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Let the consecutive positive even numbers be x and x + 2.

From the given information,

x² + (x + 2)² = 52

2x² + 4x + 4 = 52

2x² + 4x − 48 = 0

x² + 2x − 24 = 0

(x + 6) (x − 4) = 0

x = − 6, 4

Since, the numbers are positive, so x = 4.

Thus, the numbers are 4 and 6.

Let the two numbers be x and y, y being the bigger number. From the given information,

x² + y² = 208 ..... (i)

y² = 18x ..... (ii)

From (i), we get y² = 208 − x². Putting this in (ii), we get,

208 − x² = 18x

⇒ x² + 18x − 208 = 0

⇒ x² + 26x – 8x − 208 = 0

⇒ x(x + 26) − 8(x + 26) = 0

⇒ (x − 8)(x + 26) = 0

⇒ x can't be a negative number, hence x = 8

⇒ Putting x = 8 in (ii), we get y² = 18 x 8 = 144

⇒ y = 12, since y is a positive integer

Hence, the two numbers are 8 and 12.

1 + √2 Is a root of quadratic equation with rational coefficients that is the sum of the roots is rational and the product of the roots is also rational.

Since the rational roots occurs in conjugate pairs so the other root of the equation is 1 – √2

(d) x² + x + 1 = 0

Let α, β be the roots of the equations x² + x + 1 = 0. Then, 

Sum of roots = α + β = – 1, Product of roots = αβ = 1 

Now the equation whose roots are \(\frac{α}{β}\) and \(\frac{β}{α}\) is

x² - \(\big(\frac{α}{β} + \frac{β}{α}\big)x\) + \(\big(\frac{α}{β} \times \frac{β}{α}\big)=0.\)

\(\frac{α}{β}\) + \(\frac{β}{α}\) = \(\frac{α^2+β^2}{αβ}\) = \(\frac{(α+β)^2-2αβ}{αβ}\) = \(\frac{(-1)^2-2(1)}{1}\) = -1 and \(\frac{α}{β}\) x \(\frac{β}{α}\) = 1.

∴ Required equation = x² + x + 1 = 0.

Subtracting both the equations we get, 

x² – 11x + a - x² +14x + 2a = 0 

3x – a = 0 

x = \(\frac{a}{3}\)

Substituting it in first equation we get,

\(\Big(\frac{a}{3}\Big)^2\) - 11 \(\frac{a}{3}\) + a = 0

\(\frac{a^2}{9}\) - \(\frac{8a}{3}\) = 0

a² - 24a = 0

a = 24

(a)  a < 2.

If the roots of the equation x² – 2ax + a² – a – 3 = 0 are real and less than 3, then D ≥ 0 and f (3) > 0. 

⇒ 4a² – 4(a² + a – 3) ≥ 0 and (3)² – 2a (3) + a² + a – 3 > 0 

⇒ a² – a² – a + 3 ≥ 0 and 9 – 6a + a² + a – 3 > 0 

⇒ –a + 3 ≥ 0 and a² – 5a + 6 > 0 ⇒ a – 3 ≤ 0 and (a – 2) (a – 3) > 0 

⇒ a ≤ 3 and a < 2 or a > 3 ⇒ a < 2.

The given equation x² + ax + 12 = 0 has a root = 2 

So it will satisfy the equation 

4 + 2a + 12 = 0 

2a + 16 = 0 

a = – 8 

Putting value of a in second equation, it becomes 

x² + ax + q = 0 

x² – 8x + q = 0 

Roots are equal so d = 0 

⇒ b² – 4ac = 0 

⇒ 64 – 4q = 0 

⇒ q = 64/4 

⇒ q = 16

Correct option is (B) 2 only

Given equation is

\(\sqrt{2x-3}+\sqrt{3x-5}-\sqrt{5x-6}=0\)      _______________(1)

\(\Rightarrow\) \(\sqrt{2x-3}+\sqrt{3x-5}=\sqrt{5x-6}\)

\(\Rightarrow2x-3+3x-5\) \(+2\sqrt{(2x-3)(3x-5)}=5x-6\)     (By squaring both sides)

\(\Rightarrow\) \(2\sqrt{(2x-3)(3x-5)}\) \(=5x-6-5x+8\)

\(\Rightarrow\) \(2\sqrt{6x^2-19x+15}=2\)

