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USING INEQUALITY

TO SOLVE

THE EQUATION

Hoang Le Nhat Tung ,

( Student K61 Math, Hanoi, Vietnam)

¢ The equation is a familiar and frequent mathematical formula used in secondary mathematics
tests. The solution of the equation includes many methods such as: hidden sub, multiplication, ....
In this article, I would like to exploit a method that is: use inequalities.
¢ Method: Experiment of the equation (usually the only solution). Then use the common
inequalities such as Cauchy (AM-GM), Bunhiacopxki, Mincopxki ..... to evaluate the equation
and satisfy the '=' equation.
¢ Following are some examples:

Example 1. Solve the equation: x +x? —x+1 =x? —x+2
.

(1)

.

1

2

My solution. x > 0. Easy to see : x =1 is the solution (1). Therefore, x7 —x+1= [+ — 3
left-hand side equation contains the square root which should apply the AM-GM

; numbers: Vx +Vx? —x+
nonnegative

+Let(): =x!-x+2<Š

2

*3 62

3

+ 1 > 0, The

inequality to the two

(x? -x41f+1 x +3
ke
“

_2

= Val +y (+i

s > +1

-2xd4 eee

-2x+1<0e(x-1) <0

2

2

(x-1} =0
>

)x=1

<> x=1

(satisfy ) . Therefore Š = {1} is the only solution of the problem. .

x -x+lI=l

Example 2. Solve the equation :16x* +5 =6N4x° +x

(2)

My solution. Because 16x* +5 > 5 > 0 then let (2): = 4x) +x>0 © xÍ4x?+1)>0©x>0.

+ We see: x = 3 is the solution (2). Therefore , by AM-GM

inequality for three positive real

numbers we have : 6.V4x° +x = 34l2.4x(4x? +1Ì<2+4x+ (4
+1)= Ax? +4x43
+ Let (3): > 16x47 +5 <4x° +.4x4+3< 16x" — 4x? —4x+2<0
2 8x" -2x° -2x4+1<0

<> 4x3 (2x -1)+ 2x? (2x -1)- (2-1) <0
© (2x- 1y <0> (2x - 1) =0

(2x-1) =0
2=4x=4x +1

&

“x19
x==

1

(2x1) (4x? + 2x-1)< 0

(2x-1) (2x? +x4:1)< 0

(because x >0 then 2x* +x+1>0). Therefore, equation occurs if
Sx =—

|


2

(satisfy)


Therefore S = ‘st

is the only solution of the problem.

Example 3. Solve the equation : ¥x° + 2x +V¥3x-l=vx° +4x° +4x41
ve 42x>0

x(x? +2)>0

My solution.:53x—-1=0

3

x*+4x

(3)

<©‹4x>—

2 +4x+lI>0

(3) © Vix? +2)4 V3x—

=


(x+ (x? +3x+1)2 0

= f(x +e?

-.

3

+ 3x41)

(4)

+ We see that the expression on the left side (4) is the sum of two roots of the second order, noting

that:

(Vx) +P =x41 va (Vi

+2]

2

+ (3x — iy = x° +3x+1

are 2 terms in the right hand side.

Therefore by Bunhiacopxki inequality we have :

(xe? +2)+1V3x-1) < (Vx) + Phe +2) + Wax-1 | =(x+ I(x? +3x+1)

=> af x(x? +2Ì+A3x-1< V(x + 1x? +3x+1) Let (4), equation occurs when:
x>—

1

3

&

xA3x-1=1Ax?+2

x>—

1

3

&

x(x—1)=x”+2

Therefote S =

1417

4

x>—

1


3

axe

^'

( satisfy)

2x-x-2=0

| is the only solution of the problem.

