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Given: (2x – 1) (x – 3) = (x + 5) (x – 1)

⇒ 2x (x – 3) -1 (x – 3) = x (x – 1) + 5(x – 1) 

⇒ 2x² – 6x – x + 3 = x² – x + 5x – 5 

⇒ 2x² -7x + 3 – x² – 4x + 5 = 0 

⇒ x² – 11x + 8 = 0 

Hence it’s a Q.E.

(or) 

Co.eff. of x² on L.H.S. = 2 × 1 = 2 

Co.eff. of x² on R.H.S = 1 × 1 = 1 

LHS ≠ RHS Hence it is a Q.E.

Given: x² + 3x + 1 = (x – 2)² 

⇒ x² + 3x + 1 = x² – 4x + 4 

⇒ 7x – 3 = 0 is not a Q.E.

Given: (x – 2) (x + 1) = (x – 1) (x + 3) 

⇒ x (x + 1) – 2 (x +1) 

= x (x + 3) – 1 (x + 3) 

Note : Compare the coefficients of x² on both sides. If they are equal it is not a Q.E. 

⇒ x² + x – 2x – 2 = x² + 3x – x -3

⇒ x² – x – 2 = x² + 2x – 3 

⇒ 3x – 1 = 0 is not a Q.E.

Given: x² – 2x = -2(3 – x) 

⇒ x² – 2x = -6 + 2x 

⇒ x² – 4x + 6 = 0 is a Q.E.

Given,

  \(\frac{(x+3)}{(x+2)}\) = \(\frac{(3x-7)}{(2x-3)}\)

On cross-multiplying we get,

(x + 3)(2x – 3) = (x + 2)(3x – 7)

⇒ 2x² – 3x + 6x – 9 = 3x² – x – 14

⇒ 2x² + 3x – 9 = 3x² – x – 14

⇒ x² – 3x – x – 14 + 9 = 0

⇒ x² – 5x + x – 5 = 0

⇒ x(x – 5) + 1(x – 5) = 0

⇒ (x – 5)(x + l) – 0

Now, either x – 5 = 0 or x + 1 = 0

⇒ x = 5 and x = -1

Thus, the roots of the given quadratic equation are 5 and -1 respectively.

Here we have,

LHS = x² – 3x + 2

Substituting x = 2 in LHS, we get

(2)² – 3(2) + 2

⇒ 4 – 6 + 2 = 0 = RHS

⇒ LHS = RHS

Thus, x = 2 is a solution of the given equation.

Similarly,

Substituting x = -1 in LHS, we get

(-1)² – 3(-1) + 2

⇒ 1 + 3 + 2 = 6 ≠ RHS

⇒ LHS ≠ RHS

Thus, x = -1 is not a solution of the given equation.

Given,

x(x + 1) + 8 = (x + 2)(x – 2)

x² + x + 8 = x² – 4

⇒ x + 12 = 0

Now, it’s clearly seen that x + 12 is a not quadratic polynomial since its degree is 1. Thus, the given equation is not a quadratic equation.

Since roots are equal

∴ d = 0 ….(1)

4x² - 2 (k + 1) x + (k + 4) = 0

d = b² – 4ac

d = (–2 (k + 1)² – 4 (4) (k + 4)

d = (– 2k – 2)² – 16k – 64

d = 4k² + 4 + 8k – 16k – 64

(∵ (a – b)² = a² + b² – 2ab)

d = 4k² – 8k – 60

From (1), d = 0

∴ Equation will be:

4k² – 8k – 60 = 0

Dividing by 4

k² – 2k – 15 = 0

4k² – 5k + 3k – 15 = 0

k (k – 5) + 3(k – 5) = 0

(k – 5) (k + 3) = 0

K – 5 = 0 k + 3 = 0

K = 5 k = –3

d = b² – 4ac

d = (–4)² – 4 (2) (3)

d = 16 – 24

d = –8

Since, d < 0, no real roots exist for the given equation.

d = b² – 4ac

d = (–5)² – 4 (2) (–3)

d = 25 + 24

d = 49

∴ Roots are real and unique.

d = b² – 4ac

d = (10)² – 4 (1) (39)

d = 100 – 156

d = –56

Since, d < 0, no real roots exists.

Given,

(x + 2)³ = x³ – 4

On expanding, we get

⇒ x³ + 6x² + 8x + 8 = x³ – 4

⇒ 6x² + 8x + 12 = 0

Now, it’s clearly seen that 6x² + 8x + 12 is a quadratic polynomial. Thus, the given equation is a quadratic equation.

