MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §8.1 Complete The Square
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review § Any QUESTIONS About §7.7 → Complex Numbers Any QUESTIONS About HomeWork §7.7 → HW MTH 55
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 3 Bruce Mayer, PE Chabot College Mathematics The Square Root Property Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and −5, which are the solutions to this equation. Thus we see that square roots can provide quick solutions for equations of the type x 2 = k.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 4 Bruce Mayer, PE Chabot College Mathematics SQUARE ROOT PROPERTY For any nonzero real number d, and any algebraic expression u, then the Equation u 2 = d has exactly two solutions: Alternatively in a ShortHand Notation:
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 5 Bruce Mayer, PE Chabot College Mathematics Example Use SqRt Property Solve 5x 2 = 15. Give exact solutions and approximations to three decimal places. SOLUTION Isolating x 2 Using the Property of square roots The solutions are which round to and − ShortHand Notation
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 6 Bruce Mayer, PE Chabot College Mathematics Example Use SqRt Property Solve x 2 = 108 SOLN Check Use the square root principle. Simplify by factoring out a perfect square. Check Note: Remember the ± means that the two solutions are and.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 7 Bruce Mayer, PE Chabot College Mathematics Example Use SqRt Property Solve x = 32 SOLN Subtract 14 from both sides to isolate x 2 Use the square root property Simplify by factoring out a perfect square
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example Use SqRt Property Solve (x + 3) 2 = 7 SOLN Using the Property of square roots The solutions are The check is left for us to do Later Solving for x ShortHand Notation
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 9 Bruce Mayer, PE Chabot College Mathematics Example Use SqRt Property Solve 16x = 0 SOLN The solutions are The check is left for Later Recall that
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 10 Bruce Mayer, PE Chabot College Mathematics Solving Quadratic Equations To solve equations in the form ax 2 = b, first isolate x 2 by dividing both sides of the equation by a. Solve an equation in the form ax 2 + b = c by using both the addition and multiplication principles of equality to isolate x 2 before using the square root principle In an equation in the form (ax + b) 2 = c, notice the expression ax + b is squared. Use the square root principle to eliminate the square.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example Use SqRt Property Solve (5x − 3) 2 = 4 SOLN Add 3 to both sides and divide each side by 5, to isolate x. Use the square root property or
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example Use SqRt Property Solve x 2 + 8x + 16 = 17 SOLUTION: Sometimes we can factor an equation to express it as a square of a binomial. Factoring the TriNomial square root property
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 13 Bruce Mayer, PE Chabot College Mathematics Completing the Square Not all quadratic equations can be solved as in the previous examples. By using a method called completing the square, we can use the principle of square roots to solve any quadratic equation To Complete-the-Sq we Add ZERO to an expression or equation
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example Complete the Sq Solve x x + 4 = 0 SOLN: x x + 25 = – Using the property of square roots Factoring Adding 25 to both sides. The solutions are The check is left for Later
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 15 Bruce Mayer, PE Chabot College Mathematics Solving Quadratic Equations by Completing the Square 1 Write the equation in the form 1·x 2 + bx = c. Complete the square by adding (b/2) 2 to both sides. (b/2) 2 is called the “Quadratic Supplement” Write the completed square in factored form. Use the square root property to eliminate the square. Isolate the variable. Simplify as needed.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example Complete the Sq Solve by Completing the Square: 2x 2 − 10x = 9 SOLN: Divide both sides by 2. Simplify. Add to both sides to complete the square.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 17 Bruce Mayer, PE Chabot College Mathematics Example Complete the Sq Solve 2x 2 − 10x = 9 SOLN: Combine the fractions. Factor. Add to both sides and simplify the square root. Use the square root principle.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example Complete the Sq Solve by Completing the Square: 3x 2 + 7x +1 = 0 1 SOLUTION: The coefficient of the x 2 term must be 1. When it is not, multiply or divide on both sides to find an equivalent eqn with an x 2 coefficient of 1. Divide Eqn by 3
