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RRB NTPC (Railways) · Quantitative Aptitude

Algebra

Algebraic identities, linear and quadratic equations, surds, indices and graphs of linear equations.

Six concepts, and they are the six identities and root facts SSC algebra keeps asking. Surds, indices and coordinate graphs live in their own topics; this one is the algebra that turns a given sum or product into the expression the question wants, without solving for every letter.

  • RRB NTPC (Railways)
  • Hard level
  • 6 concepts
  • 43 practice questions

1Square identities, and a^2+b^2 without a and b

(a+b)^2 = a^2+2ab+b^2 and (a-b)^2 = a^2-2ab+b^2 are the same expansion with the middle sign flipped. Adding them cancels the middle term and gives (a+b)^2+(a-b)^2 = 2(a^2+b^2); subtracting them isolates 4ab.

The move the exam wants is the rearrangement a^2+b^2 = (a+b)^2 - 2ab. When the paper hands you a+b and ab and asks for a^2+b^2, you never need a and b individually — and trying to find them first is how the clock runs out.

Figure. The big square has side a+b. Its area is the a^2 block, the b^2 block, and two ab rectangles — which is exactly (a+b)^2 = a^2+2ab+b^2. The side lengths are to scale with a:b = 2:1; the labels name the pieces, not the numerical areas.

How it works

  1. Name what you were givenUnderline a+b (or a-b) and ab. Those two, not a and b, are the inputs.
  2. Square the sum (or difference)(a+b)^2 expands to a^2+2ab+b^2, so the only extra piece is 2ab.
  3. Subtract 2aba^2+b^2 = (a+b)^2 - 2ab. Flip the sign of 2ab if you started from a-b.

Find a^2+b^2 from the sum and product

If a+b = 9 and ab = 20, find a^2+b^2.

  • Identitya^2+b^2 = (a+b)^2 - 2ab
  • (a+b)^2 = 9^281
  • 2ab = 2 \times 2040
  • a^2+b^2 = 81 - 4041

Pro tip. Adding the two square identities gives 2(a^2+b^2) directly, which is the same arithmetic when both a+b and a-b are known. With only a+b and ab, the one-line rearrangement above is enough.

If a+b = 7 and ab = 10, then a^2+b^2 is
  1. 29
  2. 49
  3. 69

(a+b)^2 - 2ab = 49 - 20 = 29. Leaving the answer at 49 is squaring the sum and stopping; adding 2ab instead of subtracting it gives 69.

2Factor a^2-b^2 and a^3 \pm b^3

a^2-b^2 = (a+b)(a-b) is the factorisation that turns a difference of two squares into a product you can often evaluate by mental arithmetic. The same pattern lifts one degree: a^3-b^3 = (a-b)(a^2+ab+b^2) and a^3+b^3 = (a+b)(a^2-ab+b^2).

The sign in the linear factor matches the cube's sign; the middle term of the quadratic factor flips. Mixing those two signs is the standard wrong answer on every paper that asks for the identity rather than a number.

Figure. The shaded a^2 square loses the b^2 corner. What remains rearranges into an (a-b) by (a+b) rectangle — so a^2-b^2=(a-b)(a+b). Sides are to scale with a:b=2:1; the cube identities stay algebraic.

How to evaluate

  1. Spot the formTwo perfect squares (or cubes) with a minus — or two cubes with a plus — mean a factorisation, not a long expansion.
  2. Write the factorsFor squares: (a+b)(a-b). For cubes: linear factor keeps the sign, quadratic middle term flips it.
  3. Multiply the easy numbersChoose a and b so a+b or a-b is a round figure; the product is then one short multiplication.
Cube factorisations at a glance
FormFactorsMiddle-term sign
a^3-b^3(a-b)(a^2+ab+b^2)Same as the linear factor
a^3+b^3(a+b)(a^2-ab+b^2)Opposite the linear factor
a^2-b^2(a+b)(a-b)(no middle term)

A difference of squares without squaring

Evaluate 47^2 - 3^2 without computing either square.

