Problems need HardWork #8

Algebra Level 2

Find the sum of all real values of x x satisfying the following equation.

∣ 3 x − 1 ∣ = ∣ x − 2 ∣ |3x - 1| = |x - 2|


The answer is 0.25.

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Dev Sharma by Dev Sharma , 20, India

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6 solutions

Ravi Dwivedi
Aug 22, 2015

∣ 3 x − 1 ∣ = ∣ x − 2 ∣ |3x-1|=|x-2|

⟹ ( 3 x − 1 ) = ± ( x − 2 ) \implies (3x-1)=\pm (x-2)

Now it follows that ( 3 x − 1 ) = x − 2 (3x-1)=x-2 OR 3 x − 1 = − ( x − 2 ) 3x-1=-(x-2)

which yields x = − 1 2 , x = − 3 4 x=-\frac{1}{2}, x=-\frac{3}{4}

Putting in the given equation we find that both of these values of x x satisfy the given equation

Sum of roots = 0.25 \boxed{0.25}

13 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Simple standard approach.

nice solution...

Dev Sharma - 5 years, 9 months ago

You have a little typo: where it reads − 3 4 -\frac{3}{4} it should have been 3 4 \frac{3}{4}

Luís Sequeira - 5 years, 7 months ago

Can you tell me why didn't we took mod of (3x-1) as + - ??

Vishal Yadav - 5 years, 9 months ago

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That's the same thing as above

Ravi Dwivedi - 5 years, 9 months ago

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Thanks. Got the answer.

Vishal Yadav - 5 years, 9 months ago

How many solutions are there for n+1= 2(x+y+n)/xy . Where x y n € Z ! ! All of the viewers pls help me regarding the question ! I request u if u can post this question since I m using Phone I m unable to do so ! Pls pls!!!!

Yash Sharma - 5 years, 9 months ago
Dev Sharma
Aug 22, 2015

Squaring both sides,

(3x - 1)^2 = (x - 2)^ 2

9x^2 - 6x + 1 = x^2 - 4 x + 4

8x^2 -2x - 3 = 0

(2x + 1)(4x - 3) = 0

x = -1/2 or 3/4

Sum = -1/2 + 3/4 = 0.25

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Careful there. You squared both sides of the equation, so there might be extratenous roots. It's always good to check back whether the values you've found satisfy the equation.

Is there a way to solve this problem without squaring both sides?

Hint : Piecewise function: ∣ a ∣ = { a , a ≥ 0 − a , a < 0 |a| = \begin{cases} { a\qquad,\quad a \geq 0} \\ {-a\ \quad,\quad a < 0 } \end{cases} .

Both LHS and RHS are non negative qntys. Squaring won't give extra roots

Amartya Anshuman - 5 years, 9 months ago

Actually, Dev's solution is correct, except for the first phrase "Squaring both sides". There is no need to check for extraneous roots. Let me explain: while it is true that a = b a =b is NOT equivalent to a 2 = b 2 a^2 =b^2 , in fact it is true for all real numbers that ∣ a ∣ = ∣ b ∣ |a|=|b| if and only if a 2 = b 2 a^2=b^2 . So you could just start with ∣ 3 x − 1 ∣ = ∣ x − 2 ∣ ⇔ ( 3 x − 1 ) 2 = ( x − 2 ) 2 … |3x-1|=|x-2| \Leftrightarrow (3x-1)^2=(x-2)^2 \dots

Luís Sequeira - 5 years, 7 months ago
Pawan Mishra
Aug 25, 2015

We can break the roots as per the range under which they fall.

(A) for x>2: (3x-1)=(x-2) Root 1= -0.5

Similaily for for x < 1/3 Equation is (1-3x)=(2-x) Root = - 0.5

(B) for x lying between (1/3,2) Equation will be. (3x-1)=(2-x) Root 2 = 0.75

Adding up the roots, sum = 0.25

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Kishore S. Shenoy
Aug 23, 2015

Case 1

x > 1 3 a n d x > 2 3 x − 1 = x − 2 x = − 1 2 , which is wrong ⊗ \begin{aligned} \displaystyle x &> \frac{1}{3} ~and ~ x > 2\\\\ 3x - 1 &= x - 2\\ x &= -\frac{1}{2}, \text{ which is wrong} \otimes \end{aligned}

Case 2

x > 1 3 a n d x < 2 3 x − 1 = 2 − x x 1 = 3 4 \begin{aligned} \displaystyle x &> \frac{1}{3} ~and ~ x < 2\\\\ 3x - 1 &= 2 -x \\ x_1 &= \boxed{\dfrac{3}{4}} \end{aligned}

Case 3

x < 1 3 a n d x < 2 1 − 3 x = 2 − x x = − 1 2 \begin{aligned} \displaystyle x &< \frac{1}{3} ~and ~ x < 2\\\\ 1 - 3x &= 2 -x\\ x &= \boxed{\dfrac{-1}{2}} \end{aligned}

And the fourth case will be x < 1 3 a n d x > 2 \displaystyle x < \frac{1}{3} ~and ~ x > 2

Which is not possible.

x 1 + x 2 = 1 4 \large \boxed {x_1 + x_2 = \dfrac{1}{4}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Can you explain your cases clearly? Note that case 2 and case 3 are currently the same.

By right, for completeness, you should have 2 × 2 = 4 2 \times 2 = 4 cases, at least based on the approach that you are taking.

Your suggestions are nice.

Dev Sharma - 5 years, 9 months ago

Can you explain your cases clearly? Note that case 2 and case 3 are currently the same.

By right, for completeness, you should have 2 × 2 = 4 2 \times 2 = 4 cases, at least based on the approach that you are taking.

Calvin Lin Staff - 5 years, 9 months ago

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Oops that was a typo!

Kishore S. Shenoy - 5 years, 9 months ago
Sanat Mishra
Aug 30, 2015

It's also quite clear by drawing the graphs for the two equations and adding the values of the points where they intersect!

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Subodh Hegde
Aug 23, 2015

Squaring both the sides.... (3x-1)^2=(x-2)^2 9x^2-6x+1=x^2-4x+4 8x^2-2x-3=0 Which is a quadratic equation in the standard form ax^2+bx+c. The sum of the roots of this equation is -b/a. Comparing and substituting the values for a and b , we get -(-2/8)=1/4=0.25.

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