 1) = 3^x$

or ones like

$9^{x + 2} = 3 - 2(3^{x + 2})$ (which also uses $u = 3^x$ and gives only one solution, $x = -2$ )

or

$9^x - 3^x = \cfrac{1}{3^x} - \cfrac{1}{9^x}$ (which is trickier, but also has only one solution, $x = 0$ ).

I'll leave you to consider what substitution is needed for the second one.

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How do u solve this quadratic (1 Viewer)

Cujo10

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