Given,

16x² – 3 = (2x + 5)(5x – 3)

16x² – 3 = 10x² – 6x + 25x – 15

⇒ 6x² – 19x + 12 = 0

Now, it’s clearly seen that 6x² – 19x + 12 is a quadratic polynomial. Thus, the given equation is a quadratic equation.

Given,

x + \(\frac{1}{x}\) = x²

On multiplying by x on both sides we have,

x² + 1 = x³

⇒ x³ – x² – 1 = 0

Now, it’s clearly seen that x³ – x² – 1 is not a quadratic polynomial since its degree is 3. Thus, the given equation is not a quadratic equation.

Given,

(2x + 1)(3x + 2) = 6(x – 1)(x – 2)

⇒ 6x² + 4x + 3x + 2 = 6x² -12x – 6x + 12

⇒ 7x + 2 = -18x + 12

⇒ 25x – 10 = 0

Now, it’s clearly seen that 25x – 10 is not a quadratic polynomial since its degree is 1. Thus, the given equation is not a quadratic equation.

Given,

(x + \(\frac{1}{x}\))² = 3(x + \(\frac{1}{x}\)) + 4

⇒ x² + \(\frac{1}{x}\)² + 2 = 3x + \(\frac{3}{x}\) + 4

⇒ x⁴ + 1 + 2x² = 3x³ + 3x + 4x²

⇒ x⁴ – 3x³ – 2x² – 3x + 1 = 0

Now, it’s clearly seen that x⁴ – 3x³ – 2x² – 3x + 1 is not a quadratic polynomial since its degree is 4. Thus, the given equation is not a quadratic equation.

Given; x (2x + 3) = x² + 1

⇒ 2x² + 3x = x² + 1

⇒ 2x² + 3x − x² − 1 = 0

⇒ x² +3x − 1 = 0

∵ The highest power of x in the equation is 2;

∴ It is a quadratic equation.

Given; (x + 2)³ = x³ − 4

⇒ x³ + 6x² + 12x + 2³ = x³ − 4

⇒ x³ + 6x² + 12x + 2³− x³ + 4 = 0

⇒ 6x² + 12x + 12 = 0

∵ The highest power of x in the equation is 2;

∴ It is a quadratic equation.

Given; (x + 1) (x − 1) = (x + 2) (x + 3)

⇒ x² − 1² = x² + 5x + 6

⇒ x² − 1 − x² − 5x − 6 = 0

⇒ −5x − 7 = 0

∵ The highest power of x in the equation is 1;

∴ It is not a quadratic equation.

Given; x² + 3x + 1 = (x − 2)²

⇒ x² + 3x + 1 = x² − 4x + 4

⇒ x² + 3x + 1 − x² − 4x − 4 = 0

⇒ −x − 3 = 0

∵ The highest power of x in the equation is 1;

∴ It is not a quadratic equation.

Given; (x − 3) (2x + 1) = x (x + 2)

⇒ 2x² + x − 6x − 3 = x² + 5x

⇒ 2x² + x − 6x − 3 − x² − 5x = 0

⇒ x² − 10x − 3 = 0

∵ The highest power of x in the equation is 2;

∴ It is a quadratic equation.

Let p(x) = x² – 3x,

It’s clearly seen that p(x) = x² – 3x is a quadratic polynomial. Thus, the given equation is a quadratic equation.

The correct option is: (C) 16

Explanation:

Since x = 2 is a root of the equation

x² + bx + 12 = 0 

=> (2)² + b(2) + 12 = 0

=> 2b = -16 => b = -8

Then, the equation x² + bx + q becomes

x² - 8x + q = 0                   ....(1)

Since (1 ) has equal roots => b² - 4ac = 0

=> (-8)² - 4(1)q = 0 => q =16

Given; (x − 2) (x + 1) = (x − 1) (x + 3)

⇒ x² + x − 2x − 2 = x² + 3x − x − 3

⇒ x² − x − 2 − x² − 2x + 3 = 0

⇒ −3x + 1 = 0

∵ The highest power of x in the equation is 1;

∴ It is not a quadratic equation.

(d) 36

Suppose the total number of cows = x

\(\frac14x\) of the cows are in the forest.

\(2\sqrt{x}\) have gone to mountains. 

15 are on the banks of a river.