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example Complete the Sq Solve: 3x 2 + 7x +1 = 0 SOLN:
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example Complete the Sq Solve: 3x 2 + 7x +1 = 0 SOLN: Square Root Property Isolate x Taking the Square Root of Both Sides
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example Taipei 101 Tower The Taipei 101 tower in Taiwan is 1670 feet tall. How long would it take an object to fall to the ground from the top? Familiarize: A formula for Gravity-Driven FreeFall with negligible air-drag is s = 16t 2 where –s is the FreeFall Distance in feet –t is the FreeFall Time in seconds
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example Taipei 101 Tower Translate: We know the distance is 1670 feet and that we need to solve for t Sub 1670 for s →1670 = 16t 2 CarryOut: 1670 = 16t 2
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example Taipei 101 Tower Check: The number −10.2 cannot be a solution because time cannot be negative. Check t = 10.2 in formula: s = 16(10.2) 2 = 16(104.04) = –This result is very close to the 1670 value. State. It takes about 10.2 seconds for an object to fall to the ground from the top of the Taipei 101 tower.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 24 Bruce Mayer, PE Chabot College Mathematics Compound Interest After one year, an amount of money P, invested at 4% per year, is worth 104% of P, or P(1.04). If that amount continues to earn 4% interest per year, after the second year the investment will be worth 104% of P(1.04), or P(1.04) 2. This is called compounding interest since after the first period, interest is earned on both the initial investment and the interest from the first period. Generalizing, we have the following.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 25 Bruce Mayer, PE Chabot College Mathematics Compound Interest Formula If an amount of money P is invested at interest rate r, compounded annually, then in t years, it will grow to the amount A as given by the Formula Note that r is expressed as a DECIMAL
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 26 Bruce Mayer, PE Chabot College Mathematics Example Compound Interest Tariq invested $5800 at an interest rate of r, compounded annually. In two years, it grew to $6765. What was the interest rate? Familiarize: This is a compound interest calculation and we are already familiar with the compound-interest formula.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 27 Bruce Mayer, PE Chabot College Mathematics Example Compound Interest Translate The translation consists of substituting into the Interest formula 6765 = 5800(1 + r) 2 CarryOut: Solve for r 6765/5800 = (1 + r) 2
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 28 Bruce Mayer, PE Chabot College Mathematics Example Compound Interest Check: Since the interest rate can NOT negative, we need only to check 0.08 or 8%. If $5800 were invested at 8% compounded annually, then in 2 yrs it would grow to 5800·(1.08) 2, or $6765. The number 8% checks. State: Tariq’s interest rate was 8%.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 29 Bruce Mayer, PE Chabot College Mathematics Solving Formulas Recall that to solve a formula for a certain letter-variable, we use the principles for solving equations to isolate that letter-variable alone on one side of the Equals-Sign The Bernoulli Equation for an InCompressible Fluid:
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 30 Bruce Mayer, PE Chabot College Mathematics Example Solve SOLN Multiplying both sides by 2 Complete the Square Express LHS as Perfect Square Solve Using Square Root Principle
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 31 Bruce Mayer, PE Chabot College Mathematics Solve a Formula for a Letter – Say, b 1.Clear fractions and use the principle of powers, as needed. Perform these steps until radicals containing b are gone and b is not in any denominator. 2.Combine all like terms. 3.If the only power of b is b 1, the equation can be solved without using exponent rules. 4.If b 2 appears but b does not, solve for b 2 and use the principle of square roots to solve for b. 5.If there are terms containing both b and b 2, put the equation in standard form and Complete the Square.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 32 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work Problems From §8.1 Exercise Set 22, 44, 56, 78, 88 Solve ax 2 + bx + c = 0 by completing the square:
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 33 Bruce Mayer, PE Chabot College Mathematics All Done for Today Taipei 101 Tower Taipei, R.o.C.
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 34 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics Appendix –
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 35 Bruce Mayer, PE Chabot College Mathematics Graph y = |x| Make T-table
MTH55_Lec-48_sec_8-1a_SqRt_Property.ppt 36 Bruce Mayer, PE Chabot College Mathematics