\(\Rightarrow\) \(\sqrt{6x^2-19x+15}=1\)

\(\Rightarrow6x^2-19x+15=1\)      (By squaring both sides)

\(\Rightarrow6x^2-19x+14=0\)

\(\Rightarrow6x^2-12x-7x+14=0\)

\(\Rightarrow6x(x-2)-7(x-2)=0\)

\(\Rightarrow(x-2)(6x-7)=0\)

\(\Rightarrow x-2=0\;or\;6x-7=0\)

\(\Rightarrow x=2\;or\;x=\frac76\)

Let x = \(\frac{7}{6}\), 2x - 3 \(=2\times\frac{7}{6}-3\)

\(=\frac{7-9}3=\frac{-2}3<0\)

Hence, x = \(\frac{7}{6}\) is not in domain of given equation.

\(\therefore\) x = \(\frac{7}{6}\) is not a root of given equation.

Put x = 2 in equation (1)

\(\sqrt{2\times2-3}+\sqrt{3\times2-5}-\sqrt{5\times2-6}=0\)

\(\Rightarrow\) \(\sqrt{4-3}+\sqrt{6-5}-\sqrt{10-6}=0\)

\(\Rightarrow\) 1 + 1 - 2 = 0

\(\Rightarrow\) 0 = 0                   (Satisfied)

Hence, x = 2 is a root of given equation.

Correct option is B) 2 only

Correct option is (C) -3√2

Let \(\alpha\;and\;2\alpha\) are roots of \(x^2+ax + 4 = 0\)   _______________(1)

\(\therefore\) Product of roots = 4

\(\Rightarrow\) \(2\alpha^2=4\)

\(\Rightarrow\) \(\alpha^2=2\)

\(\Rightarrow\) \(\alpha=\pm\sqrt2\)         ______________(2)

And sum of roots = -a

\(\therefore\) \(-a=\alpha+2\alpha\)

\(=3\alpha\)

\(=3(\pm\sqrt2)\)      (From (2))

\(=\pm3\sqrt2\)

\(\therefore a=\mp3\sqrt2\)

Possible values of a are \(3\sqrt{2}\;or\;-3\sqrt{2}.\)

Correct option is C) -3√2

The correct option is: (B) (a + 1), -(a + 2)

Explanation:

Given equation is x² + x-(a + 1)(a + 2)= 0

=> x² + (a + 2)x - (a + 1)x- (a + 1) (a + 2) = 0

=> x(x + (a + 2)) -(a + 1)(x+ (a + 2)) = 0

=> (x- (a + 1))(x + (a + 2)) = 0

=> x = (a + 1) or x = - (a + 2)

Let’s assume the consecutive odd positive integer to be 2x – 1 and 2x + 1 respectively. [Keeping the common difference as 2]

Now, it’s given that the sum of their squares is 394.

Which means,

(2x – 1)² + (2x + 1)² = 394

4x² +1 – 4x + 4x² +1 + 4x = 394

By cancelling out the equal and opposite terms, we get

8x² + 2 = 394

8x² = 392

x² = 49

x = 7 and – 7

Since we need only consecutive odd positive integers, we only consider x = 7.

Now,

2x – 1 = 14 -1 = 13

2x + 1 = 14 + 1 = 15

Thus, the two consecutive odd positive numbers are 13 and 15 respectively.

Time = distance/speed

Given, train covers a distance of 90 km at a uniform speed. Had the speed been 15 km / hour more, it would have taken 30 minutes less for the journey.

Let the usual speed be ‘a’.

\(\frac{90}{a}-\frac{90}{a\,+\,15}=\frac{30}{60}\)

⇒ 90 ×(a + 15 –a) = (a² + 15a)/2

⇒ a² + 15a – 2700 = 0 

⇒ a² + 60a – 45a – 2700 = 0 

⇒ a(a + 60) – 45(a + 60) = 0 

⇒ (a + 60)(a – 45) = 0 

⇒ a = 45 km/hr

We know that (a+b)²=a²+b²+2ab

take a=y, and b=1/y

a²+b²=(a+b)²-2ab

y²+1/y²=(y+1/y)²-2x y x 1/y

y²+1/y² =(y+1/y)²-2..........................(1)