∴ \(2\sqrt{x}\) + \(\frac14x\) + 15 = x

⇒ \(2\sqrt{x}\) - \(\frac34x\) + 15 = 0

⇒ \(8\sqrt{x}\) - 3x + 60 = 0

⇒ 3y² - 8y - 60 = 0             (Let √x = y)

⇒3y² - 18y + 10y -60 = 0

⇒ 3y(y - 6) + 10(y - 6) = 0

⇒ (3y +10)(y -6) = 0 ⇒ y = \(\frac{10}{3}, 6\)

⇒ \(\sqrt{x}\) = y = 6                (Rejecting –ve value)

⇒ x = 36.

(b) 7

\(x\) - 4 = 21 x  \(\frac1x\)

⇒ x² - 4x = 21

⇒x² - 4x - 21 = 0 ⇒ (x - 7)(x + 3) = 0

⇒ x = 7, - 3  ⇒ x = 7.

Let the whole number be x.

When it is decreased by 20 ⇒(x – 20)

And, the reciprocal of the whole number is \(\frac{1}{x}\)

From the given condition, we have

(x – 20) = 69 x (\(\frac{1}{x}\))

⇒ x(x – 20) = 69

⇒ x² – 20x – 69 =0

Solving for x by factorization method, we have

⇒ x² – 23x + 3x – 69 = 0

⇒ x(x – 23) + 3(x – 23) = 0

⇒ (x – 23)(x + 3) = 0

Thus, x is either 23 Or -3

As we that a whole number is always positive, x = – 3 is not considered. Therefore, the whole number is 23.

Let x be the total number of arrows Arjun had. Then,

\(\frac{x}{2}+6+3+(4\sqrt{x}+1)=x\)

⇒ \(-\frac{x}{2}+10+4\sqrt{x}=0\) 

⇒ \(-x + 20 + 8\sqrt{x}=0\)

⇒ \(x - 8\sqrt{x}-20=0\)

⇒ y² -8y - 20 = 0              (Let √x = y)

⇒ y² - 10y + 2y - 20 = 0 

⇒ y(y -10) + 2(y-10) = 0

⇒ (y -10)(y + 2) = 0

⇒ y - 10 = 0 or y + 2 = 0 

⇒ y = 10, -2

Neglecting –ve value

y = \(\sqrt{x}\) = 10 ⇒ x = (10)² = 100.

We write, -2x  = -3x + x as 3x² \(\times\)(-1) = - 3x² = (-3x) \(\times\)x

∴ 3x² - 2x - 1 = 0

⇒ 3x² - 3x + x -1 = 0

⇒ 3x(x - 1) + 1(x - 1) = 0

⇒ (x - 1))3x + 1) = 0

⇒ x -1 = 0 or  3x + 1 = 0

⇒ x = 1 or x = -1/3

Hence, the roots of the given equation are 1 and - 1/3.

(d) (2x + 1)(3x - 4) = 2x² + 3 

We reduce each of the given parts to standard form:

(a) \(x^\frac12 + 2x +3 = 0\) ⇒ degree = 1 ⇒ Not a quad. eqn.

(b) (x - 1)(x + 4) = x² + 1 

⇒ x² + 3x - 4  = x² + 1 ⇒ 3x - 5 = 0 ⇒  Degree = 0 ⇒ Not a quad. eqn.

(c) x⁴ - 3x + 5 = 0 ⇒ Degree = 4 ⇒ Not a quad. eqn

(d) (2x + 1)(3x - 4) = 2x² + 3 

⇒ 6x² - 5x - 4 = 2x² + 3 

⇒ 4x² - 5x -4 = 0 ⇒ Degree = 2 ⇒ Hence a quad. eqn.

(d) either a = 0 or a = b

a² -ab = 0 ⇒ a(a-b) = 0

⇒ a = 0 or (a - b) = 0 ⇒ a = 0 or a = b.

(b) 5, 3

x² - 8x + 15 = 0

⇒ x² - 5x - 3x + 15 = 0

⇒ x(x - 5) -3(x - 5) = 0

⇒ (x - 5)(x - 3) = 0

⇒ x - 5 = 0 or x - 3 = 0

⇒ x = 5, 3

(a)  3, \(\frac52\).

2x² - 11x + 15 = 0

⇒ 2x² - 6x - 5x + 15 = 0

⇒ 2x (x - 3) - 5(x - 3) = 0

⇒ (x - 3)(2x - 5) = 0

⇒ x - 3 = 0 or 2x - 5 = 0 

⇒ x = 3, \(\frac52\).

