Quadratic Equations

⇒ 42 – 4 x 2 (-7)

⇒ 16 + 56 = 72 > 0

Hence, roots of quadratic equation are real and unequal.

∴ It must satisfy the quadratic equation.

i.e., b2 – 4ac = 0

⇒ b2 = 4 ac

⇒ c $=\frac{b^2}{4a}$

Product of the roots = ab = $\frac{B}{A}$ = – b

= a + b = – a and ab = -b

⇒ 2a = -b and a = -1

⇒ b = 2 and a = -1

⇒ LHS = 3x² + 13x + 14

⇒ 3(-2)² + 13(-2) + 14

⇒ 12 – 26 + 14 = 0

⇒ RHS Hence, x = -2 is a solution.

⇒ 64 – 64 = 0

⇒ x² – x – 2 + x = 0

⇒ x² – 2 = 0

D = b² – 4ac

⇒ (-4(1)(-2) = 8 > 0

∴ Given equation has two distinct real roots.

∴ x² – 0.9 = (0.3)² – 0.9 = 0.09 – 0.9 ≠ 0

Hence, 0.3 is not a root of given equation.

⇒ 2(3)² + 3 + k = 0

⇒ 18 + 3 + k = 0

⇒ k = – 21

D = 0

⇒ b² – 4ac = 0

⇒ (- 3k)² – 4 × 9 × k = 0

⇒ 9k² = 36k

⇒ k = 4

= √2x² + 5x + 2x + 5√2 = 0

x(√2x + 5) + √2 (√2x + 5) = 0

= (√2x + 5)(x + √2) = 0

∴ Either √2x + 5 = 0 or x + √2 = 0

∴ x = – $\frac{5}{\sqrt{2}}$ or x = –√2

Hence, the roots are – $\frac{5}{\sqrt{2}}$ and -√2.

(ii) We have, 2x² – x + 18 = 0

On dividing both sides by 2, we have

(ii) We have, 4x² + 4√3x + 3 = 0

Here, a = 2, b = -7 and c = 3

Therefore, D = b+ – 4ac

⇒ D = (-7)² – 4 × 2 × 3 = 49 – 24 = 25

∵ D > 0, ∴ roots exist.

So, the roots of given equation are 3 and $\frac{1}{2}.$

(ii) We have, 4x² + 4√3x + 3 = 0

Here, a = 4, b = 4√3 and c = 3

Therefore, D = b² – 4ac = (4√3)2 – 4 × 4 × 3 = 48 – 48 = 0

∴ D = 0, roots exist and are equal.

Comparing this equation with ax² + bx + c = 0, we have

a = p², b = p² – q² and c = – q²

∴ D = b² – 4ac

⇒ (p² – q)² – 4 × p² × (-q²)

⇒ (p² – q²)² + 4p²q²

⇒ (p² + q³)2 > 0

So, the given equation has real roots given by

⇒ (x + 3) (x – 6)

⇒ -20 or x² – 3x + 2 = 0

⇒ x² – 2x -x + 2 = 0

⇒ x(x – 2) -1(x – 2) = 0)

⇒ (x – 1) (x – 2) = 0

⇒ x = 1 or x = 2

Both x = 1 and x = 2 are satisfying the given equation. Hence, x = 1, 2 are the solutions of the equation.

Here, a = 3, b = – 4√3 and c = 4

Therefore,

D = b² – 4ac

⇒ (- 4√3)² – 4 × 3 × 4

⇒ 48 – 48 = 0

Hence, the given quadratic equation has real and equal roots.

(ii) We have, 2x² – 6x + 3 = 0

Here, a = 2, b = -6, c = 3

Therefore, D = b² – 4ac

= (-6)2 4 × 2 × 3 = 36 – 24 = 12 > 0

Hence, given quadratic equation has real and distinct roots.

Here, a = 2, b = k, c = 3

D = b² – 4ac = k² – 4 × 2 × 3 = k² – 24 For equal roots

D = 0

i.e., k² – 24 = 0

⇒ ķ² = 24

⇒ k = ± √24

⇒ k = + 2√6

(ii) We have, kx(x – 2) + 6 = 0

⇒ kx2 – 2kx + 6 = 0

Here, a = k, b = – 2k, c = 6

For equal roots, we have

D = 0

i.e., b2 – 4ac = 0

⇒ (-2k)2 – 4 × k × 6 = 0

⇒ 4k2 – 24k = 0

⇒ 4k (k – 6) = 0

Either 4k = 0 or k – 6 = 0

⇒ k = 0 or k = 6

But k = 0 6) because if k = 0 then given equation will not be a quadratic equation).

So, k = 6.

D = (b – c)² – 4(a – b) (c – a) = 0

⇒ b² + c² – 2bc – 4(ac – a² – bc + ab)

⇒ b² + c² – 2bc – 4ac + 4a² + 4bc – 4ab = 0

⇒ 4a² + b² + c² – 4ab + 2bc – 4ac = 0

⇒ (2a)² + (- b)² + (-c)² + 2(2a) (-b) + 2(-b) (-c) + 2(-c) 2a = 0

⇒ (2a – b – c)² = 0

⇒ 2a – b – c = 0

⇒ 2a = b + c. Hence Proved.