  • Identitya^2-b^2 = (a+b)(a-b)
  • a+b = 47+350
  • a-b = 47-344
  • 50 \times 442200

Pro tip. The same pattern scores on cubes: 9^3-1^3 = (9-1)(81+9+1) = 8 \times 91 = 728. Once the factors are written, the arithmetic is ordinary.

a^3+b^3 factorises as
  1. (a+b)(a^2-ab+b^2)
  2. (a+b)(a^2+ab+b^2)
  3. (a-b)(a^2-ab+b^2)

The linear factor keeps the plus from a^3+b^3, and the quadratic middle term flips to -ab. Keeping +ab in the quadratic is the most common mix-up; writing a-b as the linear factor is the identity for a^3-b^3.

3The x + 1/x family

If x + \frac{1}{x} = k, then squaring both sides and rearranging gives x^2 + \frac{1}{x^2} = k^2 - 2. Cubing and rearranging gives x^3 + \frac{1}{x^3} = k^3 - 3k. Both are staples of SSC algebra because the question never asks you to find x.

The method is always the same: raise the given sum to the power you need, then peel off the cross terms that the binomial produces. Jumping straight to k^2-2 or k^3-3k is the shortcut; solving a quadratic for x first is the long way to the same number.

Figure. If x + 1/x = k, squaring and peeling the cross term gives x² + 1/x² = k² − 2. Cubing and peeling gives x³ + 1/x³ = k³ − 3k. For k = 4 that is 16 − 2 = 14 and 64 − 12 = 52. The question never asks you to find x.

How it works

  1. Name kRead off k = x + 1/x from the stem and stop looking for x.
  2. Square for the second power\left(x+\frac{1}{x}\right)^2 = x^2+2+\frac{1}{x^2}, so x^2+\frac{1}{x^2} = k^2-2.
  3. Cube for the third power\left(x+\frac{1}{x}\right)^3 = x^3+3x+\frac{3}{x}+\frac{1}{x^3}, which rearranges to k^3-3k.

Evaluating a cube from x + 1/x

If x + \frac{1}{x} = 4, find the value of x^3 + \frac{1}{x^3}.

  • Identityx^3+\frac{1}{x^3} = k^3-3k
  • k = 44
  • k^3 = 4^364
  • k^3-3k = 64-1252

Pro tip. Memorise x^3+\frac{1}{x^3}=k^3-3k; it turns a messy cubing into one subtraction. The same k also gives x^2+\frac{1}{x^2}=k^2-2=14 if a follow-up asks for it.

If x + \frac{1}{x} = 3, then x^2 + \frac{1}{x^2} is
  1. 7
  2. 9
  3. 11

k^2-2 = 9-2 = 7. Leaving the answer at 9 is squaring k and forgetting to subtract 2; adding 2 instead of subtracting it gives 11.

4When a+b+c=0, the cubes collapse to 3abc

The full identity is a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca). The second factor is finite, so the whole right-hand side vanishes exactly when a+b+c=0, which forces a^3+b^3+c^3 = 3abc.

That is the exam move: test the sum first. If the three terms add to zero, replace the sum of cubes with 3abc and skip the expansion. If they do not, the shortcut does not apply and you need the full identity or direct cubing.

Figure. Three signed steps 7, -4 and -3 start at 0 and finish at 0. When a+b+c=0, the identity collapses to a^3+b^3+c^3=3abc — test the sum before expanding cubes.

How it works

  1. Add the three termsCompute a+b+c. The shortcut lives or dies on this one number.
  2. Confirm it is zeroIf the sum is anything else, stop — a^3+b^3+c^3=3abc is false in general.
  3. Replace with 3abcEvaluate 3abc with ordinary multiplication, watching the sign of each factor.

Using the a+b+c=0 identity

If a=7, b=-4, c=-3, find a^3+b^3+c^3.

  • a+b+c = 7+(-4)+(-3)0
  • Since sum is 0a^3+b^3+c^3 = 3abc
  • abc = 7 \times (-4) \times (-3)84
  • 3abc = 3 \times 84252

Pro tip. Always test whether the three terms sum to zero first — if so, the answer is just 3abc. Two negatives in the product make abc positive here; missing either sign is how 252 becomes -252.

