What is the square root of 4x^{4} + 8x³ - 4x + 1?

Correct Answer is : 2x2 + 2x - 1

Correct Answer is : 2x2 + 2x - 1

Test Index

UPSC CDS 2 2020 Math Paper

Question : 35 of 100

Marks: +1, -0

4x^{4} + 8x^{3} - 4x + 1

Solution:

Concept:

Recognize the given polynomial as a perfect square trinomial of the form

(a + b - c)^2

Explanation:

We need to find

\sqrt{4x^{4} + 8x^{3} - 4x + 1}

Rewrite the expression by adding and subtracting

4x^{2}

4x^{4} + 8x^{3} - 4x + 1 = 4x^{4} + 8x^{3} + 4x^{2} - 4x^{2} - 4x + 1

Group terms to match the square pattern:

= (2x^{2})^{2} + (2x)^{2} + (1)^{2} + 2 \cdot 2x^{2} \cdot 2x - 2 \cdot 2x \cdot 1 - 2 \cdot 2x^{2} \cdot 1

This is exactly

(2x^{2} + 2x - 1)^{2}

Therefore,

\sqrt{4x^{4} + 8x^{3} - 4x + 1} = \sqrt{(2x^{2} + 2x - 1)^{2}} = 2x^{2} + 2x - 1

(positive square root).

As a quick check, substitute

x = 1

: the expression becomes

4 + 8 - 4 + 1 = 9

, and

2(1)^{2} + 2(1) - 1 = 3

3^{2} = 9

, confirming the result.

Answer:

2x^{2} + 2x - 1

(Option D)

Go to Question:

Prev Question

Next Question