Here, A = 1 + m², B = 2mc and C = c² – a²

Since the given equation has equal roots, therefore D = 0 = B² – 4AC = 0.

⇒ (2mc)² – 4(1 + m²) (c² – a²) = 0

⇒ 4m²c² – 4(c² – a² + m²c² – m²a²) = 0

⇒ m²c² – c² + a² – m²c² + m²a² = 0. [Dividing throughout by 4]

⇒ – c² + a² (1 + m²) = 0

⇒ c² = a(1 + m²) Hence Proved

Here, A = ab, B = b² – ac, C = – bc

i.e., b² – 4ac = 0

In (2p + 1 )x² – (7p + 2)x + (7p – 3) = 0

Here, a = (2p + 1), b = -(7p + 2), c = (7p – 3)

So, 3 years ago, Rehman’s age = (x – 3) years

And 5 years from now, Rehman’s age = (x + 5) years

Now, according to question, we have

But x ≠ -3 (age cannot be negative)

Therefore, present age of Rehman = 7 years.

According to the question,

Difference of numbers, x – y = 5 ⇒ x = 5 + y …..(i)

Difference of the reciprocals,

∴ y is a natural number.

∵ y = 5

Putting the value of y in (i), we have

⇒ x = 5 + 5

⇒ x = 10

The required numbers are 10 and 5.

Hence, the numbers are 13 and 15 or -15 and -13.

According to given condition,

⇒ 150 = 3x(15 – x)

⇒ 50 = 15x – x²

⇒ x² – 15x + 50 = 0

⇒ x² – 5x – 10x + 50 = 0

⇒ x(x – 5) -10(x – 5) = 0

⇒ (x – 5)(x – 10) = 0

⇒ x = 5 or 10.

When x = 5, then 15 – x = 15 – 5 = 10

When x = 10, then 15 – x = 15 – 10 = 5

Hence, the two numbers are 5 and 10.

Therefore, Shefali’s marks in English is (30 – x).

Now, according to question,

⇒ (x + 2) (30 – x – 3) = 210

⇒ (x + 2) (27 – x) = 210

⇒ 27x – x² + 54 – 2x = 210

⇒ 25x – x² + 54 – 210 = 0

⇒ 25x – x² – 156 = 0

⇒ -(x² – 25x + 156) = 0

⇒ x² – 25x + 156 = 0

= x² – 13x – 12x + 156 = 0

⇒ x(x – 13) – 12(x – 13) = 0

⇒ (x – 13) (x – 12) = 0

Either x – 13 or x – 12 = 0

∴ x = 13 or x = 12

Therefore, Shefali’s marks in Mathematics = 13

Marks in English = 30 – 13 = 17

or Shefali’s marks in Mathematics = 12

marks in English = 30 – 12 = 18.

But x cannot be negative, so x ≠ – 45

therefore, x = 40

Hence, the uniform speed of train is 40 km/h

Then, x² + y² = 468 …(i)

Let x be the length of the side of the bigger square.

4x – 4y = 24

⇒ x – y = 6 or x = y + 6 …(ii)

Putting the value of x in terms of y from equation (ii), in equation (i), we get

(y + 6)² + y² = 468

⇒ y² + 12y + 36 + y² = 468 or 232 + 12y – 432 = 0

⇒ y² + 6y – 216 = 0

⇒ y² + 18y – 12y – 216 = 0

⇒ y(y + 18) – 12(y + 18) = 0

⇒ (y + 18)(y – 12) = 0

Either y + 18 = 0 or y – 12 = 0

⇒ y = -18 or y = 12

But, sides cannot be negative, so y = 12

Therefore, x = 12 + 6 = 18

Hence, sides of two squares are 18 m and 12 m.

Solution:

⇒ x² + 3x + 7 = 5x – 11

⇒ x² – 2x + 18 = 0 is a quadratic equation.

⇒ x² -2x -8 = x² – x – 12

⇒ x – 4 = 0 is not a quadratic equation.

⇒ 2x³ + x² – 4 = 5x² – 10

⇒ 2x³ – 4x² + 6 = 0 is not a quadratic equation.

x³ + 6x² – 7x = 6x² – 9x

x³ + 2x = 0 is not a quadratic equation.

Solution:

Let two consecutive integers be x, x + 1. Given, x² + (x + 1)² = 650.

⇒ 2x² + 2x + 1 – 650 = 0

⇒ 2x² + 2x – 649 = 0

Solution:

Let the two numbers be x and 15 – x. Given 1x+115−x=310

⇒ 10(15 – x + x) = 3x(15 – x)

⇒ 50 = 15x – x²

⇒ x² -15x + 50 = 0

Solution:

Let the numbers be x and x + 3. Given, x(x + 3) = 504

⇒ x² + 3x – 504 = 0

Solution:

Let the number be x. According to question, x² – 84 = 3(x + 8)

⇒ x² – 84 = 3x + 24

⇒ x² – 3x – 108 = 0

Solution:

Let the number be x .Accoding to question x+12=160xx+12=160x

⇒ x² + 12x – 160 = 0

Assertion Reason Answer-