Respuesta :

mixter17

Hello!

The answer is:

[tex]x_{1}=\sqrt{14}-2\\x_{2}=-\sqrt{14}-2[/tex]

Why?

First let's identify each term:

[tex]a=1\\\b=4\\c=10[/tex]

So,

[tex]x^{2}+4x=10[/tex]

Then, we need to add [tex](\frac{b}{2})^{2}[/tex]  to each side, resulting:

[tex]x^{2}+4x+(\frac{4}{2})^{2} =10+(\frac{4}{2})^{2}\\\x^{2}+4x+4 =10+4\\x^{2}+4x+4=14\\ (x+2)^{2}=14\\\sqrt{(x+2)^{2}} =\sqrt{14}\\x+2=+-\sqrt{14}[/tex]

So:

[tex]x_{1}=\sqrt{14}-2\\x_{2}=-\sqrt{14}-2[/tex]

For the second function we have:

[tex]-3x^{2}-6x=9}[/tex]

Dividing each side into -3 we have:

[tex]x^{2}+2x=-3}[/tex]

[tex]b=2[/tex]

Adding [tex](\frac{b}{2})^{2}[/tex] to each side, we have:

[tex]x^{2}+2x+(\frac{2}{2})^{2}=-3+(\frac{2}{2})^{2}[/tex]

[tex]x^{2}+2x+1=-3+1[/tex]

Then,

[tex](x+1)^{2}=-2[/tex]

[tex]\sqrt{(x+1)^{2}} =\sqrt{-2}\\1+x=+-\sqrt{-2}[/tex]

[tex]x+1=+-\sqrt{-2}\\x_{1}=-1-\sqrt{-2}\\x_{2}=-1+\sqrt{-2}[/tex]

Since there is no negative roots in the real numbers, there is no solution for the second equation.

Have a nice day!

zainebamir540

Answer to Q1:(19)

The solution of given equation are 2+√14 and 2-√14.

Step-by-step explanation:

Given equation is :

x²+4x  = 10

We have to solve above equation by completing the square.

Adding square of the half of 4 to both sides of above equation, we have

x +4x +(2)²  = 10+(2)²

(x+2)² = 10+4

(x+2)²  = 14

Taking square root to both sides of above equation, we have

x+2 = ±√14

Hence, x  = 2±√14

x  =  2+√14 x = 2-√14

Hence, the solution of given equation are 2+√14 and 2-√14.

Answer to Q2: (22)

There is no real solution of -3x²-6x-9 = 0.

Step-by-step explanation:

We have given a quadratic equation.

-3x²-6x-9 = 0

We have to find the solution of above equation by completing square.

Taking -3 common from given equation, we have

-3(x²+2x+3) = 0

x²+2x+3 = 0

Adding -3 to both sides of above equation, we have

x²+2x+3-3 = 0-3

x²+2x = -3

Adding (1)² to both sides of above equation, we have

x²+2x+(1)² = -3+(1)²

(x+1)² = -3+1

(x+1)² = -2

Taking square root to both sides of above equation we have

x+1 = ±√-2

x+1 = ±√2i  where i = √-1

x = -1±√2i

Hence, the solution of given equation are  -1±√2i.

There is no real solution of -3x²-6x-9 = 0.

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