9.4 USING SQUARE ROOT Chapter 9: Quadratic Functions

QPROPERTY TO SOLVE QUADRATIC EQUATIONSQUADRATIC EQUATIONS

Simplify square root: √72Simplify square root: √72

√72 8 485281374√72 ≈ 8.48528137472= 36 · 2√72= √36 · 2 =√36 · √2= 6 · √2Leave irrational part of root under radical signg

Simplifying a radical quotientSimplifying a radical quotient288− 288

=744 ⋅

=724

=10 1010

−=105

−=105

−=

7474 −

FIRST write each term of numerator over5

757

5=−=

FIRST write each term of numerator over the denominator!!Reduce each fraction and simplify radicalReduce each fraction and simplify radicalCan then rewrite over single denominator if you choose: doesn’t really matterif you choose: doesn t really matter

Use Square Root propertyUse Square Root property

If d th thi t b th id fIf you do the same thing to both sides of equation, it is still a valid equationI l di kiIncluding taking square rootBe sure to write ( ) around each side, so you take the square root of the entire side, not of separate terms on the side

Square root property to solveSquare root property to solve equations

2 25x2=25√ x2=± √ 25x=5, -5 : ±5

Square root property to solveSquare root property to solve equations

2 3x2=3√ x2=± √ 3x= ± √ 3

Using square root propertyUsing square root property

2 24 0x2—24=0Need to isolate the square root first!!x2=24√ x2 =± √24x =± 2√6

Using square root propertyUsing square root property

3 2 4 33x2—4=3Need to isolate the square root first!!3x2=7x2=

37 72 ±=x3 3

±x

2137=±=x

333=⋅±=x

(x+4)2—36=0(x+4) 36 0Vertex form of equation( 4)2 36 t(x+4)2=36, can use sq.rt.prop.√(x+4)2 =± √36x+4 = ±6: now subtract 4x = — 4±6Both are rational, need to evaluate!!x = — 4+6, — 4 — 6x 4+6, 4 6x= 2, — 10

(x-7)2=50(x 7) 50√(x— 7)2 =± √50

7 ± √50 i lif √50x — 7 = ± √50 : now simplify √50x — 7 = ± 5√2Add 7, but 2 stays under radical signx = 7 ± 5√2Radical expression, write two of themx = 7 + 5√2, 7 — 5√2x 7 + 5√2, 7 5√2Or use ± sign after the 7

√49 —√49 √-49√49 √49 √ 49

Thi d t l bThird: not a real numberThere is not a real number you can multiply by itself to get a negative productitself to get a negative product

When radicand is negative, there is not a real square rootreal square rootBut radicand is -1·49…

f f “ “ fDefine sq. rt. of -1 as “i“ for imaginary

i is the square root of —1i is the square root of 1Factor out — 1 from negative radicands FIRSTFIRSTThe proceed to simplify the rootWhen solving equations use both rootsWhen solving equations, use both roots± sign: plus or minusWrite + then underline it with theWrite +, then underline it with the —If results has ± radical, ok to leave ±If result is ± a value add or subtractIf result is ± a value, add or subtract value from the rest, and get two answers

√-49√ 49

√ 1 49√-1·49=√-1·√49= i ·7 = 7i

— √-24√ 24

i √24— i √24= — i √4·6 = — 2i √6

If there are terms with real partsIf there are terms with real parts and imaginary parts

It i ll d ‘ l b ’It is called a ‘complex number’2 + 5iAny rational number could be written 3 = 3 + 0iBut it is not complex is the coefficient of iis zero

Recall how to square binomialRecall how to square binomial

( 4)2(x + 4)2=x2 +8x +16(x — 3)2=x2 — 6x + 9And these can be factored to binomial sq.

What term do we need to add to binomial to make it factor to squarebinomial to make it factor to square

x2 + 10x +25 = (x + 5)2x2 + 10xx2 + 6xx2 14x

+25 = (x + 5)2

+9 = (x + 3)2

+ 49 = (x 7)2x2 — 14xx2 — 2xThese are not the same value as original

+ 49 = (x — 7)2

+ 1 = (x — 1)2

These are not the same value as original expression, because we added somethingBut if you add and subtract the same thing, it y g,will still be the same, or add the same thing to both sides of

i i ill ill b lequation, it will still be equal

Use this fact to ‘complete the square’ and apply the squaresquare , and apply the square root property to solve quad.eq.

2 10 25x2 + 10x = — 25x2 + 6x = 16x2 — 14x = — 40x2 — 2x = 15On left, add what you need to make it have a square for factorsqAdd same thing on the rightApply the square root propertyApply the square root property

Add the term to both sidesAdd the term to both sides

2 10 25 25 25x2 + 10x + 25= — 25 + 25x2 + 6x + 9 = 16 + 9x2 — 14x + 49 = — 40 + 49x2 — 2x + 1= 15 + 1

Factor on left, simplify on rightFactor on left, simplify on right

( 5)2 0(x + 5)2 = 0(x + 3)2 = 25(x — 7)2 = 9(x — 1)2 = 16( )And apply square root property

Apply square root propertyApply square root property

5 ±0x + 5 = ±0x + 3 = ±5x — 7 = ±3x — 1 = ±4Add or subtract constant term from x and from right sidegFind value of each solution when rational on rightg

Add or subtract constant termAdd or subtract constant term from x and from right side

5 5 5 ±0 5x + 5 — 5= — 5 ±0 = — 5x + 3 — 3 = — 3 ±5 = — 8, 2x — 7 + 7= 7 ±3 = 4, 10x — 1 + 1 = 1 ±4 = 5, — 3Find value of each solution because it is ± a rational number

x2 — 14x +4 = 0x 14x +4 0

2 14 4 4 0 4x2 — 14x +4 — 4 = 0 — 4x2 — 14x = — 4Complete the squarex2 — 14x + 49 = — 4 + 49 = 45Factor on left(x — 7)2 = 45(x 7) 45Apply sq.rt.prop.x 7 = ± √45 = ± 3 √5x — 7 = ± √45 = ± 3 √5

x2 — 14x +4 = 0x 14x +4 0

( 7)2 45(x — 7)2 = 45Apply sq.rt.prop.x — 7 = ± √45 = ± 3 √5 Add 7 to both sidesx — 7 + 7 = 7 ± 3 √5

Radical remains, so write this orRadical remains, so write this orx = 7 + 3 √5 , 7 — 3 √5