1001 Algebra Problems.PDF

Set 40 (Page 95)

625. d.Apply the quadratic formula with a= 1,b= 0,

and c= –7 to obtain:

x= = =

±^ = ± = ± 7 

626. a.Apply the quadratic formula with a= 2,b= 0,
     and c= –1 to obtain:

x= = =

± = ± = ±

627. a.Apply the quadratic formula with a= 4,b= 3,
     and c= 0 to obtain:

x= =

= = = 0, –^34 

628. b.Apply the quadratic formula with a= –5,
     b= 20, and c= 0 to obtain:

x= = =

= = 0, 4

629. c.Apply the quadratic formula with a= 1,
     b= 4, and c= 4 to obtain:

x= = =

= –2 (repeated solution)

630. c.Apply the quadratic formula with a= 1,
     b= –5, and c= –6 to obtain:

x= = =

= = –1, 6

631. b.Apply the quadratic formula with a= 3,b= 5,
     and c= 2 to obtain:

x= = =

= = –1, –^23 

632. a.Apply the quadratic formula with a= 5,
     b= 0, and c= –24 to obtain:

x= = =

±= ±= ±

633. a.First, put the equation into standard form
     by moving all terms to the left side of the
     equation to obtain the equivalent equation
     2 x^2 +5x+ 4 = 0. Now, apply the quadratic for-
     mula with a= 2,b= 5, and c= 4 to obtain:

x= = =

=

634. a.Apply the quadratic formula with a= 1,
     b = –2 2 , and c= 3 to obtain:

x= =

= =

= =  2 ±i

635. b.First, put the equation into standard form
     by moving all terms to the left side of the
     equation to obtain the equivalent equation
     x^2 + 2x= 0. Now, apply the quadratic formula
     with a= 1,b= 2, and c= 0 to obtain:

x= = =

–2 ± 2 ^4  = = –2, 0–2 ±2 2

–(2) ±(2)^2 –4(1)(0)

2(1)

–b±b^2 –4ac

2 a

^2 ^2 ±2i

2

^2 ^2 ±–4

2

^2 ^2 ±8 –12

2

–(–2^2 ) ±(–2^2 )^2 –4(1)(3)

2(1)

–b±b^2 –4ac

2 a

–5 ±i^7 

4

–5–7

4

–(5)±(5)^2 –4(2)(4)

2(2)

–b±b^2 –4ac

2 a

^2 ^30 

5

^4 ^30 

10

^480 

10

–(0) ±(0)^2 –4(5)(–24)

2(5)

–b±b^2 –4ac

2 a

–5 ± 1

6

–5 ±^1 

6

–(5) ±(5)^2 –4(3)(2)

2(3)

–b±b^2 –4ac

2 a

5 ± 7

2

5 ±^49 

2

–(–5)(–5)^2 – 4(1)(–6)

2(1)

–b±b^2 – 4ac

2 a

–4 ±^0 

2

–(4)±(4)^2 –4(1)(4)

2(1)

–b±b^2 – 4ac

2 a

–20 ± 20

–10

–20 ±^20 ^2

–10

–(20)±(20)^2 – 4(–5)(0)

2(–5)

–b±b^2 – 4ac

2 a

–3 ± 3

8

–3 ±^9 

8

–(3)±(3)^2 – 4(4)(0)

2(4)

–b±b^2 – 4ac

2 a

^2 

2

^2 ^2 

4

^8 

4

–(0) ±(0)^2 – 4(2)(–1)

2(2)

–b±b^2 – 4ac

2 a

^2 ^7 

2

^28

2

–(0) ±(0)^2 – 4(1)(–7)

2(1)

–b ±b^2 – 4ac

2 a

ANSWERS & EXPLANATIONS–