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- 1. Chapter IV<br />Much of the study in quadratic equation consist of different solving equation, we have equation in quadratic form, equation containing radicals and equation reducible to quadratic equation. They have their own steps and procedures to be followed in order to solve the given equation. <br /> TARGET SKILLS: <br />At the end of this chapter, students are expected to:<br />• discuss solving equation on quadratic;<br />• determine the index and its radicals;<br /> • interpret the solution of the original equation; and<br />• select appropriate method in solving quadratic equation.<br />Lesson 10<br />Equation in Quadratic Form<br />OBJECTIVES:<br />At the end of this lesson, students are expected to:<br />identify equation in quadratic form;<br />select appropriate method in solving quadratic equation; and<br />change the equation in standard form.<br />Quadratic in Form<br />An equation is quadratic in form when it can be written in this standard form<br />where the same expression is inside both ( )'s.<br /> In other words, if you have a times the square of the expression following b plus b times that same expression not squared plus c equal to 0, you have an equation that is quadratic in form. If we substitute what is in the ( ) with a variable like t, then the original equation will become a quadratic equation.<br /> Solving Equations that are Quadratic in Form<br />Step 1: Write in Standard Form, , if needed. <br />If it is not in standard form, move any term(s) to the appropriate side by using the addition/subtraction property of equality. Also, make sure that the squared term is written first left to right, the expression not squared is second and the constant is third and it is set equal to 0.<br />Step 2: Substitute a variable in for the expression that follows b in the second term. <br />In other words, substitute your variable for what is in the ( ) when it is in standard form, . I’m going to use t for my substitution, but really you can use any variable as long as it is not the variable that is used in the original equation.<br />Step 3: Solve the quadratic equation created in step 2. <br />You can use any method you want to solve the quadratic equation: factoring, completing the square or quadratic formula.<br />Step 4: Find the value of the variable from the original equation. <br />Keep in mind that you are finding a solution to the original equation and that the variable you substituted in for in step 2 is not your original variable. Use the substitution that was used to set up step 2 and then solve for the original variable.<br />Step 5: Check your solutions. <br />In some cases, you will be working with rational exponents and square roots in your problems. Those types of equations can cause extraneous solutions. Recall that an extraneous solution is one that is a solution to an equation after doing something like raising both sides of an equation by an even power, but is not a solution to the original problem. Even though not all of the quadratic in form equations can cause extraneous solutions, it is better to be safe than sorry and just check them all.<br />Example 1: Solve the equation that is quadratic in form: .<br />Standard Form, <br />*Rewriting original equation to show it is quadratic in form *Note that (y squared) squared = y to the fourth *When in stand. form, let t = the expression following b.<br /> <br />Next, we need to substitute t in for y squared in the original equation. <br /> <br />*Original equation *Substitute t in for y squared<br />Note how we ended up with a quadratic equation when we did our substitution. From here, we need to solve the quadratic equation that we have created.<br /> <br />Solve the quadratic equation: factoring, completing the square or quadratic formula. <br />*Factor the trinomial *Use Zero-Product Principle *Set 1st factor = 0 and solve *Set 2nd factor = 0 and solve <br />Let's find the value(s) of y when t = -4:<br />*Plug in - 4 for t *Use square root method to solve for y *First solution *Second solution <br />Let's find the value(s) of y when t = 1:<br />*Plug in 1 for t *Use square root method to solve for y *First solution *Second solution <br />Example 2: Solve the equation that is quadratic in form: .<br /> Standard Form, <br /> <br />*Inverse of add. 3 is sub. 3 *Equation in standard form<br />Note how when you square x to the 1/3 power you get x to the 2/3 power, which is what you have in the first term.<br />* Rewriting original equation to show it is quadratic in form *Note that (x to the 1/3 power) squared = x to the 2/3 power *When in stand. form, let t = the expression following b.