This Quadratic Equations – CA Foundation Maths Study Material is designed strictly as per the latest syllabus and exam pattern.

## Quadratic Equations – CA Foundation Maths Study Material

ax^{2} + bx + c = 0; where a ≠ 0; a,b,c, are constants form equation is called Quadratic Equation or Second degree equation.

I. If b = 0 Then ax^{2} +c = 0 is called PURE Quadratic Equation.

II. If b ≠ 0 Then the equation, ax^{2} + bx + c = 0 where a ≠ 0 is called an AFFECTED Quadratic Equation.

**Roots:**

The value of the variable “x” which satisfies the given equation is called its Solution or roots of the Quadratic Equation.

Discriminant

For Quad. Eqn. ax^{2} + bx + c = 0.

Discriminant D = b^{2} – 4ac.

Example

For Eqn. 3x^{2} + 7x + 2 = 0.

a = 3;b = 7;c = 2 Discriminant D = b^{2} – 4a c

= 7^{2} – 4.3.2 = 49 – 24 = 25.

III. Roots of Quad. Eqn. ax^{2} + bx + c = 0 are

x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-b \pm \sqrt{D}}{2 a}\)

[Remember this formula, No need to prove it.]

IV. If a and f are roots of a Quadratic Equation ax^{2} + bx + c = 0

Then α + β = \(\frac{-b}{a}\)

∴ Sum of roots = \(\)

αβ = \(\frac{c}{a}\)

V. If α and β are roots of a Quadratic Eqn. Then the eqn. is

x^{2} – (α + β)x+a P =0

⇒ x^{2} – (sum of roots) x + Product of roots = 0.

VI. Nature of Roots

Nature of roots of a Quad. Eqn. depends upon Discriminant D = b^{2} – 4ac.

(A) If D > 0, Roots Real & Unequal

(i) D a perfect square then roots are Rational & unequal

As. \(\frac{2}{3} ;-\frac{2}{3}\)

(ii) D not a perfect Square.

Then roots are irrational & unequal and Conjugated As. 2 + √3; √5

(B ) If D = 0, Then Roots are Real & equal. b

Each root = –\(-\frac{b}{2 a}\)

(C) If D < 0, Then Roots are imaginary.

VII. If one root of a quadratic Eqn. is irrational then its other root is its irrational conjugate.

Example

If one root = 3+ √5

Then other root = 3 – √5 (irrational conjugate)

[To find conjugate change the sign of irrational part.]

Note:

(i) If one root is reciprocal to the other Then c = a

(ii) If one root is equal to the other but opposite in sign. Then b = 0.

**Cubic Equations**

1. Meaning of Cubic Equation The equation having form. ax^{3} + bx^{2} + cx + d = 0, a 0,

Where a, b, c, d are real numbers , is called a cubic equation.

2. Relation Between Roots and Coefficients

If α, β, γ are the roots of the cubic equation ax^{3} + bx^{3} + cx + d = 0, a ≠ 0, then

(i) α + β + γ = \(\frac{-b}{a}\)

(ii) αβ + βγ + γα = \(\frac{c}{a}\)

(Hi) αβγ = \(\frac{-d}{a}\)

3. The Cubic equation having roots a, p, y is

x^{3} – (α + β + γ)x^{2} + (αβ + βγ + γα)x – αβγ = 0

**Previous Year Exam Questions**

Question 1.

On Solving \(\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}\) = 2\(\frac{1}{6}\), we get one value of x as:

(a) \(\frac{4}{13}\)

(b) \(\frac{1}{13}\)

(c) \(\frac{2}{13}\)

(d) \(\frac{3}{13}\)

Answer:

\(\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}\)

Tricks : RHS is in fraction (i.e. rational No.)

Numerator value of x and 1-x must be a perfect square.

∴ (a) \(\frac{4}{13}\) = x must be its answer.

(a) is correct

Detail Method It is too much time taking.

Question 2.

Find the positive value of k for which the equations : x^{2} + kx + 64 = 0 and x^{2} – 8x + k = 0 will have real roots:

(a) 12

(b) 16

(c) 18

(d) 22

Answer:

(b) For Real roots

D = b^{2} -4ac ≥ 0

or b^{2} ≥ 4ac

For eqn(1); k^{2} ≥ 4.1.64 ⇒ k ≥ 16 ……….(I)

For eqn (2) (-8)^{2} ≥ 4.1.k ⇒ k ≤ 16 ……….(II)

Clearly k = 16 satisfies (I) & (II)

∴ (b) is correct.

