UNIT 3 EXPRESSIONS AND EQUATIONS Lesson 4: Fundamental Theorem of Algebra. Instruction. Guided Practice Example 1
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1 Guded Practce 3.4. Example 1 Instructon For each equaton, state the number and type of solutons by frst fndng the dscrmnant. x + 3x =.4x x = 3x = x 9x + 1 = 6x 1. Fnd the dscrmnant of x + 3x =. The equaton x + 3x = s n the form ax + bx + c = wth a =, b = 3, and c =. Use the dscrmnant formula. b 4ac Dscrmnant formula (3) 4()( ) Substtute for a, 3 for b, and for c Smplfy. 49 The dscrmnant of x + 3x = s 49. The dscrmnant s postve, so the equaton has two real solutons. U Walch Educaton
2 Instructon. Fnd the dscrmnant of.4x x =. The equaton.4x x = s n the form ax + bx + c = wth a =.4, b = 1, and c =. Use the dscrmnant formula. b 4ac Dscrmnant formula ( 1) 4(.4)() Substtute.4 for a, 1 for b, and for c. ( 1) + Smplfy. 1 The dscrmnant of.4x x = s 1. The dscrmnant s postve, so the equaton has two real solutons. 3. Fnd the dscrmnant of 3x = x. The equaton 3x = x s not n the form ax + bx + c =, so begn by wrtng t n that form. 3x = x 3x x = 3x x + = Orgnal equaton Subtract x from both sdes. Add to both sdes. The equaton 3x x + = s n the form ax + bx + c =, wth a = 3, b =, and c =. Use the dscrmnant formula. b 4ac Dscrmnant formula ( ) 4(3)() Substtute 3 for a, for b, and for c. 4 6 Smplfy. 6 The dscrmnant of 3x = x s 6. The dscrmnant s negatve, so the equaton has two magnary solutons, whch together are a complex conjugate par. Walch Educaton U
3 Instructon 4. Fnd the dscrmnant of 9x + 1 = 6x. The equaton 9x + 1 = 6x s not n the form ax + bx + c =, so begn by wrtng t n that form. 9x + 1 = 6x 9x 6x + 1 = Orgnal equaton Subtract 6x from both sdes. The equaton 9x 6x + 1 = s n the form ax + bx + c = wth a = 9, b = 6, and c = 1. Use the dscrmnant formula. b 4ac Dscrmnant formula ( 6) 4(9)(1) Substtute 9 for a, 6 for b, and 1 for c Smplfy.. The dscrmnant of 9x + 1 = 6x s equal to, so the equaton has one real soluton, whch s a double root. U Walch Educaton
4 Instructon Example Solve by usng a property of square roots. Show that the quadratc formula gves the same solutons. Check your solutons. Explan how the graph of the related functon supports the type of solutons. 1. Determne whch property of square roots to use. There s no x-term, so you can solve the equaton by usng ths property of square roots: If x = a, then x=± a. Orgnal equaton Subtract 6 from both sdes. 1 x = Multply both sdes by. x =± 1 Apply ths property of square roots: If x = a, then x=± a. x=± 1 For any postve real number a, a= a. The solutons to are ± 1. Walch Educaton U
5 Instructon. Wrte the radcal n smplest form. x=± 1 x=± 1 x=± 1 x=± 6 x=± 6 x=± 4 1 x=± 4 1 x=± For any real numbers a and b wth b, Multply by, whch s equal to 1. For any real numbers a and b, a b= ab. Smplfy. Smplfy the denomnator. Factor 6 to show a perfect square factor. For any real numbers a and b, a b= ab. Smplfy. a a =. b b The solutons to, wrtten n smplest form, are 1 and 1. These solutons are a conjugate par and can be wrtten as + 1 and 1. U Walch Educaton
6 Instructon 3. Show that usng the quadratc formula gves the same solutons. The quadratc formula states that f ax + bx + c =, then b b ac x = ±. a Wrte the gven equaton. x + 1= 4 x = ± 4 ( )( 1 ) ( ) x = ± 4 1 x = ± 4 1 x = ± x = ± ±4 1 x = 1 ± 1 x = Multply both sdes by 1, the least common denomnator, to elmnate fractons. Substtute values for a, b, and c: a =, b =, c = 1. Smplfy. For any postve real number a, a= a. Factor 4 to show ts greatest perfect square factor, 16. For any real numbers a and b, ab = a b. Smplfy. The solutons to, found usng the quadratc formula, are 1 and 1, matchng those found n step 1. Walch Educaton U