If a=5, b=3, c=-8, then a^3+b^3+c^3 equals
  1. -360
  2. 360
  3. 120

a+b+c=0, so the sum of cubes is 3abc = 3\times5\times3\times(-8) = -360. Dropping the minus on c gives +360; computing abc without the leading 3 gives -120, and a sign slip then yields the distractor 120.

5Quadratic roots: sum, product and the formula

For ax^2+bx+c=0 the roots are x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}. You rarely need both roots written out: their sum is -\frac{b}{a} and their product is \frac{c}{a}, read straight from the coefficients.

Vieta's relations are how SSC asks most quadratic questions — "sum of roots", "product of roots", "equation with given roots" — without ever demanding the quadratic formula. Keep the signs: sum carries a minus on b, product does not.

Figure. For 2x^2-5x-3=0 the roots sit at -1/2 and 3. Their midpoint is half the sum -b/(2a); the product is c/a. Read sum and product from the coefficients before writing the quadratic formula.

How to read a quadratic

  1. Identify a, b, cWrite the equation as ax^2+bx+c=0 with the signs absorbed into b and c.
  2. Sum and productSum = -b/a, product = c/a. Do not flip either sign a second time.
  3. Formula only if askedReach for \frac{-b\pm\sqrt{D}}{2a} when a specific root is required; otherwise Vieta is enough.

Roots of 2x^2-5x-3=0

For 2x^2 - 5x - 3 = 0, find the sum and product of the roots, then the roots themselves.

  • a=2, b=-5, c=-3coefficients
  • Sum =-b/a = 5/22.5
  • Product =c/a = -3/2-1.5
  • D=b^2-4ac=25+24=49; roots (5\pm7)/43 and -0.5

Pro tip. Check: 3+(-0.5)=2.5 and 3\times(-0.5)=-1.5, which matches Vieta. If the check fails, the error is almost always a missed sign on b when reading the sum.

For x^2-5x+6=0, the product of the roots is
  1. 6
  2. -6
  3. 5

Product = c/a = 6. The sum is 5, which is the distractor that confuses the two Vieta numbers; -6 flips the product sign as if the constant term were negative.

6Nature of roots, and building the equation

The discriminant D = b^2-4ac decides the nature of the roots: D>0 gives two distinct real roots, D=0 gives equal (repeated) real roots, and D<0 gives a conjugate pair of imaginary roots. A perfect-square D makes those real roots rational when a, b, c are rational.

Questions that ask for the value of a parameter so that the roots are equal are asking you to set D=0 and solve. The reverse construction is just as common: given two roots, the monic quadratic is x^2 - (\text{sum})x + (\text{product}) = 0 — write that form rather than multiplying (x-r_1)(x-r_2) out by hand.

Figure. The sign of D=b^2-4ac is the whole classification: positive for two distinct real roots, zero for a repeated root, negative for a conjugate imaginary pair. A perfect-square D with rational coefficients makes those real roots rational.

How to set the condition

  1. Write DForm b^2-4ac with the parameter still in it.
  2. Apply the conditionEqual roots means D=0; real roots means D\ge 0; for a constructed equation, skip D and use sum and product.
  3. Solve or write the quadraticSolve the resulting equation for the parameter, or write x^2-(\text{sum})x+(\text{product})=0.
Discriminant → nature of roots
ConditionNature of rootsTypical ask
D>0Two distinct real rootsAre the roots real?
D=0Equal real rootsFind k for equal roots
D<0Imaginary (conjugate) rootsNo real root
D a perfect squareRational roots (rational coeffs)Roots rational or not

Equal roots fix the constant term

Find k so that the roots of x^2 - 4x + k = 0 are equal, and state the root.

  • D = (-4)^2 - 4(1)(k)16-4k
  • Equal roots \Rightarrow D=016-4k=0
  • k = 16/44
  • Repeated root = -b/(2a) = 4/22

Pro tip. To build a quadratic with roots 3 and -2, write x^2-(3-2)x+(3)(-2)=0, i.e. x^2-x-6=0, rather than expanding binomials. Sum and product are the only numbers you need.