Question 3.

If one root of an equation is 2 + √5, then the quadratic equation is:

(a) x^{2} + 4x – 1 = 0

(b) x^{2} – 4 x – 1 = 0

(c) x^{2} + 4x + 1 = 0

(d) x^{2} – 4x + 1 = 0

Answer:

(b) If one root = 2 + √5

Its other root = 2 – √5 (Irrational conjugate) Eqn is

x^{2} – (Sum of roots) x + Product of roots = 0

or x^{2} – (2 + √5 + 2 – √5)x + (2 + √5)(2 – √5)=0

or x^{2} – 4x + (4 – 5) = 0

or x^{2} – 4x – 1 = 0

(b) is correct.

Question 4.

A man starts his job with a certain monthly salary and earns a fixed increment every year. Ifhis salary was ₹ 1,500 after 4 years of service and ₹ 1,800 after 10 years of service, what was his starting salary and what is the annual increment in rupees? [1 Mark, May 2007]

(a) ₹ 1,300, ₹ 50

(b) ₹ 1,100, ₹ 50

(c) ₹ 1,500, ₹ 30

(d) None

Answer:

(a) Tricks : Go by choices.

For (a) Salary After 4 yrs 1300 + 4x 50 = ₹ 1500

Salary after 10 yrs = 1300 + 10x 50 = ₹ 1800

∴ (a) is correct Detail Method

Let Fixed Salary = a

Increment per yr = b

Salary after t yrs = s

s = a + bt

∴ a + b × 4 = 1500 = a + 4b = 1500 (1)

a + b × 10 = 1800 = a+ 10b = 1800 (2)

Eqn. (2) – Eqn. (1); we get

a + 10b = 1800

a + 4b = 1500

6b = 300 ∵ b = \(\frac{300}{6}\) = ₹ 50 o

From (1) a + 4 × 50 = 1500

or a = 1500 – 200 = ₹ 1300

(a) is correct

Question 5.

The sides of an equilateral triangle are shortened by 12 units, 13 units and 14 units respectively and a right angled triangle is formed. The side of the equilateral triangle is : [1 Mark, May 2007]

(a) 17 units

(b) 16 units

(c) 15 units

(d) 18 units

Answer:

Tricks : Go by choices

For (a) let each side =17 units.

∴ Sides are 17 -12 ; 17 -13

and 17 -14 = 5 ; 4 and 3 units respectively

They make a right angled triangle because sum of squares of two sides is equal to square of Largest side.

(a) is Correct

Detail Method

Let length of each side = x

Side of the right angle triangle are x – 12; x – 13 and x – 14

∴ By Pythagoras Theorem (x – 12)^{2} = (x – 13)^{2} +(x – 14)^{2}

or x^{2} – 24x + 144 = x^{2} – 26x + 169 + x^{2} – 28x + 196

or 0 = x^{2} – 30x + 221

or x^{2} – 17x – 13x + 221 = 0

or x(x – 17) – 13(x – 17) = 0

or (x – 17)(x – 13) = 0

x = 17 ; 13

No side = x – 13 =0 x=17

∴ (a) is correct.

Question 6.

The value of \(\sqrt{6+\sqrt{6+\sqrt{6+\ldots \ldots \ldots \infty}}}\) is:

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

Tricks :

(c) \(\sqrt{6+\sqrt{6+\sqrt{6+\ldots \ldots \ldots \infty}}}\) = +3

Find two factors of 6 such that their difference becomes 1.

6 = 3 × 2

(i) Give + sign for = Greater factor

(ii) – sign for Smaller factor

(c) is Correct

Question 7.

Area of a rectangular garden is 8000 square metres. Ratio in length and breadth is 5:4. A path of uniform width, runs all round the inside of the garden. If the path occupies 3200 m^{2}, what is its width? [1 Mark, Nov. 2007]

(a) 12m

(b) 6m

(c) 10m

(d) 4m

Answer:

(c) Let x is common in the ratio

∴ length = 5x and Breadth = 4x

∴ Area = 5x × 4x = 8000

or 20x^{2} = 8000

or x^{2} =400 = 20^{2}

x = 20

Length = 5x = 5 × 20 = 100m

Breadth = 4x = 4 × 20 = 80 m

Let Width of Path = ym

∴ Area of rest part = 8000 – 3200

or (100 – 2y) (80 – 2y) = 4800

Tricks : Go by choices.