7(y+1/y)-2(y²+1/y²)=9

7(y+1/y)-2[(y+1/y)²-2]=9..............from equation(1)

taking (y+1/y)=z we get

7z-2[z²-2]=9

=> 7z-2z²+4=9

=> -2z²+7z-5=0................(2)

solving equation (2) we get 

2z²-7z+5=0

=> z(2z-5)-1(2z-5)=0

=> (z-1)(2z-5)=0

=>(z-1)=0 or (2z-5)=0

z=1  or, z=5/2

Now,

y+1/y=1 or y+1/y=5/2

=> y²+1=y

=> y²-y+1=0 [this is not possible]

Now,

y+1/y=5/2

=>y²+1=5y/2

=> 2y²+2=5y

=> 2y²-5y+2=0

=>2y(y-2)-1(y-2)=0

=> (2y-1)(y-2)=0

=>2y-1=0 ,y-2=0

number of integral solution of the equation

y=1/2, y=2

Arrange the equation in quadratic form.

As x must be real, the discriminant must be ≥0

Make discriminant D ≥ 0

After solving the discriminant equation you will get the number of integral solution.

Correct answer is (B) 1/64

2x² + 2px - 15 = 0

Put x = –5

2(–5)² + 2p(–5) - 15 = 0

50 – 10p – 15 = 0

35 = 10p

p = 3.5

Equation p (x² + x) + k = 0 has equal roots i.e. d = 0

p (x² + x) + k = 0

p = 3.5

3.5 (x² + x) + k = 0

3.5x² + 3.5x + k = 0

d = b² – 4ac

d = (3.5)² – 4(3.5)(k)

d = 12.25 – 14k

Putting d = 0

∴ Equation will be:

0 = 12.25 – 14k

14k = 12.25

k = 12.25/14

k = 0.875

Since roots are equal

∴ d=0 (1)

(a — b)x² + (b — c) x + (c — a) = 0

d = b² – 4ac

d = (b–c)² – 4 (a–b) (c–a)

d = b² + c² – 2bc –4 [a (c – a) – b (c – a)]

d = b² + c² – 2bc – 4 [ac – a² – bc + ba]

From (1), d = 0

∴ Equation will be:

0 = b² + c² – 2bc – 4ac + 4a² + 4bc – 4ba

b² + c² – (2a)² 2bc + 2c (–2a) + 2(–2a)b = 0

(b + c – 2a)² = 0

(b + c – 2a) = 0

b + c = 2a

Hence proved.

3x² - 15x + x - 5 = 0

3x (x – 5) + (x – 5) = 0

(3x + 1) (x – 5) = 0

3x + 1 = 0 x – 5 = 0

x = -1/3, x = 5

Therefore, the roots of the equation are -1/3, 5.

Given, 2x² + 5/3 x  - 2 = 0

⇒ 6x² + 5x – 6 = 0

By splitting the middle term,

⇒ 6x² + 9x - 4x – 6 = 0

Taking common in the expression,

⇒ 3x (2x + 3) – 2 (2x + 3) = 0

⇒ (2x + 3) (3x - 2) = 0

∴ 2x + 3 = 0 and ∴ 3x – 2 = 0

∴ x = -3/2 and ∴ x = 2/3

Given, 2/5 x² - x - 3/5 = 0

⇒ 2x² - 5x – 3 = 0

By splitting the middle term,

⇒ 2x² - 6x + x – 3 = 0

Taking common in the expression,

⇒ 2x (x - 3) + 1 (x - 3) = 0

⇒ (2x + 1) (x - 3) = 0

∴ 2x + 1 = 0 and ∴ x – 3 = 0

∴ x = -1/2 and ∴ x = 3

Since the equation 

(a² + b²) x² + 2 (ac + bd) x + c² + d² = 0 has no real root 

D < 0 

b² – 4ac < 0 

b² < 4ac 

Here a = (a² + b²), b = 2 (ab + bd), c = c² + d² 

4(ac + bd)² – 4 (a² + b2)(c² + d²) < 0 

4a²c² + 4b²d² + 8abcd – 4(a²c² + b²c² + a²d² + b²d²) < 0 

– 4(a²d² + b²c² – 2abcd) < 0 

– 4(ad + bc)²< 0

∴ d is always negative

And ad \(\neq\) bc

Since p and q are roots of the equations then 

Sum of the roots is p + q = = – (p) = – p 

Here a = 1, b = p and c = q 

Products of the root = p × q = = q 

∴ p × q = q p = 1 

Putting value of ‘p’ in p + q = – p 

1 + q = – 1 

q = – 2

The roots of the equation are real (given) 

