Exercise 43 Page 568

In order to solve the given quadratic equation, let's use the Quadratic Formula. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a Let's start by rewriting the equation so that all of the terms are on the left-hand side.

- 3=x^2+4x

AddEqn

LHS+3=RHS+3

0=x^2+4x+3

RearrangeEqn

Rearrange equation

x^2+4x+3=0

Now, we can identify the values of a, b, and c. x^2+4x+3=0 ⇕ 1x^2+ 4x+ 3=0 We see that a= 1, b= 4, and c= 3. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 4±sqrt(4^2-4( 1)( 3))/2( 1)

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Solve for x and Simplify

CalcPow

Calculate power

x=- 4±sqrt(16-4(1)(3))/2(1)

Multiply

Multiply

x=- 4±sqrt(16-12)/2

AddTerms

Add terms

x=- 4±sqrt(4)/2

CalcRoot

Calculate root

x=- 4± 2/2

The solutions for this equation are x= - 4± 22. Let's separate them into the positive and negative cases.

x=- 4± 2/2
x_1=- 4+2/2	x_2=- 4-2/2
x_1=- 2/2	x_2=- 6/2
x_1=- 1	x_2=- 3

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=- 1 and x_2=- 3.