(c) y = 10 m Satisfies it

Width = 10m is correct

Question 8.

The value of: [1 Mark, June 2008]

(a) 1 ± √2

(b) 2 + √5

(c) 2 ± √3

(d) None

Answer:

(a) let x =

(a) is correct

Question 9.

If x^{3} – 6x^{2} + 11x -6 = 0 then find the value of (3x – 4). [1 Mark, Dec. 2008]

(a) (1, 2, 3)

(b) (-1, 2, 5)

(c) (-1, 3, 5)

(d) (2, 3, 5)

Answer:

(a) Tricks : Go by choices For(a)

Let, 3x – 4 = -1 ; 3x = 3 ; So, x = 1

Putting x = 1 in the given eqn ; we get

LHS = 13 – 6 × 12 + 11 × 1 – 6 = 0 = RHS

Similarly, 3x – 4 =2 ; So x = 2

For x = 2 ; LHS = 23 – 6 × 22 + 11 × 2 – 6

= 8 – 24 + 22 – 6 = 0 = RHS

Similarly, 3x – 4 = 5; So x = 3

For x = 3;

LHS = 33 – 6 × 32 +11 × 3 – 6

= 27 – 54 + 33 – 6 = 0 = RHS

∴ (a) is correct

Question 10.

If (2 + √3 ) is a root of a quadratic equation x^{2} + px + q = 0 then find the value of p and Question

Answer:

Tricks:

One root = 2 + √3

∴ Another root =2-73

∴ Eqn is

x^{2} – (2 + √3 + 2 – √3)x + (2 + √3)(2 – √3) = 0

or x^{2} – 4x + (4 – 3) = 0

Comparing it with x^{2} + px + q = 0

p = -4 & q = 1

∴ (c) is correct

Question 11.

If area and perimeter of a rectangle is 6000 cm2 and 340 cm respectively, then the length of rectangle is : [1 Mark, Dec. 2008]

(a) 140

(b) 120

(c) 170

(d) 200

Answer:

(b) Let length = 1 & Breadth = b

lb = 6000 ………(1)

2(l + b) = 340

or l + b = 170 …………(2)

Tricks:

Then Go by choices

For (a) if l = 140 Then b = 170 – 140 = 30

∴ lb = 140 × 30 = 4200 ≠ 6000

∴ (a) is not the answer

For (b) if 1 = 120

Then 120+ b= 170

∴ b = 170 – 120 = 50

∴ Area = lb = 120x 50 = 6000

∴ (b) is correct

Question 12.

A straight line passes through the point (3,2). Find the equation of the straight line. [1 Mark, June 2009]

(a) x + y = 1

(b) x + y = 3

(c) x + y = 5

(d) x + y = 2

Answer:

(c) It cannot be solved properly because only one condition is given For Answer.

Tricks: Go by choices

(c) Point (3 ; 2) satisfies x + y = 5

∴ (c) is correct

Question 13.

One root of the equation: x^{2} – 2(5 + m) + 3(7 + m) = 0 is reciprocal of the other. [1 Mark, June 2009]

Find the value of m.

(a) -20/3

(b) 7

(c) 1/7

(d) 117

Answer:

(a)

Let one root = α, Then other root = \(\frac{1}{α}\)

α × \(\frac{1}{\alpha}=\frac{3(7+\mathrm{m})}{1}\)

1 = 21 + 3m

or 3m = -20 m = \(\frac{-20}{3}\)

(a) is correct

Question 14.

Roots of the equation 3x^{2} – 14x – k = 0 will be reciprocal of each other if: [1 Mark, June 2010]

(a) k= -3

(b) k = 0

(c) k = 3

(d) k = 14

Answer:

(c) is correct.

Let one root = α & Another root = \(\frac{1}{α}\) (given)

∵ Product of roots = c/a

α.\(\frac{1}{α}\) = — a 3

∴ 1 = \(\frac{k}{3}\) So, K = 3

∴ (c) is correct

Question 15.

