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Solve each quadratic equation by completing the square. Give exact answers--no decimals.

Solve each quadratic equation by completing the square. Give exact answers--no decimals-example-1

2 Answers

4 votes

Answer to Q1:(19)

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

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.

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.

User Maytal
by
4.5k points
4 votes

Hello!

The answer is:


x_(1)=√(14)-2\\x_(2)=-√(14)-2

Why?

First let's identify each term:


a=1\\\b=4\\c=10

So,


x^(2)+4x=10

Then, we need to add
((b)/(2))^(2) to each side, resulting:


x^(2)+4x+((4)/(2))^(2) =10+((4)/(2))^(2)\\\x^(2)+4x+4 =10+4\\x^(2)+4x+4=14\\ (x+2)^(2)=14\\\sqrt{(x+2)^(2)} =√(14)\\x+2=+-√(14)

So:


x_(1)=√(14)-2\\x_(2)=-√(14)-2

For the second function we have:


-3x^(2)-6x=9}

Dividing each side into -3 we have:


x^(2)+2x=-3}


b=2

Adding
((b)/(2))^(2) to each side, we have:


x^(2)+2x+((2)/(2))^(2)=-3+((2)/(2))^(2)


x^(2)+2x+1=-3+1

Then,


(x+1)^(2)=-2


\sqrt{(x+1)^(2)} =√(-2)\\1+x=+-√(-2)


x+1=+-√(-2)\\x_(1)=-1-√(-2)\\x_(2)=-1+√(-2)

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

Have a nice day!

User Sergio Tanaka
by
5.2k points