Given: x(x + 4) = 12 

⇒ x² + 4x = 12 

⇒ x² + 4x – 12 = 0

⇒ x² + 6x – 2x – 12 = 0 

⇒ x(x + 6) – 2(x + 6) = 0 

⇒ (x + 6) (x – 2) = 0 

⇒ x + 6 = 0 or x – 2 = 0 

⇒ x = -6 or x = 2 

⇒ x = -6 or 2 are the roots of the given Q.E.

Given : 100x² – 20x + 1 =0 

⇒ 100x² – 10x – 10x + 1 = 0 

⇒ 10x(10x – 1) – 1(10x – 1) = 0 

⇒ (10x – 1) (10x – 1) = 0 

⇒ 10x – 1 = 0 

⇒ 10x = 1 

⇒ x = 1/10 , 1/10 are the roots of the given Q.E.

(i) LHS = (x – 2)² + 1 = x² – 4x + 4 + 1 = x² – 4x + 5

Therefore, (x – 2)² + 1 = 2x – 3 can be rewritten as

x² – 4x + 5 = 2x – 3

i.e., x² – 6x + 8 = 0

It is of the form ax² + bx + c = 0.

Therefore, the given equation is a quadratic equation.

(ii) Since x(x + 1) + 8 = x² + x + 8 and (x + 2)(x – 2) = x² – 4

Therefore, x² + x + 8 = x² – 4

i.e., x + 12 = 0

It is not of the form ax² + bx + c = 0.

Therefore, the given equation is not a quadratic equation.

(iii) Here, LHS = x (2x + 3) = 2x² + 3x

So, x (2x + 3) = x² + 1 can be rewritten as

2x² + 3x = x² + 1

Therefore, we get x² + 3x – 1 = 0

It is of the form ax² + bx + c = 0.

So, the given equation is a quadratic equation.

(iv) Here, LHS = (x + 2)³ = x³ + 6x² + 12x + 8

Therefore, (x + 2)³ = x³ – 4 can be rewritten as

x³ + 6x² + 12x + 8 = x³ – 4

i.e., 6x² + 12x + 12 = 0 or, x² + 2x + 2 = 0

It is of the form ax² + bx + c = 0.

So, the given equation is a quadratic equation.

Given: 2x² – x + 1/8 = 0 

⇒ \(\frac{16x^2 - 8x +1}{8}\) = 0 

⇒ 16x² – 8x + 1 =0 

⇒ 16x² – 4x – 4x + 1 = 0 

⇒ 4x(4x – 1) – l(4x – 1) = 0 

⇒ (4x – 1) (4x – 1) – 0 

⇒ 4x – 1 = 0

⇒ 4x = 1

⇒ x = 1/4, 1/4 are the roots of given Q.E.

Given: 3x² – 5x + 2 = 0 

⇒ 3x² – 3x – 2x + 2 = 0 

⇒ 3x(x – 1) – 2(x – 1) = 0 

⇒ (x – 1) (3x – 2) = 0 

⇒ x – 1 = 0 or 3x – 2 = 0 

⇒ x = 1 or 2/3 , 

⇒ x = 1 or 2/3 are the roots of the given Q.E.

Take (x – 4) = a, then the given Q.E. reduces to 

3a² – 5a = 12 

⇒ 3a² – 5a – 12 = 0 

⇒ 3a² – 9a + 4a – 12 = 0 

⇒ 3a(a – 3) + 4(a – 3) = 0

⇒ (a – 3) (3a + 4) = 0 

⇒ a – 3 = 0 or 3a + 4 = 0 

⇒ a = 3 or a = \(\frac{-4}{3}\)

but a = x – 4 

x – 4 = 3 (or) x – 4 = \(\frac{-4}{3}\)

⇒ x = 7 or x = 4 – \(\frac{-4}{3}\) = \(\frac{8}{3}\)

∴ x = 7 or \(\frac{8}{3}\) are the roots of the given Q.E.

Given: x – \(\frac{3}{x}\) = 2 

⇒ \(\frac{x^2-3}{x}\)= 2 

⇒ x² – 3 = 2x 

⇒ x² – 2x – 3 = 0 

⇒ x² – 3x + x – 3 = 0 

⇒ x(x – 3) + 1(x – 3) = 0 

⇒ (x – 3) (x + 1) = 0 

⇒ (x – 3) = 0 or (x + 1) = 0 

⇒ x = 3 or x = -1 

⇒ x = 3 or -1 are the roots of the given Q.E.

