EXAMPLE 3 Find lengths using Theorem 10.16

Use the figure at the right to find RS.

SOLUTION

256 = x2 + 8x

0 = x2 + 8x – 256

RQ2 = RS RT

162 = x (x + 8)

x –8 + 82 – 4(1) (– 256)

2(1)=

x = – 4 + 4 17

Use Theorem 10.16.

Substitute.

Simplify.

Write in standard form.

Use quadratic formula.

Simplify.

EXAMPLE 3 Find lengths using Theorem 10.16

Use the positive solution, because lengths cannot be negative.

= – 4 + 4 17So, x 12.49, and RS 12.49

GUIDED PRACTICE for Example 3

Find the value of x.

SOLUTION

x2 = 1 (1 + 3)

x = 2

Use Theorem 10.16.

Simplify.

Simplify.

4.

= 4x2

GUIDED PRACTICE for Example 3

Find the value of x.

SOLUTION

5.

49 = 25 + 5x

24 = 5x

72 = 5 (x + 5) Use Theorem 10.16.

Simplify.

Write in standard form.

Simplify.x = 245

GUIDED PRACTICE for Example 3

Find the value of x.

SOLUTION

6.

144 = x2 + 10x

0 = x2 + 10x – 144

122 = x (x + 10)

x –10 + 102 – 4(1) (– 144)

2(1)=

Use Theorem 10.16.

Simplify.

Write in standard form.

Use quadratic formula.

Simplify.x = 8

GUIDED PRACTICE for Example 3

Determine which theorem you would use to find x. Then find the value of x.

7.

Theorem 10.16

SOLUTION

255 = x2 + 14x

0 = x2 + 14x – 255

152 = x (x + 14)

x –14 + 142 – 4(1) (– 255)

2(1)=

Use Theorem 10.16.

Simplify.

Write in standard form.

Use quadratic formula.

Simplify.x = – 7 + 274

GUIDED PRACTICE for Example 3

Determine which theorem you would use to find x. Then find the value of x.

8.

Use Theorem 10.14.

x (18) = (9) (16) Substitute.

18x = 144 Simplify.

SOLUTION

Simplify.x = 8

GUIDED PRACTICE for Example 3

Determine which theorem you would use to find x. Then find the value of x.

8.

Use Theorem 10.15.

SOLUTION

18 (18 + 22) = x (x + 29) Substitute.

720 = 29x + x2 Simplify.

0 = x2 + 29x – 720

x –29 + 292 – 4(1) (– 720)

2(1)=

Write in standard form.

Use quadratic formula.

Simplify.x = 16

GUIDED PRACTICE for Example 3

SOLUTION

9.In the diagram for Theorem 10.16, what must be true about EC compared to EA ?

EC < EA