How do you solve $2a^2 - 9a = 5$ ?

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Answer 1 · 2016-03-22T20:15:29

$a=-1/2" or "a=-5$

Given: $" "2a^2-9a=5$

Write as: $" "2(a^2-9/2a-5/2)=0$

Divide both sides by 2

$a^2-9/2a-5/2=0$

As there is no coefficient in front of $a^2$ there are two possible starting point

$(-a+?)(-a+?)" or "(a+?)(a+?)$

Lets investigate positive $a$

Notice that there are halves in this equation and we have a constant of $5/2$

Lets test $1/2xx5$ giving

$(a+1/2)(a-5)-> a^2+1/2 a -5a-5/2$

But $1/2 a-5a -> 1/2 a-10/2 a= -9/2 a" "$ Which is what we need.
$color(blue)((a+1/2)(a-5)-> a=-1/2" or "a=-5)$

How do you solve 2a^2 - 9a = 52a^2 - 9a = 5?