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[email protected] • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§7.6 2Var§7.6 2VarRadical EqnsRadical Eqns



Review §Review §

Any QUESTIONS About• §7.6 → Radical Equations

Any QUESTIONS About HomeWork• §7.6 → HW-35

7.6 MTH 55



Radical EquationsRadical Equations

A Radical Equation is an equation in which at least one variable appears in a radicand.

Some Examples:



Solve Eqns with 2+ Rad. TermsSolve Eqns with 2+ Rad. Terms

1. Isolate one of the radical terms.

2. Use the Exponent Power Rule

3. If a radical remains, perform steps (1) and (2) again.

4. Solve the resulting equation.

5. Check the possible solutions in the original equation.



Example Example Solve Solve

SOLUTION

3 2 1.x x

3 2 1x x

3 2 1x x

2 23 2 1x x

3 2 2 2 1x x x

4 2 2x




SOLUTION

3 2 1.x x

2 2x

222 2x

4 2x 6 x

Check 6 by Inspection → 3−2=1 Thus The number 6 checks

and it IS the solution




SOLN

2 2 2 3.x x

2 22 2 2 3x x One radical is isolated.

We square both sides.

22 4 4 2 3 2 3x x x

4 2 3 9x x

2 24 2 3 ( 9)x x

2 4 4 2 3 2 3x x x

216(2 3) 18 81x x x 232 48 18 81x x x 20 14 33x x

0 ( 3)( 11)x x 3 or 11x x

Square both sides.

Factoring

Using the principle of zero products



Example Example Solve Solve 2 2 2 3.x x

2 2 2 3x x 2 2 2 3x x

2 23 32( ) 3

1 2 6 3 1 2 9 1 2 3 1 1

2 21 21 ) 311(

9 2 22 3

3 2 25 3 2 5 3 3

Check: x = 3 x = 11

The numbers 3 and 11 check and

are then confirmed as solutions.



Example Example Solve Solve x + 7 + –x + 6 = 5.

Start by isolating one radical on one side of the equation by subtracting

from each side. Then square both sides.

x + 7 + –x + 6 = 5

–x + 6

x + 7 = 5 – –x + 6

2x + 7 = 5 – –x + 6 2

= 25 – 10 –x + 6 + (–x + 6)x + 7

Twice the product of 5 and .–x + 6




This equation still contains a radical, so square both sides again.

Before doing this, isolate the radical term on the right.

= 25 – 10 –x + 6 + (–x + 6)x + 7

= 31 – x – 10 –x + 6x + 7

= –10 –x + 62x – 24 Subtract 31 and add x.

= –5 –x + 6x – 12 Divide by 2.

= –5 –x + 6(x – 12)2 Square both sides again. 2




= –5 –x + 6(x – 12)2 Square both sides again. 2

This equation still contains a radical, so square both sides again.

= (–5)2 –x + 6x2 – 24x + 144 (ab)2 = a2 b2 2

= 25 ( –x + 6 )x2 – 24x + 144

= –25x + 150x2 – 24x + 144 Distributive property

= 0x2 + x – 6 Standard form

= 0(x + 3)(x – 2) Factor.




= 0(x + 3)(x – 2)

x + 3 = 0 or x – 2 = 0 Zero-factor property

x = –3 or x = 2

Now finish solving the equation.

Finally CHECK for Extraneous Solutions




Check each potential solution, –3 and 2, in the original equation.

–3 + 7 + –(–3) + 6 = 5?

If x = –3, then If x = 2, then

2 + 7 + –(2) + 6 = 5?

x + 7 + –x + 6 = 5? x + 7 + –x + 6 = 5?

4 + 9 = 5? 9 + 4 = 5?

5 = 5 5 = 5

The solution set is { −3, 2 }.



The Principle of Square RootsThe Principle of Square Roots

Recall the definition of the PRINCIPAL Square Root

For any NONnegative real number n, If x2 = n, then



Recall The Pythagorean TheoremRecall The Pythagorean Theorem

In any right triangle, if a and b are the lengths of the legs and c is the length of the hypotenuse, then

a2 + b2 = c2

aLeg

LegHypotenusec b



Example Example Pythagorus Pythagorus

How long is a guy wire if it reaches from the top of a 14 ft pole to a point on the ground 8 ft from the pole?

SOLUTION

14

8

d

2 2 214 8d 196 64 260 • We now use the principle of

square roots. Since d represents a length, it follows that d is the positive square root of 260:

260 ftd

16.125 ft.d Diagram



Isosceles Right TriangleIsosceles Right Triangle

When both legs of a right triangle are the same size, we call the triangle an isosceles right triangle. If one leg of an isosceles right triangle has length a then

c

a

a

22

22

222

2

2

ac

ac

aac



Lengths for Isosceles Rt Triangles Lengths for Isosceles Rt Triangles

The length of the hypotenuse in an isosceles right triangle is the length of a leg times 2.

45o

a

a45o

2a



Example Example Isosceles Rt. Tri. Isosceles Rt. Tri.

The hypotenuse of an isosceles right triangle is 8 ft long. Find the length of a leg. Give an exact answer and an approximation to three decimal places.

SOLUTION

45o

a

a45o

88 = 2a

8

2a

4 2 .a

Exact answer:Approximation:

4 2 fta 5.657 ft.a

( after Rationalizing Divisor).

22

222

82

8

a

aa



30°-60°-90° Triangle30°-60°-90° Triangle

A second special triangle is known as a 30°-60°-90° right triangle, so named because of the measures of its angles



Lengths for 30/60/90 Rt Triangles Lengths for 30/60/90 Rt Triangles

The length of the longer leg in a 30/60/90 right triangle is the length of the shorter leg times The hypotenuse is twice as long as the shorter leg.

3.

a

2a3a

60o

30o



Example Example 30°-60°-90° 30°-60°-90° TriangleTriangle The shorter leg of a 30/60/90 right

triangle measures 12 in. Find the lengths of the other sides. Give exact answers and, where appropriate, an approximation to three decimal places.

SOLUTION

12

2a3a

60o

30o• The hypotenuse is twice as long as the shorter leg, so we have

c = 2a = 2(12) = 24 in.



Example Example 30°-60°-90° 30°-60°-90° TriangleTriangle SOLUTION The length of the longer leg

is the length of the shorter leg times This yields

12

2a3a

60o

30o

3.

3b a = 12 3 in.

Exact answer:

Approximation:

24 in., 12 3 in.c b

20.785 in.b



WhiteBoard WorkWhiteBoard Work

Problems From §7.6 Exercise Set• 24, 34, 38, 48, 60

Astronomical Unit = Sun↔Earth Distance

= 149 598 000 km



All Done for TodayAll Done for Today

The SolarStar System



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

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srsrsr 22



Graph Graph yy = | = |xx||

Make T-tablex y = |x |

-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6

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-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

file =XY_Plot_0211.xls



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M55_§JBerland_Graphs_0806.xls -5

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-10 -8 -6 -4 -2 0 2 4 6 8 10

M55_§JBerland_Graphs_0806.xls

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