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​ALGEBRA 1 STATE TEST PRACTICE

Equivalent equations and solutions to equation

12/21/2014

 
Use the information provided to answer Part A and Part B.
Consider the equation (x² + 3)² + 21 = 10x² + 30.
Part A
Let u = x² + 3.  Which equation is equivalent to
(x² + 3)² + 21 = 10x² + 30 in terms of u?
A.)  u² + 10u + 51 = 0
B.)  u² - 10u + 51 = 0
C.)  u² + 10u + 21 = 0
D.)  u² - 10u + 21 = 0
Part B
What are the solutions of the equation
(x² + 3)² + 21 = 10x² + 30?  Select all that apply.
A.)  -4            E.)  2
B.)  -3            F.)  3
C.)  -2            G.)  4
D.)  0

Part A
If u = x² + 3, then replace the x² + 3 with u
(x² + 3)² + 21 = 10(x² + 3)   {factored 10 out of right side}
u² + 21 = 10u   {replaced (x² + 3) with u}
u² - 10u + 21 = 0   {subtracted 10u from each side}
D.)  u² - 10u + 21 = 0

Part B
Solve the equation u² - 10u + 21 = 0 for u
u² - 10u + 21 = 0
(u - 7)(u - 3) = 0   {factored into two binomials}
u - 7 = 0   or   u - 3 = 0   {set each factor equal to 0}
u = 7   or   u = 3   {solved each equation for u}

If u = x² + 3, then
x² + 3 = 7   or   x² + 3 = 3   {substituted x² + 3 in for u}
x² = 4   or   x² = 0   {subtracted 3 from each side of each equation}
x = 2 or -2 or 0   {took the square root of each side of each equation}

C.)  -2
D.)  0
E.)  2


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