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3 years ago

Solve 6v(2v + 3)Please solve and explain this

2 answers:

8 0

$6v(2v+3)=6v\cdot2v+6v\cdot3=12v^2+18v$

$6v\cdot2v=6\cdot2\cdot v\cdot v=12\cdot v^2=12v^2\\\\6v\cdot3=6\cdot3\cdot v=18\cdot v=18v$

Dmitry_Shevchenko [17]3 years ago

5 0

6v(2v+3)
1) 6v*2v=12v^2
2) 6v*3=18v
3)put them together like this: 12v^2+18v this is your answer. If the sign had been a minus in front of the 3 your answer would have been a subtraction equation.

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Which equation has the solutions x=1+or-square root of 5?

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We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

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Complete the square. Remember to balance the equation by adding the same constants to each side.

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therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

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

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