Question

$3 {x}^{2} + 5x - 7 \\ a = 3, b = 5, c = (-7) \\ x = \frac{- b \: \frac{ + }{} \sqrt{ {b}^{2} - 4ac} }{2a} \\ x = \frac{- 5 \: \frac{ + }{} \sqrt{ {5}^{2} - 4(3)( - 7)} }{2a} \\ x = \frac{- 5 \: \frac{ + }{} \sqrt{25 + 84} }{6} \\ x = \frac{- 5 \: \frac{ + }{} \sqrt{ 109} }{6} \\ x = \frac{- 5 + \sqrt{ 109} }{6} \\ x = \frac{- 5 - \sqrt{ 109} }{6}$
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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

The last way to solve a quadratic equation is the Quadratic Formula. This formula is derived from completing the square for the equation

ax2+bx+c=0

(see #13 from the Problem Set in the previous concept). We will derive the formula here.

Investigation: Deriving the Quadratic Formula

Walk through each step of completing the square of

ax2+bx+c=0

.

1. Move the constant to the right side of the equation.

ax2+bx=−c

2. “Take out”

a

from everything on the left side of the equation.

a(x2+bax)=−c

3. Complete the square using

ba

.

(b2)2=(b2a)2=b24a2

4. Add this number to both sides. Don’t forget on the right side, you need to multiply it by

a

(to account for the

a

outside the parenthesis).

a(x2+bax+b24a2)=−c+b24a

5. Factor the quadratic equation inside the parenthesis and give the right hand side a common denominator.

a(x+b2a)2=b2−4ac4a

6. Divide both sides by

a

.

(x+b2a)2=b2−4ac4a2

7. Take the square root of both sides.

x+b2a=±b2−4ac−−−−−−−√2a

8. Subtract

b2a

from both sides to get

x

by itself.

x=−b±b2−4ac−−−−−−−√2a

This formula will enable you to solve any quadratic equation as long as you know

a,b

, and

c

(from

ax2+bx+c=0

).

Example A

Solve

9x2−30x+26=0

using the Quadratic Formula.

Solution: First, make sure one side of the equation is zero. Then, find

a,b,

and

c

.

a=9,b=−30,c=26

. Now, plug in the values into the formula and solve for

x

.

x=−(−30)±(−30)2−4(9)(26)−−−−−−−−−−−−−−√2(9)=30±900−936−−−−−−−−√18=30±−36−−−−√18=30±6i18=53±13i

Example B

Solve

2x2+5x−15=−x2+7x+2

using the Quadratic Formula.

Solution: Let’s get everything onto the left side of the equation.

2x2+5x−153x2−2x−13=−x2+7x+2=0

Now, use

a=3,b=−2,

and

c=−13

and plug them into the Quadratic Formula.

x=−(−2)±(−2)2−4(3)(−13)−−−−−−−−−−−−−−−√2(3)=2±4+156−−−−−−√6=2±160−−−√6=2±410−−√3

Example C

Solve

x2+20x+51=0

by factoring, completing the square, and the Quadratic Formula.

Solution: While it might not look like it, 51 is not a prime number. Its factors are 17 and 3, which add up to 20.

x2+20x+51(x+17)(x+13)x=0=0=−17,−3

Now, solve by completing the square.

x2+20x+51x2+20xx2+20x+100(x+10)2x+10x=0=−51=−51+100=49=±7=−10±7→−17,−3

Lastly, let’s use the Quadratic Formula.

a=1,b=20,c=51

.

x=−20±202−4(1)(51)−−−−−−−−−−−√2(1)=−20±400−204−−−−−−−−√2=−20±196−−−√2=−20±142=−17,−3

Notice that no matter how you solve this, or any, quadratic equation, the answer will always be the same.

Guided Practice

1. Solve

−6x2+15x−22=0

using the Quadratic Formula.

2. Solve

2x2−x−15=0

using all three methods.

Answers

1.

a=−6,b=15,

and

c=−22

x=−15±152−4(−6)(−22)−−−−−−−−−−−−−−√2(−6)=−15±225−528−−−−−−−−√−12=−15±i303−−−√−12=54±303−−−√12i

2. Factoring:

ac=−30

. The factors of -30 that add up to -1 are -6 and 5. Expand the

x−

term.

2x2−6x+5x−152x(x−3)+5(x−3)(x−3)(2x+5)x=0=0=0=3,−52

Complete the square

2x2−x−152x2−x2(x2−12x)2(x2−12x+116)2(x−14)2(x−14)2x−14x=0=15=15=15+18=1218=12116=±114=14±114→3,−52

Quadratic Formula

x=1±12−4(2)(−15)−−−−−−−−−−−−√2(2)=1±1+120−−−−−−√4=1±121−−−√4=1±114=124,−104→3,−52