Solving an Equation in Different Ways. We have learned several different ways to solve an equation in this section. Some equations can be tacked by more than one nethod. For example, the equation x−x−2=0 is of quadratic type: We can solve it by letting x=u and x=u2 and factoring. Or we could solve for x, squre each side, and then solve the resulting qudratic equation. Solve the following equations using both methods indicated, and show that you get the same final answers. a) x−x−2=0 Quadratic type; solve for the radical, and square b) 12(x−3)2+10x−3+1=0 Quadratic type; multiply by LCD

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Solving an Equation in Different Ways. We have learned several different ways to solve an equation in this section. Some equations can be tacked by more than one nethod. For example, the equation  x−x−2=0  is of quadratic type: We can solve it by letting x=u and x=u2 and factoring. Or we could solve for x, squre each side, and then solve the resulting qudratic equation. Solve the following equations using both methods indicated, and show that you get the same final answers.  a) x−x−2=0 Quadratic type; solve for the radical, and square  b) 12(x−3)2+10x−3+1=0 Quadratic type; multiply by LCD

Zachary Pickett · Accepted Answer

Step 1 Here we have to solve the equations in two different ways, a) x−x−2=0 b) 12(x−3)2+10(x−3)+1=0 Step 2 First method: a) x−x−2=0 let x=u ⇒x=u2 ⇒u2−u−2=0 ⇒u2−2u+u−2=0 ⇒u(u−2)+1(u−2)=0 ⇒(u−2)(u+1)=0 ⇒u=2 and u=−1 ⇒x=2 and x=−1 ⇒x=4 and x=1 x=4, 1 second method: a) x−x−2=0 x−2=2 taking square both of sides (x−2)2=(x)2x2+4−4x=x x2−4x−x+4=0 x2−5x+4=0 x2−4x−x+4=0 x(x−4)−1(x−4)=0 (x−4)(x−1)=0 x=4 and x=1 ⇒x=4, 1 for second equation : First method: b) 12(x−3)2+10(x−3)+1=0 let (x−3)=u 12u2+10u+1=0 12+10u+u2u2=0 u2+10u+12=0 compare with au2+bu+c=0 a=1, b=10, c=12

Solving an Equation in Different Ways. We have learned several differe

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