Algebra Know-It-ALL
We need more points for reference. We can find two more points on the graph by plugging in values of x smaller and larger than ...
686 Worked-Out Solutions to Exercises: Chapters 21 to 29 Chapter 25 Here’s the binomial-cubed equation that we’ve been told to ...
We can morph the mixed products to get (31/2× 9 1/2)x^3 − (31/2× 144 1/2)x^2 + (31/2× 144 1/2)x − (121/2× 9 1/2)x^2 + (121/2× 14 ...
688 Worked-Out Solutions to Exercises: Chapters 21 to 29 We can multiply through by −1 to get 105 x^3 + 211 x^2 + 112 x+ 12 = 0 ...
Here’s the general binomial-trinomial equation once again: (a 1 x+b 1 )(a 2 x^2 +b 2 x+c)= 0 Using the product of sums rule, w ...
690 Worked-Out Solutions to Exercises: Chapters 21 to 29 We found the real roots x=−5/3 or x= 7/4 Let’s put −5/3 in place of x, ...
5 − 9 21 104 80 − 45 ## − 9 − 24 ## 5 − 9 21 104 80 − 45 − 120 # − 9 − 24 ## (^5) − 9 21 104 80 − 45 − 120 # − 9 − 24 − 16 # 5 − ...
692 Worked-Out Solutions to Exercises: Chapters 21 to 29 Because d= 0, we know that the quadratic we get by setting the trinomia ...
In the trinomial, the coefficient of x^2 is 1, the coefficient of x is 6, and the stand- alone constant is 9. We must find a num ...
694 Worked-Out Solutions to Exercises: Chapters 21 to 29 (a) The real root is found by solving x+ 3 = 0 That root is x=−3. Becau ...
and break it down to (x+ 2)^5 (− 3 x+ 1)^5 = 0 (c) Here is the original equation: (4x^2 − 9)^3 = 0 Here, the coefficient of x^2 ...
696 Worked-Out Solutions to Exercises: Chapters 21 to 29 and 2 x− 3 = 0 Those roots are x=−3/2 or x= 3/2. Because each binomial ...
The solution set is X= {−4, 4, −36}. The multiplicity of each root is the same as the power to which its binomial is raised in t ...
698 Worked-Out Solutions to Exercises: Chapters 21 to 29 All the coefficients, as well as the stand-alone constant, are integer ...
The numbers in the bottom row alternate in sign. This indicates that −3 is a lower bound for the real roots. Here is an outline ...
700 Worked-Out Solutions to Exercises: Chapters 21 to 29 Let a be the coefficient of x^2 , let b be the coefficient of x, and le ...
Next, let’s check (0,1) in the original two-variable quadratic equation: 2 x^2 −y+ 1 = 0 2 × 02 − 1 + 1 = 0 2 × 0 − 1 + 1 = 0 0 ...
702 Worked-Out Solutions to Exercises: Chapters 21 to 29 Let a be the coefficient of x^2 , let b be the coefficient of x, and le ...
Next, let’s check [j,(1−j3)] in the original quadratic equation: 2 x^2 − 3 x−y+ 3 = 0 2 j^2 − 3 j− (1 −j3)+ 3 = 0 2 × (−1)+ (−j3 ...
704 Worked-Out Solutions to Exercises: Chapters 21 to 29 x-value into either of the original quadratic functions. Let’s use the ...
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