How do you solve $3 y^{2} - 48 = 0$ $3 y ^2 - 48 = 0$ ?

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How do you solve $3 y^{2} - 48 = 0$ $3 y ^2 - 48 = 0$ ?

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Answer 1 · 2016-05-20T14:59:24

$y = \pm 4$ $y = +-4$

Explanation:

We see that there are only two terms in the given problem. One contains the unknown $y$ $y$ and other is a constant.
$3 y^{2} - 48 = 0$ $3y^2- 48=0$

Step 1. Keepning the term which contains $y$ $y$ on the LHS, move the contant term to RHS of the equation. We get

$3 y^{2} = 48$ $3y^2= 48$

Step 2. Divide both sides by the common facotor $3$ $3$ .
$\frac{3 y^{2}}{3} = \frac{48}{3}$ $(3y^2)/3 = 48/3$
Simplify

$y^{2} = 16$ $y^2 = 16$

Step 3. To solve for $y$ $y$ , Take square root of both sides
$\sqrt{y^{2}} = \sqrt{16}$ $sqrt(y^2) = sqrt16$
Simplify and remember to place $= \pm$ $=+-$ sign in front of one of the two terms. Selected RHS

$y = \pm 4$ $y = +-4$

Answer 2 · 2016-05-20T15:01:25

$y = 4, - 4$ $y=4, -4$

Explanation:

$3 y^{2} - 48 = 0$ $3y^2-48=0$

$3$ $3$ is a common factor between the two terms.
Take $3$ $3$ and write the remaining terms in brackets:

$3 (y^{2} - 16) = 0$ $3(y^2-16) = 0$

Now, the product of $3$ $3$ and $(y^{2} - 16)$ $(y^2-16)$ is $0$ $0$ .

This means that $atleast ONE of the two terms is 0$ $color(red)("atleast ONE of the two terms is " 0)$ .

$\Rightarrow 3 = 0 or y^{2} - 16 = 0$ $=> 3=0 " or " y^2-16=0$

Obviously, $3 \neq 0$ $3!=0$

Then, $y^{2} - 16 = 0$ $y^2-16=0$

Or $y^{2} - 4^{2} = 0$ $y^2-4^2=0$

This is in the form, $a^{2} - b^{2}$ $color(red)(a^2-b^2)$

And we know that, $a^{2} - b^{2} = (a - b) (a + b)$ $color(red)(a^2-b^2=(a-b)(a+b))$

$\Rightarrow y^{2} - 4^{2} = 0$ $=>y^2-4^2=0$

$\Rightarrow (y - 4) (y + 4) = 0$ $=>(y-4)(y+4)=0$

Again, $atleast ONE of the two terms is 0$ $color(red)("atleast ONE of the two terms is " 0)$ .

$\Rightarrow y - 4 = 0 or y + 4 = 0$ $=>y-4=0 " or "y+4=0$

$\Rightarrow y = 4 or y = - 4$ $=>y=4 " or "y=-4$

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How do you solve $3 y^{2} - 48 = 0$ $3 y ^2 - 48 = 0$ ?

2 Answers

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