in Quantitative Aptitude retagged by
2,323 views
8 votes
8 votes

​​​​​​If $\left( x  – \dfrac{1}{2} \right)^2 – \left( x- \dfrac{3}{2} \right) ^2 = x+2$, then the value of $x$ is:

  1. $2$
  2. $4$
  3. $6$
  4. $8$
in Quantitative Aptitude retagged by
by
2.3k views

3 Answers

9 votes
9 votes
Best answer
Given that, $\left(x – \dfrac{1}{2}\right)^{2} –  \left(x – \dfrac{3}{2}\right)^{2} = x + 2$

$\implies \left(x – \dfrac{1}{2} – x + \dfrac{3}{2}\right) \left(x – \dfrac{1}{2} + x - \dfrac{3}{2}\right) = x + 2 \quad [\because a^{2} – b^{2} = (a – b)(a + b) ]$

$\implies  2x - 2 = x + 2$

$\implies x = 4$

So, the correct answer is $(B).$
selected by
2 votes
2 votes

Option B

$$(x – \frac{1}{2})^2 – (x – \frac{3}{2})^2 = x + 2$$

 

$$(\frac{2x-1}{2})^2 – (\frac{2x-3}{2})^2 = x+2$$

 

$$\frac{(2x-1)^2 – (2x – 3)^2}{4} = x + 2$$

 

$$(2x – 1 + 2x – 3)(2x – 1 – (2x – 3)) = 4(x + 2)$$

 

$$(4x – 4)(2) = 4(x + 2)$$

 

$$4(x – 1)(2) = 4(x + 2)$$

 

$$2x – 2 = x + 2$$

 

$$x = 4$$

 

 

Of course simple substitution would be faster.

1 vote
1 vote
on expanding the given equation,we get:

$(x^2+1/4-2*x*1/2)-(x^2+9/4-2*x*3/2)=x+2$

$x^2+1/4-x-x^2-9/4+3x=x+2$

$3x-x+1/4-9/4=x+2$

$2x-2=x+2$

$x=4$
Answer:

Related questions