Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.
1. [tex]$x^2-3 x-4=0$[/tex]
2. [tex]$-x^2+6 x-9=0$[/tex]
3. [tex]$x^2-x-12=0$[/tex]
4. [tex]$x^2+4 x-3=0$[/tex]
5. [tex]$x^2-10 x=-21$[/tex]
6. [tex]$6 x^2-13 x-15=0$[/tex]
7. NUMBER THEORY Find two numbers that have a sum of 2 and a product of -15.
Answer (2)
The quadratic equations were solved primarily by factoring, and where factoring was not feasible, the quadratic formula was employed. The solutions include integral roots and approximations where necessary. The number theory problem resulted in the numbers 5 and -3, satisfying the given conditions.
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Factored or used the quadratic formula to find the roots of the quadratic equations.
Estimated non-integer roots to the nearest tenth.
Solved a system of equations to find two numbers with a given sum and product.
The solutions are: 16. x = 4 , − 1 , 17. x = 3 , 18. x = 4 , − 3 , 19. x ≈ 0.6 , − 4.6 , 20. x = 3 , 7 , 21. x = 3 , − 0.8 , 22. The two numbers are 5 and -3.
Explanation
Problem Analysis We are given several quadratic equations to solve by graphing. If the roots are integers, we will identify them exactly. If not, we will estimate them to the nearest tenth. We also have a number theory problem to solve.
Solving Equation 16 For equation 16, x 2 − 3 x − 4 = 0 , we can factor the quadratic as ( x − 4 ) ( x + 1 ) = 0 . Thus, the roots are x = 4 and x = − 1 .
Solving Equation 17 For equation 17, − x 2 + 6 x − 9 = 0 , we can multiply by -1 to get x 2 − 6 x + 9 = 0 . This factors as ( x − 3 ) 2 = 0 . Thus, the root is x = 3 (a repeated root).
Solving Equation 18 For equation 18, x 2 − x − 12 = 0 , we can factor the quadratic as ( x − 4 ) ( x + 3 ) = 0 . Thus, the roots are x = 4 and x = − 3 .
Solving Equation 19 For equation 19, x 2 + 4 x − 3 = 0 , this quadratic does not factor easily. We can use the quadratic formula to find the roots: x = 2 a − b ± b 2 − 4 a c = 2 ( 1 ) − 4 ± 4 2 − 4 ( 1 ) ( − 3 ) = 2 − 4 ± 16 + 12 = 2 − 4 ± 28 = 2 − 4 ± 2 7 = − 2 ± 7 . Since 7 ≈ 2.6 , the roots are approximately − 2 + 2.6 = 0.6 and − 2 − 2.6 = − 4.6 .
Solving Equation 20 For equation 20, x 2 − 10 x = − 21 , we rewrite it as x 2 − 10 x + 21 = 0 . This factors as ( x − 3 ) ( x − 7 ) = 0 . Thus, the roots are x = 3 and x = 7 .
Solving Equation 21 For equation 21, 6 x 2 − 13 x − 15 = 0 , this quadratic does not factor easily. We can use the quadratic formula to find the roots: x = 2 a − b ± b 2 − 4 a c = 2 ( 6 ) 13 ± ( − 13 ) 2 − 4 ( 6 ) ( − 15 ) = 12 13 ± 169 + 360 = 12 13 ± 529 = 12 13 ± 23 . Thus, the roots are x = 12 13 + 23 = 12 36 = 3 and x = 12 13 − 23 = 12 − 10 = − 6 5 ≈ − 0.8 .
Solving Problem 22 For problem 22, we need to find two numbers a and b such that a + b = 2 and ab = − 15 . From the first equation, b = 2 − a . Substituting into the second equation, a ( 2 − a ) = − 15 , which gives 2 a − a 2 = − 15 , or a 2 − 2 a − 15 = 0 . This factors as ( a − 5 ) ( a + 3 ) = 0 . Thus, a = 5 or a = − 3 . If a = 5 , then b = 2 − 5 = − 3 . If a = − 3 , then b = 2 − ( − 3 ) = 5 . So the two numbers are 5 and -3.
Final Answers Final Answers:
x = 4 , − 1
x = 3
x = 4 , − 3
x ≈ 0.6 , − 4.6
x = 3 , 7
x = 3 , − 0.8
The two numbers are 5 and -3.
Examples
Quadratic equations are incredibly useful in physics, engineering, and economics. For example, when designing a bridge, engineers use quadratic equations to calculate the curve of suspension cables. Similarly, in economics, quadratic functions can model cost and revenue curves to determine the break-even points for a business. Understanding how to solve these equations helps in making informed decisions and predictions in these fields.