<br />Next, we need to substitute t in for x to the 1/3 power in the original equation. <br />*Original equation *Substitute t in for x to the 1/3 power<br /> <br />You can use any method you want to solve the quadratic equation: factoring, completing the square or quadratic formula.<br />*Factor the trinomial *Use Zero-Product Principle *Set 1st factor = 0 and solve *Set 2nd factor = 0 and solve <br /> <br />Let's find the value(s) of x when t = 3:<br />*Plug in 3 for t *Solve the rational exponent equation *Inverse of taking it to the 1/3 power is raising it to the 3rd power<br /> <br />Let's find the value(s) of x when t = -1:<br /> <br />*Plug in -1 for t *Solve the rational exponent equation *Inverse of taking it to the 1/3 power is raising it to the 3rd power<br />Let's double check to see if x = 27 is a solution to the original equation.<br />*Plugging in 27 for x *True statement<br /> <br />Since we got a true statement, x = 27 is a solution.<br />Let's double check to see if x = -1 is a solution to the original equation.<br /> <br />*Plugging in -1 for x *True statement<br />Since we got a true statement, x = -1 is a solution.<br />There are two solutions to this equation: x = 27 and x = -1.<br />Exercises:<br /><ul><li>a4 + 2a2 – 5 = 0
- 2. x2 – 3x + 2 = 0
- 3. s6 + 8s3 – 6 = 0
- 4. n2 – 6n + 10 = 0
- 5. g8 + 2g4 – g = 0</li></ul>-317500-619941Name: ___________________ Section: _______<br />Instructor: ________________ Date: _______ Rating: ____<br />Instruction: Solve the equation that is in quadratic form.<br /><ul><li>a8 + 2a4 – 8 = 0</li></ul> _____________________________________________________<br /><ul><li>l2 + 4l2 – 6 = 0</li></ul> _____________________________________________________<br /><ul><li>e4 – 8e2 – 3 = 0
- 6. _____________________________________________________
- 7. l6 – 10l – 5 = 0
- 8. _____________________________________________________
- 9. i10 – 8i5 – 4 = 0
- 10. _____________________________________________________
- 11. s6 – 5s3 – 25 = 0
- 12. _____________________________________________________
- 13. -349954-516441h2/4 + 8h1/4 – 12 = 0
- 14. _____________________________________________________
- 15. a6- 5a4 – 15 = 0
- 16. _____________________________________________________
- 17. n8 + 12n2 – 8 = 0
- 18. _____________________________________________________
- 19. e9 – 3n3 – 10 = 0
- 20. _____________________________________________________
- 21. x2/3 – 2x 1/3 = 8
- 22. _____________________________________________________
- 23. x3/6 – 3x1/2 = 9</li></ul> _____________________________________________________<br /><ul><li>y2- 8y = 5</li></ul> _____________________________________________________<br /><ul><li>y4 + 2y2 = 6</li></ul> _____________________________________________________<br /><ul><li>x6 – 9x2 + 8 = 0</li></ul> _____________________________________________________<br />Lesson 11<br />Equation Containing Radicals<br />OBJECTIVES:<br />At the end of this lesson, students are expected to:<br />determine the index and its radicals;<br />positively respond to the note to be remembered; and<br />perform isolation of one radical if there are two radicals in the equation.<br />In the radicals nb which is read the “nth root of b,” the positive integer n is called the index or order of the radical, and b is called its radicand. When n is 2, 2 is no longer written, just simply write b instead of 2b to indicate the square root of b. thus 3b is read as “cube root of b”; 4b as “4th root of b”.<br />Note: <br />In order to solve for x, you must isolate x.<br />In order to isolate x, you must remove it from under the radial.<br />If there are two radicals in the equation, isolate one of the radicals.<br />Then raise both sides of the equation to a power equal to the index of the isolated radical.<br />Isolate the the remaining radical.<br />Raise both sides of the equation to a power equal to the index of the isolated radical.<br />You should now have a polynomial equation. Solve it.<br />Remember that you did not start out with a polynomial; therefore, there may be extraneous solutions. Therefore, you must check your answers. <br />Example 1: <br />First make a note of the fact that you cannot take the square root of a negative number. Therefore, the term is valid only if and the second term is valid if <br />Isolate the term. <br />Square both sides of the equation.<br />Isolate the term. <br />Square both sides of the equation. <br />Check the solution by substituting 9 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer. <br />Left side: <br />Right Side:1 <br />Since the left side of the original equation does not equal the right side of the original equation after we substituted our solution for x, then there is no solution.