Exercise 2 Page 191

To solve the given equation by factoring, we will start by writing all the terms on the left-hand side. Then, we will factor out the greatest common factor. We have rewritten the left-hand side as a product of two factors. Now, we will apply the Zero Product Property to solve the equation. From Equation (I), we found that one solution is x=0. To find other solutions, we will solve Equation (II). To do so, we need to define another variable. If we let z=x^2, we can rewrite Equation (II) in terms of the z-variable. x^4-7x^2+12=0 ⇔ z^2-7z+12=0 Note that the above equation in terms of z is a quadratic equation. Thus, we can solve it using the Quadratic Formula. az^2+bz+c=0 ⇔ z=- b±sqrt(b^2-4ac)/2a To do so, we first need to identify a, b, and c. z^2-7z+12=0 ⇔ 1z^2+( - 7)z+ 12=0 We see that a= 1, b= - 7, and c= 12. Let's substitute these values into the formula and solve for z.

z=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

z=- ( - 7)±sqrt(( - 7)^2-4( 1)( 12))/2( 1)

▼

Solve for z

NegNeg

- (- a)=a

z=7±sqrt((- 7)^2-4(1)(12))/2(1)

CalcPow

Calculate power

z=7±sqrt(49-4(1)(12))/2(1)

IdPropMult

Identity Property of Multiplication

z=7±sqrt(49-4(12))/2

MultNegPos

(- a)b = - ab

z=7±sqrt(49-48)/2

SubTerms

Subtract terms

z=7±sqrt(1)/2

CalcRoot

Calculate root

z=7± 1/2

The solutions for this equation are z= 7± 12. Let's separate them into the positive and negative cases.

x=7± 1/2
z=7+1/2	z=7-1/2
z=8/2	z=6/2
z=4	z=3

We found that the solutions for z^2-7z+12=0 are z=3 and z=4. This means that x^2=3 and x^2=4. Let's solve these equations!

lcx^2=3 & (I) x^2=4 & (II)

(I), (II): sqrt(LHS)=sqrt(RHS)

lx=± sqrt(3) x=± 2

We found in total five solutions for x. x=- sqrt(3), x=sqrt(3), x=- 2, x=2, x=0