Let α and β be the two roots according to the given condition 

α = α² 

β = β² 

Sum of the roots = α + β = α² + β² 

Product of the roots = α β = α² β² 

There are only two number who does not change on squaring them that is 0 and 1 

So the number of equations could be 2 by being the roots as 

(0,1) and (1,0)

For quadratic equation to have real roots, 

d≥ 0 

b² – 4a ≥ 0 

b² ≥ 4a 

For a = 1, 4a = 4, b = 2, 3, 4 (3 equations) 

With values of (a,b) as (1,2), (1,3), (1,4) 

a = 2, 4a = 8, b = 3, 4 (2 equations) 

With values of a,b as (2,3), (2,4) 

a = 3, 4a = 12, b = 4 (1 equation) 

With value of (a,b) as (3,4) 

a = 4, 4a = 16, b = 4 (1 equation) 

With values of (a,b) as (4,16) 

Thus, total 7 equations are possible.

(i) 2x² + kx + 3 = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac 

If D = 0, roots are real and equal 

2x² + kx + 3 = 0

⇒ D = k² – 4 × 2 × 3 = 0 

⇒ k² = = 24 

⇒ k = 2√6

(ii) kx(x - 2) + 6 = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac 

If D = 0, roots are real and equal

kx(x - 2) + 6 = 0

⇒ kx² – 2kx + 6 = 0 ⇒ D 

= 4k² – 4 × 6 × k = 0 

⇒ 4k(k – 6) = 0 

⇒ k = 0, 6 but k can’t be 0 a it is the coefficient of x², thus k = 6

(iii) x² - 4kx + k = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

x² - 4kx + k = 0

⇒ D = 16k² – 4k = 0 

⇒ 4k(4k – 1) = 0 

⇒ k = 0, 1/4

(iv) \(k\text{x}(\text{x} - 2\sqrt{5})+10=0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

\(k\text{x}(\text{x} - 2\sqrt{5})+10=0\)

⇒ kx² – 2√5kx + 10 = 0

⇒ D = 4 × 5k² – 4 × k × 10 = 0 

⇒ k² = 2k 

⇒ k = 2

(v) px(x - 3) + 9 = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

px(x - 3) + 9 = 0

⇒ px² – 3px + 9 = 0 

⇒ D = 9p² – 4 × 9 × p = 0 

⇒ p = 4

(vi) 4x² + px + 3 = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

4x² + px + 3 = 0

⇒ D = p² – 4 × 4 × 3 = 0 

⇒ p² = 48 

⇒ p = 4√3

(i) (k + 1) x² + 2 (k + 3) x + (k + 8) = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

(k + 1) x² + 2 (k + 3) x + (k + 8) = 0

⇒ D = 4(k + 3)² – 4(k + 1)(k + 8) = 0 

⇒ 4k² + 36 + 24k – 4k² – 32 – 36k = 0 

⇒ 12k = 4 

⇒ k = 1/3

(ii) x² - 2kx + 7x + 1/4 = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

x² – 2kx + 7x + 1/4 = 0 

⇒ D = (7 – 2k)² – 4 × 1/4 = 0 

⇒ 49 + 4k² – 28k – 1 = 0 

⇒ k² – 7k + 12 = 0 

⇒ k² – 4k – 3k + 12 = 0 

⇒ k(k – 4) – 3(k – 4) = 0

⇒ (k – 3)(k – 4) = 0 

⇒ k = 3, 4

(iii) (k + 1)x² - 2(3k + 1)x + 8k + 1 = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

(k + 1)x² - 2(3k + 1)x + 8k + 1 = 0

⇒ D = 4(3k + 1)² – 4(k + 1)(8k + 1) = 0

⇒ 4 × (9k² + 6k + 1) – 32k² – 4 – 36k = 0

⇒ 36k² + 24k + 4 – 32k² – 4 – 36k = 0

⇒ 4k(k – 3) = 0

⇒ k = 0, 3

(iv) 5x² - 4x + 2 + k (4x² - 2x - 1) = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