<br />You can also check the answer by graphing the equation: <br />.The graph represents the right side of the original equation minus the left side of the original equation.. The x-intercept(s) of this graph is (are) the solution(s). Since there are no x-intercepts, there are no solutions. <br />Exercises: <br />Solve each of the following equation.<br />x+1 = 3<br />x2-2x = x + 1<br />5x-1 = 11-x<br />3x+1 = 5<br />5+3x = 10<br />-419100-470912Name: ___________________ Section: _______<br />Instructor: ________________ Date: _______ Rating: ____<br />Instruction: Solve each of the following equation.<br />3x+1 =5<br /><ul><li> _____________________________________________________</li></ul>1-2x =3<br /><ul><li> _____________________________________________________</li></ul>5+3x=10<br /><ul><li> _____________________________________________________</li></ul>2x+1=3<br /><ul><li> _____________________________________________________</li></ul>32x+1=2<br /><ul><li> _____________________________________________________</li></ul>32x=4<br /><ul><li>_____________________________________________________</li></ul>-405130-35433032(x+1)=1<br /><ul><li> _____________________________________________________</li></ul>3x-3=2<br /><ul><li> _____________________________________________________</li></ul>x+1=3x+2<br /><ul><li> _____________________________________________________</li></ul>2x-5= x+2<br /><ul><li> _____________________________________________________</li></ul>3x-6=2 x2<br /><ul><li> _____________________________________________________</li></ul>32x-3= 3x+1<br /><ul><li> _____________________________________________________</li></ul>53x-2=3x-5<br /><ul><li> _____________________________________________________</li></ul>23x-8=4x+1<br /><ul><li> _____________________________________________________</li></ul>x-x+5=x-1<br /><ul><li> _____________________________________________________</li></ul>Lesson 12<br />Equations Reducible to Quadratic Equations<br /> OBJECTIVES:<br />At the end of this lesson, students are expected to:<br /><ul><li>interpret the solution of the original equation;
- 24. organize the equation if it is quadratic equation; and
- 25. solve the equation by factoring or quadratic formula.</li></ul>A variety of equations can be transformed into quadratic equations and solved by methods that we have discussed in the previous section. We will consider fractional equations, equations involving radicals and equation that can be transformed into quadratic equations by appropriate substitutions. Since the transformation process may introduce extraneous roots which are not solutions of the original equation, we must always check the solution in the original equation.<br />Example: Solve 1 1 7<br /> x+2 + x+3 = 12<br />Solution: First note that neither -2 nor -3 can be a solution since at either of these points the equation is meaningless.<br />Multiplying by the LCD, 12(x+2) (x+3), we get<br />12(x+3) + 12(x+2) = 7(x+2) (x+3)<br />24x + 60 = 7(x2 + 5x + 6)<br />or<br />7x2 + 11x – 18 = 0<br />Factoring, we get, (7x + 18)(x – 1) = 0<br />x = 1 or -18 <br /> 7<br />If x = 1, _1_ _1_ _1_ _1_ _7_<br /> 1+2 1+3 = 3 + 4 = 12<br />Therefore x = 1 is a solution.<br />If x = -18, __1__ + __1__<br /> 7 -18/7 + 2 -18/7 + 3<br />= __7__ + __7__<br />-18 + 14 -18 + 21<br />= _-7_ + _7_ = _7_<br /> 4 3 12<br />Therefore, x = _-18_ is a solution.<br /> 7<br />Example 2. 3x + 4 = x + 16 - 2<br />Solution: squaring both sides of the equation, we obtain,<br />(3x+4) 2 = x + 6 + 4<br />2x – 16 = 4 x + 16 <br />Dividing both sides by 2 gives, x – 8 = -2 x + 16 <br />Squaring both sides of the equation we get <br /> (x - 8) = (–2x + 16) 2<br />x2 - 16x + 64 = 4 (x +16)<br />x2 – 20x = 0 <br />x(x – 20) =0<br />x = 0 or x = 20 <br />Check: if x = 0, 30+ 4 = 20+16- 2 64 = 36 - 2 <br />8 = 6 – 2 <br />8 ≠ 4<br />Therefore x = 20 is not a solution of the original equation.<br />Thus the only root of 3x+4 = x+16 - 2 is 0.<br />Many equations are not quadratics equations. However, we can transform them by means of appropriate substitutions into quadratics equations and then solve these by techniques that we know.<br />Example: Solve:<br />a. 2x-2 – 7x-1 + 3 = 0<br /><ul><li>x4 – 2x2 – 2 = 0
- 26. x4x+12 – x4x+1- 2=0</li></ul>Solutions <br />a. Let u = x-1. Then u2 = (x-1)2 = x-2 and our equation becomes <br />2u2 – 7u + 3 = 0, a quadratic equation in u.<br />To solve the equation, we factor the left-hand side.<br />(2u – 1)( u – 3) = 0<br />U = ½ or u = 3<br />Since u = x-1, x-1= 1x=12 or x-1 = 3, from which<br />x = 2 or x = 13<br />Check: if x = 2, 2(2-2) – 7(2-1) + 3 = 24-72+3=0<br />Thus x = 2 is solution<br />If x = 1/3, 2(1/3)-2 – 7(1/3)-1 + 3 = 2(3)2 – 7(3) + 3<br />So, x = 1/3 is a solution.<br /><ul><li>Let u = x2. Then u2 = x4 and the given equation becomes a quadratic equation in u.
