How do you solve the equation ${(3 x + 2)}^{2} - 49 = 0$ $(3x+2)^2-49=0$ ?

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How do you solve the equation ${(3 x + 2)}^{2} - 49 = 0$ $(3x+2)^2-49=0$ ?

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Answer 1 · 2017-02-19T18:16:55

$x = \frac{5}{3}$ $x=5/3$ and $x = - 3$ $x=-3$

Explanation:

${(3 x + 2)}^{2} - 49 = 0$ $(3x+2)^2-49=0$ can be written as

${(3 x + 2)}^{2} - 7^{2} = 0$ $(3x+2)^2-7^2=0$

As $a^{2} - b^{2}$ $a^2-b^2$ can be factorized as $(a + b) (a - b)$ $(a+b)(a-b)$ , this can be factorized as

$(3 x + 2 + 7) (3 x + 2 - 7) = 0$ $(3x+2+7)(3x+2-7)=0$

Hence, either $3 x + 2 + 7 = 0$ $3x+2+7=0$ i.e. $3 x = - 9$ $3x=-9$ and $x = - 3$ $x=-3$

or $3 x + 2 - 7 = 0$ $3x+2-7=0$ i.e. $3 x = 7 - 2 = 5$ $3x=7-2=5$ and $x = \frac{5}{3}$ $x=5/3$

Answer 2 · 2017-02-19T18:22:35

Alternative to get $x = \frac{5}{3}$ $x=5/3$ and $x = - 3$ $x=-3$

Explanation:

You can also treat this as a completing the square problem where the step of making the square is already done for you.

Move the 49 to the other side:

${(3 x + 2)}^{2} = 49$ $(3x+2)^2=49$

Square root both sides:

$3 x + 2 = (\frac{+}{-}) 7$ $3x+2=(+/-)7$

Treat it as two problems and solve for x

$3 x + 2 = 7$ $3x+2=7$

$3 x = 5$ $3x=5$

$x = \frac{5}{3}$ $x=5/3$

That's the first answer

$3 x + 2 = - 7$ $3x+2=-7$

$3 x = - 9$ $3x=-9$

$x = - 3$ $x=-3$

That's the second answer.

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How do you solve the equation ${(3 x + 2)}^{2} - 49 = 0$ $(3x+2)^2-49=0$ ?

2 Answers

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How do you solve the equation (3x+2)2−49=0(3x+2)^2-49=0?

2 Answers

Explanation:

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How do you solve the equation ${(3 x + 2)}^{2} - 49 = 0$ $(3x+2)^2-49=0$ ?