GATE CSE 2021 Set 2 | GA Question: 4

Question

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

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

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 $(B).$

Answer 2

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.

Answer 3

on expanding the given equation,we get:

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

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

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

$2x-2=x+2$

$x=4$