Positive value of‘k’ for which the roots at equation 12x^{2} + kx + 5 = 0 are in ratio 3 :2, is: [1 Mark, Dec. 2010]

(a) 5/12

(b) 12/5

(c) \(\frac{5 \sqrt{10}}{2}\)

(d) 5\(\sqrt{10}\)

Answer:

Let a is common in the ratio.

∴ Roots are 3α and 2α

Sum of roots = \(-\frac{b}{a}\)

∴ 3α + 2α = 5α = \(-\frac{k}{12}\)

So, α = \(-\frac{k}{60}\)

Product of roots = 3α × 2α = \(\frac{c}{a}\)

∴ 6α^{2} = \(\frac{5}{12}\)

⇒ 6.\(\left(-\frac{k}{60}\right)^2=\frac{5}{12}\), So, k^{2} = 250

k = 5\(\sqrt{10}\)

∴ (d) is correct

Question 16.

If one root of the equation x^{2} – 3x + k = 0 is 2, then value of k will be : [1 Mark, Dec. 2010]

(a) -10

(b) 0

(c) 2

(d) 10

Answer:

(c) is correct

2 is a root of given eqn.

∴ 2^{2} – 3 × 2 + k = 0

or -1 + k = 0

∴ k = 2

∴ (c) is correct

Question 17.

It roots of equation x^{2} + x + r = 0 are ‘α’ and ‘β’ and α^{3} + β^{3} = -6 . Find the value of ‘r’? [1 Mark, June 2011]

(a) \(\frac{-5}{3}\)

(b) \(\frac{7}{3}\)

(c) \(\frac{-4}{3}\)

(d) 1

Answer:

(a) is correct.

α + β = \(\frac{-1}{1}\) = -1 & = αβ = r

∴ α^{3} + β^{3} = -6

or (α + β)^{3} – 3αβ(α + β) = -6

or (-1)^{3} – 3r(-1) = -6

or -1 + 3r = -6;

or 3r = -5

r = -5/3

∴ (a) is correct

Question 18.

If one root of the Equation px^{2} + qx + r = 0 is r then other root of the Equation will be: [1 Mark, Dec. 2011]

(a) 1/q

(b) 1/r

(c) 1/p

(d) \(\frac{1}{p+q}\)

Answer:

(c) Let α is another root.

rα = \(\frac{r}{p}\)

∴ α = \(\frac{1}{p}\)

Question 19.

If the ratio of the root of the Equation 4x^{2} – 6x + p = 0 is 1:2 then the value of p is: [1 Mark, Dec. 2011]

(a) 1

(b) 2

(c) -2

(d) -1

Answer:

(b) Let a is common in the ratio

∴ α + 2α = \(\frac{-(-6)}{4}\) ⇒ α = \(\frac{1}{2}\)

∴ α.2α = \(\frac{p}{4}\)

∴ p = 8α^{2} = 8.\(\frac{1}{4}\) = 2

Question 20.

If p & q are the root of the Equation x^{2} – bx + c =0, then what is the Equation whose roots are (pq + p + q) and (pq – p – q) ? [1 Mark, Dec. 2011]

Answer:

(a) Eqn is x^{2} – (p + q)x + pq = 0

∴ b = p + q ; C = pq

New roots are pq + (p + q) = c + b

& pq – (p + q) = c – b

∴ Eqn is

x^{2} – (c + b + c – b)x + (c + b)(c – b) = 0

or x^{2} – 2cx + c^{2} – b^{2} = 0

Question 21.

If arithmetic mean between roots of a quadratic equation is 8 and the geometric mean between them is 5, the equation is: [1 Mark, June 2012]

O) x^{2} – 16x – 25 = 0

(b) x^{2} – 16x + 25 = 0

(c) x^{2} – 16x + 5 = 0

(d) None of these

Answer:

(b) Let α and β are two roots.

(α + β)/2 = 8 and \(\sqrt{αβ}\) = 5

⇒ α + β =16 & αβ = 25

Eqn. is x^{2} – 16x + 25 = 0

Question 22.

The minimum value of the function x^{2} – 6x +10 is

(a) 1

(b) 2

(c) 3

(d) 10

Answer:

(a) coeff. of x^{2} = 1 > o; function is minimum (Formula) i

Minimum value = \(\frac{4 a c-b^2}{4 a}\)

\(\frac{4.1 .10-(-b)^2}{4 \times 1}=\frac{4}{4}\) = 1

Question 23.

