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

Question

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

1 Answer

Impact of this question

2018 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-22T20:15:29

$a = - \frac{1}{2} or a = - 5$ $a=-1/2" or "a=-5$

Explanation:

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

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

Divide both sides by 2

$a^{2} - \frac{9}{2} a - \frac{5}{2} = 0$ $a^2-9/2a-5/2=0$

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

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

Lets investigate positive $a$ $a$

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

Lets test $\frac{1}{2} \times 5$ $1/2xx5$ giving

$(a + \frac{1}{2}) (a - 5) \to a^{2} + \frac{1}{2} a - 5 a - \frac{5}{2}$ $(a+1/2)(a-5)-> a^2+1/2 a -5a-5/2$

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

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 2a2−9a=52a^2 - 9a = 5?

1 Answer

Explanation:

Related questions

Impact of this question

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