Algebraic Expressions and Equations Problems

What Are Algebraic Expressions and Equations?

An algebraic equation is of the form [say some variable] P = 0. On the other hand, the algebraic expression contains integer constants, variables and other mathematical tools of operation such as addition, subtraction, multiplication and division. In this section, we will discuss the quadratic expressions and equations in detail, and some algebraic expressions and equations problems covered in previous years’ IIT JEE Exams.

The parts of the algebraic expression which are separated by operations are called terms. Monomial, binomial, trinomial expressions and polynomial expressions are different types of algebraic expressions. This topic is wide and contains various concepts such as simplifying algebraic expressions, the relationship between expressions, common algebraic equations and inequalities and many more.

Important Algebraic Formulae

a² – b² = (a + b) (a – b)
(a + b)² = a² + 2ab + b²
(a – b)² = a² – 2ab + b²
(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc
(a – b – c)² = a² + b² + c² – 2ab – 2ac + 2bc
(a + b)³ = a³ + 3a²b + 3ab² + b³ or
(a + b)³ = a³ + b³ + 3ab(a + b)
(a – b)³ = a³ – 3a²b + 3ab² – b³
a³ – b³ = (a – b)(a² + ab + b²)
a³ + b³ = (a + b)(a² – ab + b²)
(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
(a – b)⁴ = a⁴ – 4a³b + 6a²b² – 4ab³ + b⁴
a⁴ – b⁴ = (a – b)(a + b)(a² + b²)
a⁵ – b⁵ = (a – b)(a⁴ + a³b + a²b² + ab³ + b⁴)
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2xz
(x + y – z)² = x² + y² + z² + 2xy – 2yz – 2xz
(x – y + z)² = x² + y² + z² – 2xy – 2yz + 2xz
(x – y – z)² = x² + y² + z² – 2xy + 2yz – 2xz
x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – xz)

Laws of Exponents

(a^m)(aⁿ) = a^m+n
(ab)^m = a^mb^m
(a^m)ⁿ = a^mn

Fractional Exponents

a⁰ = 1
a^m/aⁿ = a^m-n
1/a^m = a^-m

Quadratic Expressions and Equations

Quadratic expressions: An expression with its degree 2 is a quadratic expression. The equation of the form ax² + bx + c = 0 can be called a quadratic equation. Here a, b and c are constants, and a is not equal to 0. The constants a, b and c are called coefficients. A quadratic equation with one unknown is called univariate. The discriminant of a quadratic equation is D = b² – 4ac.

The nature of roots is determined by the discriminant.

Related topics

Theory of Equations

Quadratic Inequalities

Solved Examples for IIT JEE

Example 1: Find the number of real values of x for which the equality ∣3x² + 12x + 6∣ = 5x + 16 holds good.

Solution:

Equation is |3x² + 12x + 6|= 5x + 16 …….(i)

When 3x² + 12x + 6 ≥ 0

⇔ x²+ 4x ≥ −2

⇔ |x + 2|² ≥ 4 − 2

⇔ |x + 2| ≥ (√2)²

⇔ x + 2 ≤ √−2 or x + 2 ≥ √2 …….(ii)

Then, (i) becomes 3x² + 12x + 6 = 5x + 16

⇔3x² + 7x −10 = 0

⇒x = 1, −10 / 3

But x = −10 / 3 does not satisfy (ii).

When 3x² + 12x + 6 < 0 ⇒ x²+ 4x < −2

|x + 2| ≤ √2 ⇒ [−√2 −2] ≤ x ≤ [−2 + √2] ? …….(iii)

Then, (i) becomes ⇒ 3x² + 12x + 6 = −(5 x + 16)

⇒3x² + 17x + 22 = 0

⇒x =−2,−11 / 3

But x = −11 / 3 does not satisfy (iii).

So, 1 and – 2 are the only solutions.

Example 2: What is the set of all real numbers x for which x²−|x + 2|+ x > 0?

Solution:

Case I:

When x + 2 ≥ 0 i.e. x ≥ −2, then given inequality becomes x²−|x + 2|+ x > 0

x²−2 > 0 ⇒|x| > √2

x<√−2 or x > √2

As x ≥ −2, in this case, the part of the solution set is [−2, √−2) ∪ (√2, ∞).

Case II: When x + 2 ≤ 0, i.e., x ≤ −2, then given inequality becomes x² + (x + 2) + x > 0

⇒ x² + 2x + 2 > 0 ⇒ (x + 1)² + 1 > 0, which is true for all real x.

Hence, the part of the solution set in this case is (−∞, −2].

Combining the two cases, the solution set is (−∞, −2) ∪ ([−2, √−2] ∪ (√2, ∞) = (−∞, √−2) ∪ (√2, ∞).

Example 3: If the roots of the equation 8x³ − 14x² + 7x − 1 = 0 are in GP, then find the roots.

Solution:

Let the roots be αβ, α, αβ, where β ≠ 0.

Then the product of roots is α³ = −[−1 / 8] = [1 / 8] ⇒ α = 1 / 2 and hence β = 1 / 2, so roots are 1, 1 / 2, 1 / 4

Hint: By inspection, we get the numbers 1, 1 / 2, and 1 / 4 satisfying the given equation.

Example 4: If a, b, c are real and x³ − 3b²x + 2c³ is divisible by x − a and x − b, then what is a?

Solution:

As f(x) = x³ − 3b²x + 2c³ is divisible by x − a and x − b,

Therefore, f(a) = 0 ⇒ a³ − 3b²a + 2c³ = 0 …..(i) and

f(b)=0

b³ − 3b³ + 2c³ = 0 …..(ii)

From (ii), b = c

From (i), a³ − 3ab² +2b³ = 0 (Putting b=c) ⇒ (a−b) (a² + ab − 2b²) = 0 ⇒ a = b or a² + ab = 2b

Thus, a = b = c or a² + ab = 2b² and b = ca² + ab = 2b² is satisfied by a = −2b.

But b = c , a² + ab − 2b² and b = c is equivalent to a = −2b = −2c.

Example 5: What are the roots of the equation x⁴ − 2x³ + x = 380?

Solution:

Given equation is x⁴ − 2x³ + x = 380.

Using remainder theorem, we get (x − 5) (x + 4) (x² − x + 19)=0

x − 5=0, x − 4 = 0 and x² − x + 19 = 0

x = 5, x = −4 and x = [−1 ± 5√−3] / 2

Example 6: If the sum of the two roots of the equation 4x³ + 16x² − 9x − 36 = 0 is zero, then the roots are ______________.

Solution:

Given equation 4x³ + 16x² − 9x − 36 = 0,

Putting x = −4

Hence, x = −4 is the root of the equation.

Now, reduced equation is 4x² (x + 4) − 9 (x + 4) = 0

(x + 4) (4x² − 9)=0

x = −4, x = ± 3 / 2

Thus, roots are −4, −3/2, 3/2.

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