How do you solve $3 x^{2} - 2 x - 5 = 0$ $3x^2 - 2x - 5 = 0$ ?

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How do you solve $3 x^{2} - 2 x - 5 = 0$ $3x^2 - 2x - 5 = 0$ ?

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Answer 1 · 2016-05-09T07:08:32

$x = - 1, 1 \frac{2}{3}$ $x=-1, 1 2/3$

Explanation:

$3 x^{2} - 2 x - 5 = 0$ $3x^2-2x-5=0$

Let us factorize using the split the middle term method.

On inspection we observe that middle term can be split in two parts whose product is equal to the product of first and third term.
Two terms are $- 5 x and 3 x$ $-5x and 3x$
The equation becomes
$3 x^{2} + 3 x - 5 x - 5 = 0$ $3x^2+3x-5x-5=0$ , pairing the two and taking out common factors
$\Rightarrow (3 x^{2} + 3 x) - (5 x + 5) = 0$ $=>(3x^2+3x)-(5x+5)=0$ ,
don't forget to change the sign of $- 5$ $-5$ once placed inside the parenthesis.
$\Rightarrow 3 x (x + 1) - 5 (x + 1) = 0$ $=>3x(x+1)-5(x+1)=0$
$\Rightarrow (x + 1) (3 x - 5) = 0$ $=>(x+1)(3x-5)=0$
Either $(x + 1) = 0$ $(x+1)=0$ or $(3 x - 5) = 0$ $(3x-5)=0$
We obtain $x = - 1, \frac{5}{3}$ $x=-1, 5/3$
or $x = - 1, 1 \frac{2}{3}$ $x=-1, 1 2/3$

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How do you solve $3 x^{2} - 2 x - 5 = 0$ $3x^2 - 2x - 5 = 0$ ?

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How do you solve $3 x^{2} - 2 x - 5 = 0$ $3x^2 - 2x - 5 = 0$ ?