5x² - 4x + 2 + k (4x² - 2x - 1) = 0

⇒ (5 + 4k)x² – (4 + 2k)x + 2 – k = 0

⇒ D = (4 + 2k)² – 4 × (5 + 4k)(2 – k) = 0

⇒ 16 + 4k² + 16k + 16k² – 12k – 40 = 0 

⇒ 20k² – 4k – 24 = 0 

⇒ 5k² - k - 6 = 0 

⇒ 5k² – 6k + 5k – 6 = 0 

⇒ k(5k – 6) + (5k – 6) = 0 

⇒ (k + 1)(5k – 6) = 0 

⇒ k = -1, 6/5

(v) (4 - k)x² + (2k + 4)x + (8k + 1) = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

(4 - k)x² + (2k + 4)x + (8k + 1) = 0

⇒ D = (2k + 4)² – 4 × (4 – k)(8k + 1) = 0

⇒ 4k² + 16 + 16k + 32k² – 16 – 124k = 0

⇒ 36k² – 108k = 0

⇒ 36k(k – 3) = 0 

⇒ k = 0, 3

(i) 9x² - 24x + k = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

9x² - 24x + k = 0

⇒ D = 576 – 4 × 9 × k = 0

⇒ k = 576/36 = 16

(ii) 4x² - 3kx + 1 = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

4x² - 3kx + 1 = 0

⇒ D = 9k² – 4 × 4 × 1 = 0

⇒ 9k² = 16

⇒ k = 4/3

(iii) x² - 2(5 + 2k)x + 3(7 + 10k) = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

x² - 2(5 + 2k)x + 3(7 + 10k) = 0

⇒ D = 4(5 + 2k)² – 4 × 3(7 + 10k) = 0 

⇒ 100 + 16k² + 80k – 84 – 120k = 0 

⇒ 16k² – 40k + 16 = 0 

⇒ 2k² – 5k + 2 = 0 

⇒ 2k² – 4k – k + 2 = 0 

⇒ 2k(k – 2) – (k – 2) = 0 

⇒ (2k – 1)(k – 2) = 0 

⇒ k = 2, 1/2

(iv) (3k + 1) x² + 2 (k + 1) x + k = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac 

If D = 0, roots are real and equal

(3k + 1) x² + 2 (k + 1) x + k = 0

⇒ D = 4(k + 1)² – 4k(3k + 1) = 0 

⇒ 4k2 + 8k + 4 – 12k² – 4k = 0 

⇒ 2k² – k – 1 = 0 

⇒ 2k² – 2k + k – 1 = 0 

⇒ 2k(k – 1) + (k – 1) = 0 

⇒ (2k + 1)(k – 1) = 0 

⇒ k = 1, -1/2

(v) kx² + kx + 1 = -4x² - x

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

kx² + kx + 1 = -4x² - x

⇒ (k + 4)x² + (k + 1)x + 1 = 0

D = (k + 1)² – 4(k + 4) = 0

⇒ k² + 2k + 1 – 4k – 16 = 0

⇒ k² – 2k – 15 = 0

⇒ k² – 5k + 3k – 15 = 0

⇒ k(k – 5) + 3(k – 5) = 0

⇒ (k + 3)(k – 5) = 0 

⇒ k = 5, -3

Here, a= 2, b= -3, c= 5

D = b² – 4ac

= (-3)² -4(2)(5)

= 9 – 40

= -31< 0

It’s seen that D < 0 and hence, the given equation does not have any real roots.

Here, a = \(\frac{3}{5}\) \(\), b = \(\frac{-2}{3},\) c = 1

D = (\(\frac{-2}{3}\))² -4(\(\frac{3}{5}\))(1)

= \(\frac{4}{9} – \frac{12}{5}\)

= \(\frac{-88}{45}\) < 0

It’s seen that D < 0 and hence, the given equation does not have any real roots.