- 27. u2 – 2u – 2 = 0</li></ul>By the quadratic formula;<br />u = 2± -22-41-221 = 2±21+22=1±3<br />u = 1 + 3 or u = 1-3<br />Since u = x2 and u = 1-3 < 0, we have to discard this solution.<br />u = x2 = 1+3 implies<br />x = ± 1+ 3<br />It is simple to verify that both values of x satisfy the original equation. The roots of x4 – x2 – 2 = 0 are 1+ 3 and - 1+ 3.<br />Let u = x4x+1 . This substitution yields a quadratic equation in u.<br />u2 – u – 2 = 0<br />(u – 2)(u + 1) = 0 <br />u = 2 0r u = ˉ1<br />u = x4x+1 = 2 implies x = 2(4x + 1)<br /> or x = ˉ27<br />u = x4x+1 = 1 implies x = ˉ4x – 1<br /> or x = ˉ15<br />Again, it can easily be verified that both solutions check in the original equation.<br /> The roots are ˉ27 and ˉ15.<br />Exercise:<br />Solve the following equation by transforming each into a quadratic equation by a suitable substitution.<br />y - y - 20 = 0<br />r6 + 26 r3 – 27 = 0<br />(x – 4)3 + (x – 4)3/2 – 6 = 0<br />-440223-252096Name: ___________________ Section: _______<br />Instructor: ________________ Date: _______ Rating: ____<br /> <br />Instruction: Solve the following equation.<br />1x+1+ 43x+6= 23<br /><ul><li> _____________________________________________________</li></ul>1x+ 1x + 1= 920<br /> _____________________________________________________<br />x+2x+3= 2x-33x-7<br /><ul><li> _____________________________________________________</li></ul>x-1x+ xx-1=3<br /><ul><li> _____________________________________________________</li></ul>3x-12 15= 16-3xx+6+ x+35<br /><ul><li> _____________________________________________________</li></ul>3x+2= 2 6-x2x<br /><ul><li> _____________________________________________________</li></ul>-443592-202169x-2x-3x-1(x-2)+3x+1x+1 =5<br /><ul><li> _____________________________________________________</li></ul>x-2+1=x+1<br /><ul><li> _____________________________________________________</li></ul>x+4=2x+7<br /><ul><li> _____________________________________________________</li></ul>x+6=2x+6<br /><ul><li> _____________________________________________________</li></ul>x-3x+4=2x+2<br /><ul><li> _____________________________________________________</li></ul>3x+4=x+16-3<br /><ul><li> _____________________________________________________</li></ul>3x+2+6x=4<br /><ul><li> _____________________________________________________</li></ul>5y+19+y=7<br /><ul><li> _____________________________________________________</li></ul>x+3-x-2=1x+3<br /><ul><li> _____________________________________________________</li></ul>Solve for x.<br /><ul><li>x + 4 = 4
- 28. 3x + 4 - 5 = 0
- 29. 4x = 45
- 30. 3x - 5 = x+9
- 31. x²+15x = x + 2
- 32. Reduce to quadratic equation.
- 33. x4 – 5x + 4 = 0
- 34. 4(x + 3) + 5x+3 = 21
- 35. x2/3 – 5x1/3 – 6 = 0
- 36. (x2 + 4x)2 – (x2 + 4x) = 20
- 37. 2x4 – 9x2 + 7 = 0
- 38. Solve for x.
- 39. 6x+1 + 11-x = 2x
- 40. 1x-2 + 3x-3 + 52-x = 0
- 41. 2x+13 + 22x+1 = 53
- 42. x-1x+2 + x+3x-2 = 2
- 43. x-1x+2 + x+3x²-4 = 3x-15(x-2)

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