Given: 2x² + x – 6 = 0 

⇒ 2x² + 4x – 3x – 6 = 0 

⇒ 2x(x + 2) – 3(x + 2) = 0 

⇒ (x + 2) (2x – 3) = 0 

⇒ (x + 2) or 2x – 3 = 0 

⇒ x = -2 or 2x = 3 

⇒ x = -2 or 3/2 are the roots of the given Q.E.

Given: x² – 3x – 10 = 0 

x² – 5x + 2x- 10 = 0

⇒ x(x – 5) + 2 (x – 5) = 0 

⇒ (x – 5) (x + 2) = 0 

⇒ x – 5 = 0 or x + 2 = 0 

⇒ x = 5 or x = -2 

⇒ x = 5 or -2 are the roots of the given Q.E.

Let us first split the middle term – 5x as –2x –3x [because (–2x) × (–3x) =

6x² = (2x²) × 3].

So, 2x² – 5x + 3 = 2x² – 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1)

Now, 2x² – 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0.

So, the values of x for which 2x² – 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0,

i.e., either 2x – 3 = 0 or x – 1 = 0.

Now, 2x – 3 = 0 gives x=3/2 and x – 1 = 0 gives x = 1.

So, x=3/2 and x = 1 are the solutions of the equation.

In other words, 1 and 3/2 are the roots of the equation 2x² – 5x + 3 = 0.

Verify that these are the roots of the given equation.

We have

6x² – x – 2 = 6x² + 3x – 4x – 2

= 3x (2x + 1) – 2 (2x + 1)

= (3x – 2)(2x + 1)

The roots of 6x² – x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0

Therefore, 3x – 2 = 0 or 2x + 1 = 0,

i.e., x =2/3 or x =-1/2

Therefore, the roots of 6x² – x – 2 = 0 are 2/3  and -1/2.

We verify the roots, by checking that 2/3 and -1/2 satisfy 6x² – x – 2 = 0.

Equation x² + k (4x + k – 1) + 2 = 0 has equal roots 

d = 0 

d = b² – 4ac = 0 

Here a = 1, b = 4k, c = k² – k + 2 

⇒ 16 k² – 4(k² – k + 2) = 0 

⇒ 12 k² + 4k – 8 = 0 

⇒ 3k2 + k – 2 = 0 

⇒ (3k – 2) (k + 1) = 0 

⇒ K = 2/3, – 1

Correct option is (D) 5

Given quadratic equation is \(x^2+5x+5=0\)

\(\therefore\) Discriminant \(=5^2-4\times1\times5\)

= 25 - 20 = 5

Correct option is D) 5

Correct option is (A) ± 3

For equal roots, we have

Discriminant = 0

\(\Rightarrow6^2-4\times k\times k=0\)

\(\Rightarrow36-4k^2=0\)

\(\Rightarrow4k^2=36\)

\(\Rightarrow k^2=9\)

\(\Rightarrow\) k = \(\pm\,\,3\)

Correct option is A) ± 3

In the given equation kx² + 6x + 4k = 0

Sum of the roots = product of the roots (given) 

– b/a = c/a 

Here a = k, b = 6 and c = 4k

\(-6/k=4k/k\)

k = \(-3/2\)

[(x + 1) (x + 4)] [(x + 2) (x + 3)] + 1 = 0            (∵ 1 + 4 = 2 + 3 = 5) 

⇒ (x² + 5x + 4) (x² + 5x + 6) + 1 = 0 

Let x² + 5x = y. Then, (y + 4) (y + 6) + 1 = 0 

⇒ y² + 10y + 24 + 1 = 0 ⇒ y² + 10y + 25 = 0 ⇒ (y + 5)² = 0 ⇒ y = – 5

∴ x² + 5x = – 5 ⇒ x² + 5x + 5 = 0. ⇒ x = \(\frac{-5±\sqrt{25-20}}{2}\) = \(\frac{-5±\sqrt{2}}{2}\).

Important Properties of Inequalities.

1. An inequality will still hold after each side has been increased, diminished, multiplied or divided by the same positive quantity. 

2. In an inequality any term may be transposed from one side to the other if its sign is changed. 

3. Both the sides of an inequality can be multiplied or divided by the same negative number by reversing the sign of inequality.