# Types of Algebra Questions

In this article, we will have a look at the various types of questions that we encounter based on the topic of Algebra.

But you must be aware of the basic concepts and formulae of Algebra before we do so. If you aren’t, then have a look at the previous article.

## Type 1

##### Q. If a(x + y) = b (x - y) = 2ab, then what is the value of $2(x^2 + y^2)$?

(a) $a^2 – b^2$
(b) $8(a^2 + b^2)$
(c) $4(a^2 – b^2)$
(d) $4(a^2 + b^2)$

Explanation:

Given: a (x + y) = b (x - y) = 2ab

So, a (x + y) = 2ab
Or (x + y) = 2b
Squaring both sides, we get:
$(x + y)^2 = (2b)^2$
or $x^2 + y^2 + 2xy = 4b^2$ … (i)

Similarly, (x - y) = 2a
Squaring both sides, we get:
$(x - y)^2 = (2a)^2$
or $x^2 + y^2 - 2xy = 4a^2$ … (ii)

Adding equations (i) and (ii), we get:
$2x^2 + 2y^2 = 4a^2 + 4b^2$
or $2(x^2 + y^2) = 4 (a^2 + b^2)$

## Type 2

##### Q. If x + y + z = 0, then what is the value of $(3y^2 + x^2 + z^2) / (2y^2 - xz)$
(a) 1   (b) 2    (c) 3.5   (d) 5.5

Explanation:

Let x = 1, y = -1 and z = 0
Therefore, x + y + z = 1 - 1 + 1 = 0

Then, $(3y^2 + x^2 + z^2 ) / (2y 2 - xz) = [3(-1)^2 + 1^2 + 0^2] / [2(-1)^2 – 1 × 0] = 4/2 = 2$

## Type 3

##### Q. The sum of three consecutive integers is equal to their product. How many such possibilities are there?
(a) 1   (b) 2    (c) 3   (d) 0

Explanation:

Let the 3 consecutive numbers be x - 1, x, and x + 1

According to the question,
(x – 1) + x + (x + 1) = (x - 1) × x × (x + 1)
⇒ 3x = x × ($x^2$ - 1)
⇒ 3x = $x^3$ – x
⇒ $x^3$ – 4x = 0
⇒ x ($x^2$ – 4) = 0
⇒ x = 0, 2 or -2

Therefore, there are only 3 such possibilities.

Wrong Method:

Let the 3 consecutive numbers be x, x + 1 and x + 2

According to the question,
x + (x + 1) + (x + 2) = x × (x + 1) × (x + 2)
⇒ 3x + 3 = x × (x + 1) × (x + 2)
⇒ 3(x + 1) = x × (x + 1) × (x + 2)
⇒ 3 = x(x + 2) [We eliminated x + 1 from both sides, which is wrong]
⇒ $x^2$ + 2x – 3 = 0
⇒ $x^2$ + 3x – x – 3 = 0
⇒ x (x + 3) - 1 (x + 3) = 0
⇒ (x - 1) (x + 3) = 0
⇒ x = 1 or -3

As per this, there are only 2 such possibilities.