EXAMPLE 1

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# EXAMPLE 1 - PowerPoint PPT Presentation

x = 2 a – b + b 2 – 4 a c – 3 + 3 2 – 4( 1 )( – 2 ) x = 2( 1 ) 17 17 – 3 + –3 + x = x = 2 2 ANSWER The solutions are 0.56 and 17 –3 – x = –3.56 . 2 EXAMPLE 1 Solve an equation with two real solutions Solve x 2 + 3 x = 2. x 2 + 3 x = 2 Write original equation.

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## EXAMPLE 1

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x =

2a

–b+b2– 4ac

–3+32– 4(1)(–2)

x =

2(1)

17

17

–3+

–3+

x =

x =

2

2

The solutions are

0.56and

17

–3–

x =

–3.56.

2

EXAMPLE 1

Solve an equation with two real solutions

Solvex2 + 3x = 2.

x2 + 3x = 2

Write original equation.

x2 + 3x – 2 = 0

Write in standard form.

a = 1, b = 3, c = –2

Simplify.

EXAMPLE 1

Solve an equation with two real solutions

CHECK

Graph y = x2 + 3x – 2 and note that the x-intercepts are about 0.56 and about –3.56.

30+(–30)2– 4(25)(9)

x =

30+ 0

2(25)

x =

50

3

x =

5

3

5

The solution is

EXAMPLE 2

Solve an equation with one real solutions

Solve25x2 – 18x = 12x – 9.

25x2 – 18x = 12x – 9.

Write original equation.

25x2 – 30x + 9 = 0.

Write in standard form.

a = 25, b = –30, c = 9

Simplify.

Simplify.

EXAMPLE 2

Solve an equation with one real solutions

CHECK

Graph y = –5x2 – 30x + 9 and note that the only x-intercept is 0.6 = .

3

5

–4 +42 – 4(–1)(–5)

x =

–4 +–4

2(–1)

x =

–2

–4 + 2i

x =

–2

The solution is 2 + i and 2 – i.

EXAMPLE 3

Solve an equation with imaginary solutions

Solve–x2 + 4x = 5.

–x2 + 4x = 5

Write original equation.

–x2 + 4x – 5 = 0.

Write in standard form.

a = –1, b = 4, c = –5

Simplify.

Rewrite using the imaginary unit i.

x = 2 +i

Simplify.

?

–(2 + i)2 + 4(2 + i) = 5

?

–3 – 4i + 8 + 4i = 5

5 = 5

EXAMPLE 3

Solve an equation with imaginary solutions

CHECK

Graph y = 2x2 + 4x – 5. There are no x-intercepts. So, the original equation has no real solutions. The algebraic check for the imaginary solution 2 + iis shown.

5+i

115

10

1

2

for Examples 1, 2, and 3

GUIDED PRACTICE

Use the quadratic formula to solve the equation.

x2= 6x – 4