Math, asked by angel614, 9 hours ago

show your solutions follow the steps and then answer​

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Answers

Answered by tennetiraj86
2

Step-by-step explanation:

1)

Given equation is x²+2x = -1

=> x²+2x+1 = 0

The standard form = x²+2x+1 = 0

2)

Given equation is y²+3 = y+5

=> y²+3-y-5 = 0

=> y²-y+(3-5) = 0

=> y²-y-2 = 0

The standard form = y²-y-2 = 0

3)

Given equation is -(y+2)(y-1) = 3

=> -[y(y-1)+2(y-1)] = 3

=> -(y²-y+2y-2) = 3

=> -(y²+y-2) = 3

=> -(y²+y-2) -3 = 0

=> -(y²+y-2+3) = 0

=> -(y²+y+1) = 0

=> y²+y+1 = 0

The standard form is y²+y+1 = 0

4)

Given equation is x(x+9)+15 = -1

=> x²+9x+15 = -1

=> x²+9x-15+1 = 0

=> x²+9x-14 = 0

The standard form is x²+9x-14 = 0

5)

Given equation is x+(6/x) = -5

=> (x²+6)/x = -5

=> x²+6 = -5x

=> x²+6+5x = 0

=> x²+5x+6 = 0

The standard form is x²+5x+6 = 0

6)

Given equation is 5/(3n-8) = n/(n+2)

On applying cross multiplication then

=> n(3n-8) = 5(n+2)

=> 3n²-8n = 5n+10

=> 3n²-8n -5n -10 = 0

=> 3n²-13n-10 = 0

The standard form is 3n²-13n-10 = 0

Used formulae:-

The standard quadratic equation is ax²+bx+c = 0

Answered by olbmpp459
0

Step-by-step explanation:

1. x^ + 2x + 1 = 0

2. y^ + y + 8 = 0

3. y^ - 2y = 0

4. x^ + 9x + 16 = 0

5. x^ + 4x + 6 = 0

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