7 Instructon 4. Check the solutons. Substtute each soluton nto the orgnal equaton,. Check 1 : Check 1 : = = = = ( )+ = ( )+ = = = = = = =. Graph the related quadratc functon f( x)= x + on a graphng calculator. On a TI-83/84: Step 1: Press [Y=]. Step : At Y 1, type n [(][1][ ][][)][X,T,θ,n][x ][+][(][6][ ][][)]. Step 3: Press [WINDOW] to change the vewng wndow. Step 4: At Xmn, enter [( )][9]. Step : At Xmax, enter [9]. Step 6: At Xscl, enter [1]. (contnued) U Walch Educaton
8 Instructon Step 7: At Ymn, enter [( )][6]. Step 8: At Ymax, enter [6]. Step 9: At Yscl, enter [1]. Step 1: Press [GRAPH]. On a TI-Nspre: Step 1: Press the [home] key. Step : Arrow over to the graphng con and press [enter], Step 3: Enter the equaton [(][1][ ][][)][ ][x ][+][(][6][ ][][)] and press [enter]. Step 4: Change the vewng wndow by pressng [menu], then arrow down to number 4: Wndow/Zoom, and clck the center button of the navgaton pad. Step : Choose 1: Wndow Settngs by pressng the center button. Step 6: Enter n the approprate XMn value, [ ][9], then press [tab]. Step 7: Enter the approprate XMax value, [9], then press [tab]. Step 8: Leave the XScale set to auto. Press [tab] twce to navgate to YMn and enter [ ][6]. Step 9: Press [tab] to navgate to YMax. Enter [6]. Press [tab] twce to leave YScale set to auto and to navgate to OK. Step 1: Press [enter] Walch Educaton U3-3.4.
9 Instructon 6. Explan how the graph of the related quadratc functon, f( x)= x +, supports the type of solutons. The graph of f( x)= x + has no x-ntercepts, whch verfes that the equaton has no real solutons. Example 3 Solve x + 6x 13 = by usng the quadratc formula. Check your solutons. Explan how the graph of the related functon supports the type of solutons. 1. Substtute values for the equaton nto the quadratc formula. b b ac The quadratc formula states that f ax + bx + c =, then x = ±. a x + 6x 13 = s wrtten n the form ax + bx + c = ; therefore, a = 1, b = 6, and c = 13. x + 6x 13 = Orgnal equaton ( )( ) ( ) x = ± x = 6 ± 36 x = 6 ± 16 x = 6 ± 16 x = 6 ± 4 x = 6 ± 4 x= 3± 4 Substtute values for a, b and c nto the formula: a = 1, b = 6, c = 13. Smplfy. For any postve real number a, a= a. Smplfy. The solutons of x + 6x 13 = are 3 + and 3. U Walch Educaton
10 Instructon. Check the solutons. Substtute the solutons nto the orgnal formula. Check 3 + : Check 3 : x + 6x 13 = x + 6x 13 = ( 3+ ) + 6( 3+ ) 13= ( 3 ) + 6( 3 ) 13= ( 3+ ) ( 3+ )+ 6( 3+ ) 13= ( 3 ) ( 3 )+ 6( 3 ) 13= ( )+ 6( 3+ ) 13= ( )+ 6( 3 ) 13= ( )+ 6( 3+ ) 13= ( )+ 6( 3 ) 13= = = = = = = = = 3. Explan how the graph of the related quadratc functon, f(x) = x + 6x 13, supports the type of solutons. y x f(x) = x + 6x 13 8 The graph of f(x) = x + 6x 13 has no x-ntercepts, whch verfes that the equaton x + 6x 13 = has no real solutons, and has two magnary solutons. Walch Educaton U
11 Example 4 Instructon Solve x + 9 = 8x by usng the quadratc formula. Check your solutons. Explan how the graph of the related functon supports the type of solutons. 1. Substtute values for the equaton nto the quadratc formula. b b ac The quadratc formula states that f ax + bx + c =, then x = ±. a x + 9 = 8x Orgnal equaton x + 8x + 9 = x = ± ( )( 9 ) ( ) x = 8 ± x = 8± 8 4 x = 8± 8 4 x = 8± 4 x = 8 ± 4 4 x= ± Add 8x to both sdes to get the form ax + bx + c =. Substtute values for a, b, and c nto the formula: a =, b = 8, c = 9. Smplfy. 4 For any postve real number a, a= a. Smplfy. The solutons of x + 9 = 8x are + and.. Check the solutons. Substtute the solutons nto x + 8x + 9 =. (contnued) U Walch Educaton
12 Instructon Check + : x + 8x + 9 = = = = = = = = Check : x + 8x + 9 = = = = = = = = Walch Educaton U
13 Instructon 3. Explan how the graph of the related quadratc functon, f(x) = x + 8x + 9, supports the type of solutons. y 4 f(x) = x + 8x x The graph of f(x) = x + 8x + 9 has no x-ntercepts, whch verfes that the equaton x + 8x + 9 = has no real solutons, and has two magnary solutons. U Walch Educaton
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