Ifoneofthe roots of the equation x^{2} + px + a is √3 + 2, then the value of ‘p’ and ‘a’ is: [1 Mark, June 2012]

(a) -4, -1

(b) 4, -1

(c) -4, 1

(d) 4, 1

Answer:

(c) Roots are 2 + √3 and 2 – √3 (conjugate of 2 + √3)

Eqn is

x^{2} – (Sum of roots) x + product of roots = 0

x^{2} – 4x + (4 – 3) = 0

x^{2} + px + a = 0

∴ P = -4 ; a = 1

Question 24.

Roots of equation 2x^{2} + 3x + 7 = 0 are α and p. The value of αβ^{-1} + βα^{-1} is: [1 Mark, Dec. 2012]

(a) 2

(b) 3/7

(c) 7/2

(d) -19/14

Answer:

Question 25.

The quadratic equation x^{2} – 2kx + 16 = 0 will have equal roots when the value of ‘k’ is: [1 Mark, Dec. 2012]

(a) ±1

(b) ±2

(c) ±3

(d) ±4

Answer:

(d) Let roots are α; α

∴ α + α = \(\frac{-(-2 k)}{1}\) ⇒ α = k

α.α = k.k = 16 ⇒ k^{2} = 16 ⇒ k = ±4

Question 26.

If α, β are roots of x^{2} + 7x + 11 = 0 then the equation whose roots as (α + β)^{2} &.(α – β)^{2} is: [1 Mark, June 2013]

(a) x^{2} – 54x + 245 = 0

(b) x^{2} – 14x + 49 =0

(c) x^{2} – 24x + 144 = 0

(d) x^{2} – 50x + 49 = 0

Answer:

(a) is correct

α + β = \(-\frac{b}{a}=-\frac{7}{1}=\) = -7

αβ = \(\frac{c}{a}=\frac{11}{1}\) = 11

(α – β)^{2} -(α + β)^{2} – 4αβ

= (-7)^{2} – 4 × 11 = 5

Required eqn. is

x^{2} – {(α + β)^{2} + (α – β)^{2}}x + (α + β)^{2}(α – β)^{2} =0

or x^{2} – (49+5)x + 49 × 5 = 0

or x^{2} – 5x + 245 = 0

∴ (a) is correct

Question 27.

If b^{2} – 4ac is perfect square but not equal to zero then the roots of the equation ax^{2} + bx + c = 0 are: [1 Mark, Dec. 2013]

(a) Real and equal

(b) Real irrational and equal

(c) Real rational and unequal

(d) Imaginary

Answer:

(c) b^{2} -4ac > 0 & perfect square Roots are real rational and unequal

Question 28.

Divide 80 into two parts so that their products is maximum then the numbers are: [1 Mark, Dec. 2013]

(a) 15, 65

(b) 25, 55

(c) 35, 45

(d) 40, 40

Answer:

(d) is correct

Let 1st part = x

∴ 2nd part 80 -x

Let y = x(80 – x) = -x^{2} + 80x

Here co. eff. of x^{2} < 0

y is maximum at

x = \(\frac{-b}{2 a}=-\frac{80}{2(-1)}\) = 40

∴ Numbers are (40 ; 40)

Tricks : Go by choices.

Question 29.

The roots of equation y^{3} + y^{2} – y – 1 = 0 are

(a) 1,1,-1

(b) -1,-1,1

(c) 1,1,1

Answer:

(b) is correct

y^{3} + y^{2} – y – 1 = 0

or y^{2}(y + 1) – 1(y + 1) = 0

or (y + 1)(y^{2} – 1) = 0

or(y + 1)(y + 1)(y – 1)=0

y = -1 ; -1; 1.

Tricks : Go by choices.

Question 30.

If α, β are the roots of the quadratic equation 2x^{2} – 4x = 1 then the value \(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\) =

(a) -11

(6) 22

(c) -22

(d) 11

Answer:

2x^{2} – 4x – 1 = 0

Let α and β are its roots

Question 31.

If α, β be the roots of a quadratic equation if α + β = -2, αβ = -3 Find quadratic equation: [1 Mark, Dec. 2015]

(a) x^{2} + 2x – 7 = 0

(c) x^{2} + 2x – 3 = 0

(c) x^{2} – 2x – 3 = 0

(d) x^{2} – 2x + 7 = 0

Answer:

(b) is correct Quadratic Eqn. is

x^{2} – (α + β)x + αβ = 0

x^{2} – (-2)x + (-3) = 0

∴ x^{2} + 2x – 3 = 0

Question 32.