The quadratic with roots 3 and -2 is
  1. x^2 - x - 6 = 0
  2. x^2 + x - 6 = 0
  3. x^2 - x + 6 = 0

Sum =1 and product =-6, so x^2 - (\text{sum})x + (\text{product}) = x^2 - x - 6 = 0. Flipping the linear sign treats the sum as -1; flipping the constant treats the product as +6.

Notes

  • Key Identities: (a+b)^2 = a^2+2ab+b^2 and (a-b)^2 = a^2-2ab+b^2 let you find a^2+b^2 = (a+b)^2 - 2ab without knowing a,b individually.
  • The x + 1/x Family: If x+\frac{1}{x}=k, then x^2+\frac{1}{x^2}=k^2-2 and x^3+\frac{1}{x^3}=k^3-3k — a staple of SSC algebra questions.
  • Sum of Cubes Identity: a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca), so if a+b+c=0 then a^3+b^3+c^3 = 3abc.
  • Quadratic Roots: For ax^2+bx+c=0, sum of roots =-\frac{b}{a} and product =\frac{c}{a}; the discriminant D=b^2-4ac decides real (D\ge0) vs imaginary (D<0) roots.
  • Nature of Roots Trap: A perfect-square discriminant means rational, equal roots occur only when D=0; questions often ask for the condition on a parameter to make D=0.

Formulas

  • (a+b)^2 + (a-b)^2 = 2(a^2+b^2)
  • a^2-b^2 = (a+b)(a-b)
  • a^3 \pm b^3 = (a\pm b)(a^2 \mp ab + b^2)
  • If x+\frac{1}{x}=k: x^2+\frac{1}{x^2}=k^2-2, x^3+\frac{1}{x^3}=k^3-3k
  • Quadratic roots: x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}, sum =-\frac{b}{a}, product =\frac{c}{a}
  • If a+b+c=0 then a^3+b^3+c^3 = 3abc

Exam traps & shortcuts

  • Whenever you see x+\frac1x = k, jump straight to k^2-2 or k^3-3k for higher powers instead of solving for x.
  • Spot a+b+c=0 in an expression and immediately replace a^3+b^3+c^3 with 3abc to skip expansion.
  • To build a quadratic with given roots, write x^2-(\text{sum})x+(\text{product})=0 rather than multiplying factors out.

Reference tables

Every line here should be reconstructible from the concept above it, not merely recalled.

Formula sheet
RelationIdentity / ruleWatch for
(a+b)^2+(a-b)^22(a^2+b^2)Middle terms cancel
a^2+b^2(a+b)^2-2abNeed sum and product, not a and b
a^2-b^2(a+b)(a-b)Factor before squaring
a^3\pm b^3(a\pm b)(a^2\mp ab+b^2)Linear sign matches; middle flips
x+1/x=kx^2+1/x^2=k^2-2; x^3+1/x^3=k^3-3kDo not solve for x
a+b+c=0a^3+b^3+c^3=3abcTest the sum before cubing
Sum / product of roots-b/a and c/aMinus only on the sum
Equal rootsD=b^2-4ac=0Solves for the parameter

Recap

Read only this the night before.

Squares
a^2+b^2=(a+b)^2-2ab. Adding the \pm expansions cancels the middle term and doubles a^2+b^2.
Factor
a^2-b^2=(a+b)(a-b). For cubes, the linear factor keeps the sign and the middle term of the quadratic flips.
x+1/x
Jump to k^2-2 or k^3-3k. Solving for x first is the slow route to the same answer.
a+b+c=0
Then a^3+b^3+c^3=3abc. Always test the sum before expanding.
Vieta
Sum =-b/a, product =c/a. Build the monic quadratic as x^2-(\text{sum})x+(\text{product})=0.
Discriminant
D>0 distinct real, D=0 equal, D<0 imaginary. Equal-root parameter questions are D=0.

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