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Anvisha [2.4K]
3 years ago
15

PLEASE HELP ME NOW URGENT

Mathematics
2 answers:
timama [110]3 years ago
8 0

ANSWER

y   = 3 \pm \sqrt{21}

EXPLANATION

The quadratic equation is:

{y}^{2}  - 6y - 12 = 0

Group variable terms:

{y}^{2}  - 6y = 12

Add the square of half, the coefficient of y to both sides.

{y}^{2}  - 6y  + ( - 3) ^{2} = 12 + ( - 3) ^{2}

{y}^{2}  - 6y  + 9= 12 + 9

The LHS us now a perfect square trinomial:

{(y - 3)}^{2}= 21

Take square root:

y - 3 =  \pm \sqrt{21}

y = 3 \pm \sqrt{21}

The first choice is correct.

Pie3 years ago
7 0

3±√21. The equation y^{2}-6y-12=0 has two possible solutions 3+√21 y 3-√21.

If we have a general quadratic equation ay^{2} +by+c=0 we can solves the equation by completing the square. First, we divide the quadratic equation by a, we obtain y^{2} +\frac{b}{a} y+\frac{c}{a} =0.

For this problem, we have y^{2}-6y-12=0

We can skipped division in this example since the coefficient of x^{2} is 1.

Move the term c to the right side of the equation

y^{2}-6y=12

Completing the square on the left side of the equation and balance this by adding the same number to the right side of the equation, with b = -6.

(\frac{b}{2})^{2} =(\frac{-6}{2})^{2}=(-3)^{2} =9

y^{2}-6y+9=12+9

(y-3)^{2}=21

Take the square root on both sides of the equation:

y - 3 = ±√21

Add 3 from both sides:

y = 3 ± √21

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