Example 4. Solve the equation : /x—2 +4/4-x =x? 6x+11
.
My solution

x=2>0

:

(5)

ôâ2
4-x>0

- We see the expression on the left side as the sum of the two quadratic modes and: (x— 2)+(4— x)= 2.
From there, by Bunhiacopxki inequality we have:

Wa-2 444-2) =|[6x=2+W4=s} | <

nh

heals}
—

2|(x-2)+ (4- x)|= 4.2.2 = 16

=1*-2+4Jl4-x <4Jl6 =2
+ Let (5)—= x?—6x+11<2© x?-6x+9<0© (x-3}'. <0— (x-3} =0 and equality
ïf occur
simultaneously:

x-2=4-x
3=0
<> x=3. Therefore S = {3} is the only solution of the problem.
xX

=

Example 5. Solve the equation : Vx? + 12x+ 61+ Vx? —14x+113 = 4338
My solution

(6)

:

(x? +12x+61>0

x*-14x+l1l3>0_

&

(x+ 6) +252

|(x-7Ì+64>0

y(x+6) 45° +(7-x) +8? = 338

(6)

. So that means consciousness

VxeER.

(7)


+ We see, the left side of (7) is the tangent of two roots of form 2: Va’ +b* +V¥c’ +d’ , This suggests
that Mincopxki's inequality is in the form Va? +b? +Vc? +d? >J(a+c) +(0+d)

(= sign occurs

when : ad = bc)

+ Therefore: f(x +6) +5? +4(7—x) +8? > J(x+64+7-x) +(54+8) = 13? +13? = 338. And let
(7), equality occurs if : (x+6)8 =5(7—x)< 8x4 48 = 35-5x 6 13x=-13 @x=-l.
Therefore Š = {1} is the only solution of the problem.

Example 6. Solve the equation : V8x2 —16x+104+V2x? —4x44 =V7— 2x? + 2x

8x7 -16x+1020

My sohumion.

42x”—=4x+4>0_
T-x° +2x 20

(8)

|8(x-1} +2>0

©42(x-1Ÿ +2>0©1-2V2(x-1) <8

(8) <> (4x? — 4x +1)+ (4x? 12% +9) 4 f(x? — 4x $ 4)4? = /8- (x? — 2x41)

©aj(x-1Ÿ +-2x} +4 (2-x) +2? =8-(x-1)

(9)

+ The left side of (9) is the tangent of two roots of type 2 Va* +b* +V¥c* +d* , by Mincopxki inequality
we have :

Hea

ete

(3-2x+x/)

= f(x +1) +
+ Let (9) > J8-(x-1)

=Al2x? =4xz+10 =a|2(x—1} +8
> x8

> V8 = 8-(x-1) a

=0

=]

<0
=> (x-1)
x=l

ln l)x = (3-2x)(2-x)~ [ro _x=2x)-7x+6 {eo

6

=0 and equality occurs if

=1.

Therefore S = {1} is the only solution of the problem.

Example 7. Solve the equation : 2V.x*° +14

6 _ 3x” -6x+10

(10)

My solution. x` +1>0 © x>—1. We see: x = 2 is the solution (10), The left-hand side equation
contains square root.. By AM-GM inequality and Bunhiacopxki we have :
2N\x`+

=24l(x+1)W?-x+1)<(x+1)+(?—x+1}=a?+2

4

* *18 2]
2

:

ser}

2\qx`+1l
>)

[x4416
5

+2x
4

8 (a5) |= at các si6= (ở v4}

2\x`+1l+4

>)

[x4416
2


=> Wx +1+

6 < 2x7 —-2x+6


+ Let (10) > 3x* —6x+10 < 2x? -2x4 60 xX —4x4+4<00
(x-2) <05(x-2/) =0 and
x+l=z`-x+I

.
.
equality occurs if

x' +16
5
=2x
x-2=0

©

x(x — 2) =0
2
2
(x — 4) =0<>x=2

.
(satisfy )

x=2

Therefore S = {2} is the only solution of the equation.
* Application exercises. Solve the equations :

a)_

bị
c_

2N2x`-x=3x?-3x+2
4

1

Q1. DĂ?+4)=3x+2
2

A2xÌ-2x?+x+243x-2x?=xf-x)+3

d) 4l2(x*+1Ì+3ÄÝx

= x2 +4
8

e) 3Ï —1......
f) V2x? 4+2x454 2x? -10x+17 =6