Value of k for which roots are equal of given equation 4x^{2} – 12x + k = 0. [1 Mark, Dec. 2015]

(a) 144

(6) 9

(c) 5

(d) None of these

Answer:

(b) is correct 4x^{2} – 12x + k = 0

D = b^{2} – 4.ac = 0

= (-12)^{2} = 4 × 41

Or, 144 = 16k ∴ k =9

Question 33.

If difference between the roots ofthe equation x^{2} – kx + 8 = o is 4 then the value of K is: [1 Mark, June 2016]

(a) 0

(b) ±4

(c) ±8√3

(d) ±4√3

Answer:

(d) is correct.

let α, β are roots of x^{2} – kx + 8 = 0

∴ α + β = -b/a = \(-\frac{(-k)}{1}\) = k & α. β = c/a = 8/1 = 8

(α – β)^{2} = (α + β)^{2} – 4αβ = 4^{2}

⇒ k^{2} – 4 × 8 = 16

or k^{2} = 48 ⇒ k = ±\(\sqrt{16 \times 3}\) ⇒ k = ±4√3

(d) is correct.

Question 34.

If α, β be the roots of x^{2} + x + 5 = 0 then \(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\) = ___. [1 Mark, June 2017]

(a) \(\frac{16}{5}\)

(b) 2

(c) 3

(d) \(\frac{14}{5}\)

Answer:

(d) is correct.

Question 35.

If the sum of two numbers is 13 and the sum of their squares is 85 then the numbers are : [1 Mark, Dec. 2017]

(a) 6, 7

(b) 4, 9

(c) 10, 3

(d) 5, 8

Answer:

(a)

Tricks : GBC (Go by choices)

6 +7 = 13 (True)

& 6^{2} + 7^{2} = 36 + 49 = 85 (True)

Question 36.

The difference between the roots of the equation x^{2} – 7x – 9 = 0 is : [1 Mark, Dec. 2017]

(a) 7

(b) \(\sqrt{85}\)

(c) 9

(d) 2\(\sqrt{85}\)

Answer:

(b)

Let α and J3 are roots.

α + β = \(-\frac{b}{a}=-\frac{-7}{1}\) = 7

αβ = \(\frac{c}{a}=\frac{-9}{1}\) = -9.

∴ (α – β)^{2} = (α + β)^{2} – 4

αβ = 7^{2} – 4(-9) = 85

α.β = \(\sqrt{85}\)

Question 37.

The roots of the cubic equation x^{3} + 7x^{2} – 21x – 27 = 0 is ___ : [1 Mark, Dec. 2017]

(a) -1, 3, 9

(6) 1, -3, 9

(c) -1, 3, -9

(d) -1, -3, 9

Answer:

(c)

TRICKS: – Go by choices.

Question 38.

If the roots of the equation kx^{2} – 3x -1= 0 are the reciprocal of the roots of the equation x^{2} + 3x – 4 = 0 then K =

(a) 4

(b) -4

(c) 3

(d) -3

Answer:

(a)

∵ x^{2} + 3x – 4 = 0

or; x^{2} – 4x + x – 4 = 0

or; x(x – 4) + 1(x – 4) = 0

or; (x – 4)(x + 1) = 0

x = 4; -1

Eqn. having roots \(\frac{1}{2}\) & \(\frac{1}{-1}\) = \(\frac{1}{4}\) & – 1 is.

or x^{2} – (\(\frac{1}{4}\) – 1) + \(\frac{1}{4}\)(-1) = 0

or x^{2} + \(\frac{3}{4}\)x – \(\frac{1}{4}\) = 0

Multiplying by 4 ; we get

4x^{2} + 3x -1 = 0

Comparing it with kx^{2} + 3x -1 = 0

We get K = 4

Tricks : Eqn. having roots the reciprocal of the roots of ax^{2} + bx + c = 0 is cx^{2} + bx +a = 0 i.e. 1st and last term interchanges.

Question 39.

