2 Solve a quadratic equation by finding square roots EXAMPLE 1Solve a quadratic equation by finding square rootsSolve x2 – 8x + 16 = 25.x2 – 8x + 16 = 25Write original equation.(x – 4)2 = 25Write left side as a binomial squared.x – 4 = +5Take square roots of each side.x = 4 + 5Solve for x.The solutions are = 9 and 4 –5 = – 1.ANSWER
3 EXAMPLE 2Make a perfect square trinomialFind the value of c that makes x2 + 16x + c a perfect square trinomial. Then write the expression as the square of a binomial.SOLUTIONSTEP 1162=8Find half the coefficient of x.STEP 2Square the result of Step 1.82 = 64STEP 3Replace c with the result of Step 2.x2 + 16x + 64
4 EXAMPLE 2Make a perfect square trinomialANSWERThe trinomial x2 + 16x + c is a perfect square when c = 64. Then x2 + 16x + 64 = (x + 8)(x + 8) = (x + 8)2.
5 GUIDED PRACTICEfor Examples 1 and 2Solve the equation by finding square roots.x2 + 6x + 9 = 36.ANSWER3 and –9.x2 – 10x + 25 = 1.ANSWER4 and 6.x2 – 24x = 100.ANSWER2 and 22.
6 GUIDED PRACTICEfor Examples 1 and 2Find the value of c that makes the expression a perfect square trinomial.Then write the expression as the square of a binomial.4.x2 + 14x + cANSWER49 ; (x + 7)25.x2 + 22x + cANSWER121 ; (x + 11)26.x2 – 9x + c81492ANSWER; (x – )2.
7 ( ) EXAMPLE 3 Solve ax2 + bx + c = 0 when a = 1 Solve x2 – 12x + 4 = 0 by completing the square.x2 – 12x + 4 = 0Write original equation.x2 – 12x = –4Write left side in the form x2 + bx.x2 – 12x + 36 = –4 + 36Add–122()=(–6)36to each side.(x – 6)2 = 32Write left side as a binomial squared.x – 6 =Take square roots of each side.x =Solve for x.x =Simplify:32=1624The solutions are 6 + 4and 6 – 42ANSWER
8 EXAMPLE 3Solve ax2 + bx + c = 0 when a = 1CHECKYou can use algebra or a graph.Algebra Substitute each solution in the original equation to verify that it is correct.Graph Use a graphing calculator to graphy = x2 – 12x + 4. The x-intercepts are about – 4 2 and
9 ( ) EXAMPLE 4 Solve ax2 + bx + c = 0 when a = 1 Solve 2x2 + 8x + 14 = 0 by completing the square.2x2 + 8x + 14 = 0Write original equation.x2 + 4x + 7 = 0Divide each side by the coefficient of x2.x2 + 4x = –7Write left side in the form x2 + bx.Add42()=to each side.x2 – 4x + 4 = –7 + 4(x + 2)2 = –3Write left side as a binomial squared.x + 2 = –3Take square roots of each side.x = – –3Solve for x.x = –2 + i 3Write in terms of the imaginary unit i.
10 EXAMPLE 4Solve ax2 + bx + c = 0 when a = 1The solutions are –2 + i3and –2 – i3 .ANSWER
11 EXAMPLE 5Standardized Test PracticeSOLUTIONUse the formula for the area of a rectangle to write an equation.
12 ( ) EXAMPLE 5 Standardized Test Practice 3x(x + 2) = 72 3x2 + 6x = 72 Length Width = Area3x2 + 6x = 72Distributive propertyx2 + 2x = 24Divide each side by the coefficient of x2.Add2()=1to each side.x2 – 2x + 1 =(x + 1)2 = 25Write left side as a binomial squared.x + 1 = + 5Take square roots of each side.x = –1 + 5Solve for x.
13 EXAMPLE 5Standardized Test PracticeSo, x = –1 + 5 = 4 or x = – 1 – 5 = –6. You can reject x = –6 because the side lengths would be –18 and –4, and side lengths cannot be negative.The value of x is 4. The correct answer is B.ANSWER
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