Advanced algebra questions for SSC CGL

Once you are comfortable with identities, SSC CGL starts combining them with other ideas. You may be given x = 2 + √3 and asked for x² + 1/x²; or told that two equations have infinitely many solutions and asked for a constant; or asked for the minimum value of an expression. These questions reward a toolkit of techniques rather than a list of formulas.

Linear equations in two variables

For a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

Worked example: For what value of k do 2x + 3y = 5 and 4x + ky = 10 have infinitely many solutions?
2/4 = 3/k = 5/10, so k = 6.

Worked example: Solve 3x + 2y = 12 and x − y = −1.
From the second equation, x = y − 1. Substituting: 3(y − 1) + 2y = 12 → 5y = 15 → y = 3, x = 2.

Quadratic equations

For ax² + bx + c = 0 with roots α and β:

• sum α + β = −b/a;
• product αβ = c/a;
• the discriminant D = b² − 4ac decides the nature of the roots (D > 0: real and distinct; D = 0: equal; D < 0: not real).

Worked example: If α and β are the roots of x² − 5x + 6 = 0, find α² + β².
α + β = 5, αβ = 6. α² + β² = 25 − 12 = 13.

Expressions with surds

When x = a + √b, its reciprocal often simplifies neatly.

Worked example: If x = 2 + √3, find x + 1/x.
1/x = 1/(2 + √3) = (2 − √3)/(4 − 3) = 2 − √3. So x + 1/x = 4.
Then x² + 1/x² = 16 − 2 = 14.

Worked example: If x = √5 + 2, find x − 1/x.
1/x = √5 − 2, so x − 1/x = 4.

Conditions that force equal values

• If a² + b² + c² = ab + bc + ca, then a = b = c.
• If (a − 2)² + (b + 3)² = 0 for real a and b, each square must be zero: a = 2, b = −3.

Worked example: If a² + b² + c² − 2a + 4b − 6c + 14 = 0, find a + b + c.
Complete the squares: (a − 1)² + (b + 2)² + (c − 3)² = 1 + 4 + 9 − 14 = 0. So a = 1, b = −2, c = 3, and a + b + c = 2.

Minimum and maximum values

• For ax² + bx + c with a > 0, the minimum value is c − b²/4a, at x = −b/2a.
• For positive x, x + 1/x ≥ 2 (equality at x = 1).

Worked example: Find the minimum value of x² − 6x + 14.
(x − 3)² + 5 ≥ 5. Minimum = 5.

Componendo and dividendo

If a/b = c/d, then (a + b)/(a − b) = (c + d)/(c − d).

Worked example: If (x + 3)/(x − 3) = 5/3, find x.
Applying componendo and dividendo: 2x/6 = 8/2 → x/3 = 4 → x = 12.
Check: 15/9 = 5/3. ✓

Ratio-based expressions

Worked example: If x : y = 3 : 4, find (2x + 3y)/(3x − y).
Let x = 3, y = 4: (6 + 12)/(9 − 4) = 18/5.

Strategy

1. Look for structure — identity, surd conjugate, perfect square.
2. Substitute simple values that satisfy the condition.
3. Use the options — check which one works.
4. Skip long algebra in Tier 1 if time is short; return in Round 2.

Practice

1. For what value of k do 3x + ky = 9 and 6x + 4y = 18 have infinitely many solutions?
2. If α and β are roots of x² − 7x + 10 = 0, find 1/α + 1/β.
3. If x = 3 + 2√2, find x + 1/x.
4. Find the minimum value of x² + 8x + 20.
5. If a² + b² + c² = ab + bc + ca and a = 4, find b + c.
6. If x : y = 2 : 5, find (3x + y)/(x + 2y).

Answers: 1. 2. 2. 7/10. 3. 6. 4. 4. 5. 8. 6. 11/12.