(i) kx² + 4x + 1 = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

kx² + 4x + 1 = 0

⇒ D = 16 – 4k = 0

⇒ k = 4

(ii) \(kx^2-2\sqrt{5}x+4=0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

\(kx^2-2\sqrt{5}x+4=0\)

⇒ D = 4 × 5 – 4 × 4k = 0

⇒ k = 5/4

(iii) 3x² - 5x + 2k = 0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac 

If D = 0, roots are real and equal

3x² - 5x + 2k = 0

⇒ D = 25 – 4 × 3 × 2k = 0

⇒ k = 25/24

(iv) 4x² + kx + 9 =0

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D = 0, roots are real and equal

4x² + kx + 9 =0

⇒ D = k² – 4 × 4 × 9 = 0

⇒ k² – 144 = 0

⇒ k = 12

(v) 2kx² - 40x + 25 = 0

For a quadratic equation, ax² + bx + c = 0

D = b² – 4ac

If D = 0, roots are real and equal

2kx² - 40x + 25 = 0

⇒ 1600 – 4 × 2k × 25 = 0 

⇒ k = 8

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

D = (-6)² -4(2)(3)

= 36 – 24

= 12 > 0

It’s seen that D > 0 and hence, the given equation have real and distinct roots.

(i) \((x-2a)(x-2b)=4ab\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\((x-2a)(x-2b)=4ab\)

⇒ x² – (2a + 2b)x + 4ab = 4ab 

⇒ x² – (2a + 2b)x = 0 

D = (2a + 2b)² – 0 = (2a + 2b)²

Roots are real and distinct

(ii) \(9a^2b^2x^2-24abcdx+16c^2d^2=0,a\neq0,b\neq0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\(9a^2b^2x^2-24abcdx+16c^2d^2=0,a\neq0,b\neq0\)

⇒ D = 576a²b²c²d² – 4 × 16 × 9 × a²b²c²d² = 0

Roots are real and equal

(iii) \(2(a^2+b^2)x^2+2(a+b)x+1=0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac 

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\(2(a^2+b^2)x^2+2(a+b)x+1=0\)

⇒ D = 4(a + b)² – 4 × 2 × (a² + b²)

⇒ D = 4(a + b)² – 4 × 2 × (a² + b²)

Roots are not real

(iv) \((b+c)x^2-(a+b+c)x+a=0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\((b+c)x^2-(a+b+c)x+a=0\)

⇒ D = (a + b + c)² – 4a(b + c)

⇒ D = a² + b² + c² – 2ab – 2ac + 2bc

⇒ D = (a – b – c)²

Thus, roots are real and unequal

(i) \(2x^2-3x+5=0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\(2x^2-3x+5=0\)

⇒ D = 9 – 4 × 5 × 2 = -31

Roots are not real.

(ii) \(2x^2-6x+3=0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\(2x^2-6x+3=0\)

⇒ D = 36 – 4 × 2 × 3 = 12 

Roots are real and distinct.

(iii) \(\frac{3}{5}x^2-\frac{2}{3}x+1=0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac 

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\(\frac{3}{5}x^2-\frac{2}{3}x+1=0\)

⇒ D = 4/9 – 4 × 3/5 × 1 = -88/45

Roots are not real.

(iv) \(3x^2-4\sqrt{3}x+4=0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\(3x^2-4\sqrt{3}x+4=0\)

⇒ D = 48 – 4 × 3 × 4 = 0

Roots are real and equal

(v) \(3x^2-2\sqrt{6}x+2=0\)

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real 

If D > 0, roots are real and unequal 

If D = 0, roots are real and equal

\(3x^2-2\sqrt{6}x+2=0\)

⇒ D = 24 – 4 × 3 × 2 = 0

Roots are real and equal.

Given

15x² - 28 = x

⇒ 15x² - x - 28 = 0

⇒ 15x² - (21x - 20x) - 28 = 0

⇒ 15x² - 21x + 20x - 28 = 0

⇒ 3x (5x - 7) + 4(5x - 7) = 0

⇒ (3x + 4) (5x - 7) = 0

⇒ 3x + 4 = 0 or 5x - 7 = 0

⇒ x = -4/3 or x = 7/5

Hence, the roots of the equation are -4/3 and 7/5.

Given: kx(x – 2) + 6 = 0 

kx² – 2kx + 6 = 0 

As this Q.E. has equal roots, 

b² – 4ac = 0 

Here 

a = k; b = -2k; c = 6 

∴ b² – 4ac = (-2k)² – 4(k)(6) = 0

⇒ 4k² – 24k = 0 

⇒ 4k(k – 6) = 0 

⇒ 4k = 0 (or) k – 6 = 0 

⇒ k = 0 (or) 6 

But k = 0 is trivial 

∴ k = 6.

