Solving Linear & Quadratic Equations

Solve the linear equation for x:

2x + 5 = 12

• 7

• 3

• 3.5

• 8.5

Tags

CCSS.HSA.REI.A.1

CCSS.HSA.REI.B.3

Solve the linear equation for x:

6(2x + 1) = 18

• 3

• 17/12

• 1

• 2

Tags

CCSS.HSA.REI.A.1

CCSS.HSA.REI.B.3

Solve the linear equation for x:

4x + 1 = 7x - 5

• -1

• 2

• 6/11

• 8

Tags

CCSS.HSA.REI.A.1

CCSS.HSA.REI.B.3

Solve the linear equation for x:

2(3x + 4) + 2 = 4 + 3x

• 2

• -2/3

• 10

• -2

Tags

CCSS.HSA.REI.A.1

CCSS.HSA.REI.B.3

Solve the linear equation for k:

   $\frac{1}{7}k-6=\frac{3}{7}k=4$  

•  $k=\frac{35}{2}$  

•  $k=7$  

•  $k=-35$  

Tags

CCSS.HSA.REI.A.1

CCSS.HSA.REI.B.3

Solve by finding square roots and simplify completely:

 $2x^2=20$ 

•  $x=\pm10$  

•  $x=\sqrt{10}$  

•  $x=\pm\sqrt{10}$  

• No real solutions

Factor the quadratic expression:

x²-6x-40

• (x-10)(x+4)

• (x+10)(x-4)

• (x-8)(x+5)

• (x+8)(x-5)

Factor the quadratic expression:

x²-11x+24

• (x-3)(x-8)

• (x-2)(x-12)

• (x+3)(x+8)

• (x+2)(x+12)

Solve the quadratic equation by factoring and setting each factor = 0

:

5x² - 35x + 60 = 0

• x = 3, -4

• x = -3, 4

• x = 3, 4

• x = -3, -4

When does a quadratic equation have no real solutions?

• When you cannot factor it

• When it begins with a negative leading coefficient

• When there is a square root of a negative number

• When it only has two terms

Tags

CCSS.HSA.REI.B.4