What are the zero(s) of: $3 x^{2} - 5 x - 4 = 0$ $3x^2 - 5x -4 = 0$ ?

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What are the zero(s) of: $3 x^{2} - 5 x - 4 = 0$ $3x^2 - 5x -4 = 0$ ?

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Answer 1 · 2018-05-22T21:50:30

$x = \frac{5}{6} \pm \frac{\sqrt{73}}{6}$ $x = 5/6+-sqrt(73)/6$

Explanation:

Given:

$3 x^{2} - 5 x - 4 = 0$ $3x^2-5x-4 = 0$

The difference of squares identity can be written:

$A^{2} - B^{2} = (A - B) (A + B)$ $A^2-B^2 = (A-B)(A+B)$

Complete the square and use this with $A = (6 x - 5)$ $A=(6x-5)$ and $B = \sqrt{73}$ $B=sqrt(73)$ as follows:

$0 = 12 (3 x^{2} - 5 x - 4)$ $0 = 12(3x^2-5x-4)$

$0 = 36 x^{2} - 60 x - 48$ $color(white)(0) = 36x^2-60x-48$

$0 = {(6 x)}^{2} - 2 (6 x) (5) + 25 - 73$ $color(white)(0) = (6x)^2-2(6x)(5)+25-73$

$0 = {(6 x - 5)}^{2} - {(\sqrt{73})}^{2}$ $color(white)(0) = (6x-5)^2-(sqrt(73))^2$

$0 = ((6 x - 5) - \sqrt{73}) ((6 x - 5) + \sqrt{73})$ $color(white)(0) = ((6x-5)-sqrt(73))((6x-5)+sqrt(73))$

$0 = (6 x - 5 - \sqrt{73}) (6 x - 5 + \sqrt{73})$ $color(white)(0) = (6x-5-sqrt(73))(6x-5+sqrt(73))$

Hence:

$6 x = 5 \pm \sqrt{73}$ $6x = 5+-sqrt(73)$

Hence:

$x = \frac{5}{6} \pm \frac{\sqrt{73}}{6}$ $x = 5/6+-sqrt(73)/6$

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What are the zero(s) of: $3 x^{2} - 5 x - 4 = 0$ $3x^2 - 5x -4 = 0$ ?

1 Answer

Explanation:

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