If the roots of the equation x^{2} – 15x^{2} + kx – 45 = 0 are in A.P., find value of k: [1 Mark, June 2018]

(a) 56

(b) 59

(c) -56

(d) -59

Answer:

∵ Roots are in A.R

Let roots are a – d; a; a + d

So, (a – d)+a + (a + d) = 15

or; 3a = 15

or; a = 5

And Product of roots

(a – d ). a . (a + d ) = 45

or (5 – d);5. (5 + d) = 45

or 25 – d^{2} = 9

or; d^{2} = 25 – 9 = 16

or; d = \(\sqrt{16}\) = 4

Hence; roots are

a – d, a, a + d = 5 – 4; 5; 5 + 4

= 1; 5 ; 9.

The value of K

= Sum of product of two roots in a order

= (1 × 5) + (5 × 9) + (9 × 1)

= 5 + 45 + 9 = 59

(b) is correct.

Tricks : If α; β and γ are the roots of a cubic Eqn.

So ; Cubic Eqn. is

x^{3} – (α + β + γ)x^{2} + (αβ + βγ + γα)x – αβγ = 0

Given Eqn. is

x^{3} – 15x^{2} + k . x – 45 = 0

Comparing it,

α + β + γ = 15 ⇒ 1 + 5 + 9 = 15

& αβγ = 45 ⇒ 1 × 5 × 9 = 45

[Apply Hit and Trial method which can satisfy both]

Hence ; we can say ;

α = 1; β = 5; γ = 9;

k = αβ + βγ + γα= 1 × 5 + 5 × 9 + 9 × 1 = 59

Question 40.

If α + β = -2 and αβ = -3 where a and are the roots of the equation, which is [1 Mark, May 2018]

(a) x^{2} – 2x – 3 = 0

(b) x^{2} + 2x – 3 = 0

(c) x^{2} + 2x + 3 = 0

(d) x^{2} – 2x + 3 = 0

Answer:

(b)

Quadratic Eqn. having roots α and β is

x^{2} – (α + β)x + αβ = 0

or ; x^{2} – (- 2)x + (- 3)= 0

or; x^{2} + 2x – 3 = 0

(b) is correct.

Question 41.

If α and β are the roots of the equation x^{2} + x + 5 = 0 equal to: [1 Mark, May 2018]

(a) \(\frac{16}{5}\)

(b) 3

(c) 7

(d) 2

Answer:

(c)

Given Eqn. is x^{2} + x + 5 = 0.

(c) is correct.

Question 42.

When two roots of quadratic equation are α, \(\frac{1}{α}\) then what will be the quadratic equation: [1 Mark, Nov. 2018]

(a) αx^{2} – (α^{2} + 1)x + a = 0

(b) αx^{2} – αx^{2} + 1 = 0

(c) αx^{2} – (α^{2} + 1)x + 1 = 0

(d) None of these

Answer:

(a)

Tricks : For (a) α. \(\frac{1}{α}\) = 1 = \(\frac{c}{a}=\frac{\alpha}{\alpha}\) = 1 (True)

Question 43.

Let α and β be the roots of x^{2} + 7x + 12 = 0 Then the value of \(\left(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\right)\) will be: [1 Mark, Nov. 2018]

(a) \(\frac{49}{144}+\frac{144}{49}\)

(b) \(\frac{7}{12}+\frac{12}{7}\)

(c) \(-\frac{91}{12}\)

(d) None of these

Answer:

(c)

x^{2} + 7x + 12 = 0

or x^{2} + 4x +3.x+ 12 = 0

or x(x + 4)+3(x + 4) = 0

or (x + 4) (x + 3) = 0

x = – 3, 4

let α = -3; β = -4

Question 44.

Find the condition that one root is double the of ax^{2} + bx + c = 0 [1 Mark, June 2019]

(a) 2b^{2} = 3ac

(b) b^{2} = 3ac

(c) 2b^{2} = 9ac

(d) None

Answer:

(c)

Tricks

Let 1st root = 1

Then 2nd root = 2

The Eqn. is x^{2} – (1 + 2) + 1 × 2 = 0

or x^{2} – 3x + 2 = 0

Comparing it with ax^{2} + bx + c = 0 we get;

a = 1; b = -3 ; c = 2

Go by choices (GBC)

(a) 2b^{2} = 3ac

2.(-3)^{2} =3.1.2 (False)

(c) 2b^{2} =9.ac

2. (-3)^{2} = 9.1.2

⇒ 18 = 18 (True)

Hence, Option (c) is (true)