Given : 2x² + kx + 3 = 0 has equal roots 

∴ b² – 4ac = 0 

Here a = 2; b = k; c = 3 

b² – 4ac = (k)² – 4(2)(3) = 0 

⇒ k² – 24 = 0 

⇒ k² = 24 

⇒ k = √24 = ± 2√6

abx² + (b² - ac)x - bc = 0

⇒ abx² + b²x - acx - bc = 0

⇒ bx (ax + b) - c (ax + b) = 0

⇒ bx - c = 0 or ax + b = 0

⇒ x = c/b or x = -b/a

Hence, the roots of the equation are c/b and - b/a

The given equation may be written as 8x(2x - 3) = 0

This gives x = 0 or x =3/2.

x=0, 3/2, are the required solutions.

Given,

x² + kx + 4 = 0

It’s of the form of ax² + bx + c = 0

Where, a =1, b = k, c = 4

For the given quadratic equation to have real roots D = b² – 4ac ≥ 0

D = (k)² – 4(1)(4) ≥ 0

⇒ k² – 16 ≥ 0

⇒ k ≥ 4 and k ≤ -4

Considering the least positive value, we have

⇒ k = 4

Thus, the least value of k is 4 for the given equation to have real roots.

\((4-k)\text{x}^2+(2k-4)\text{x}+(8k + 1)=0\)

For the above expression to be a perfect square, D = b² – 4ac = 0

⇒ (2k + 4)² – 4 × (4 – k)(8k + 1) = 0 

⇒ 4k² + 16k + 16 + 32k² – 124k – 16 = 0 

⇒ 36k² – 108k = 0 

⇒ 36k(k – 3) = 0 

⇒ k = 0, 3

The given equation (2p + 1)x² – (7p + 2)x + (7p – 3) = 0 is in the form of ax² + bx + c = 0

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

For the equation to have real and equal roots, the condition is

D = b² – 4ac = 0

⇒ (-(7p + 2))² – 4(2p +1)( 7p – 3) = 0

⇒ (7p + 2)² – 4(14p² + p – 3) = 0

⇒ 49p² + 28p + 4 – 56p² – 4p + 12 = 0

⇒ -7p² + 24p + 16 = 0

Solving for p by factorization,

⇒ -7p² + 28p – 4p + 16 = 0

⇒ -7p(p – 4) -4(p – 4) = 0

⇒ (p – 4) (-7p – 4) = 0

Either p – 4 = 0 ⇒ p = 4 Or, 7p + 4 = 0 ⇒ p = -4/7,

So, the value of k can either be 4 or -4/7

Now, using k = 4 in the given quadratic equation we get

(2(4) + 1)x² – (7(4) + 2)x + (7(4) – 3) = 0

9x² – 30x + 25 = 0

⇒ (3x – 5)² = 0

Thus, x = 5/3 is the root of the given quadratic equation.

Next, on using k = 1 in the given quadratic equation we get

(2(-4/7 ) + 1)x² – (7(-4/7 ) + 2)x + (7(-4/7 ) – 3) = 0

x² – 14x + 49 = 0

⇒ (x – 7)² = 0

Thus, x – 7 = 0 ⇒ x = 7 is the root of the given quadratic equation.

The given equation (3k +1)x² + 2(k +1)x + 1 = 0 is in the form of ax² + bx + c = 0

Where a = (3k +1), b = 2(k + 1), c = 1

For the equation to have real and equal roots, the condition is

D = b² – 4ac = 0

⇒ (2(k + 1))² – 4(3k +1)(1) = 0

⇒ (k +1)² – (3k + 1) = 0 [After dividing by 4 both sides]

⇒ k² + 2k + 1 – 3k – 1 = 0

⇒ k² – k = 0

⇒ k(k – 1) = 0

Either k = 0 Or, k – 3 = 0 ⇒ k = 1,

So, the value of k can either be 0 or 1

Now, using k = 0 in the given quadratic equation we get

(3(0) + 1)x² + 2(0 + 1)x + 1 = 0

x² + 2x + 1 = 0

⇒ (x + 1)² = 0

Thus, x = -1 is the root of the given quadratic equation.

Next, on using k = 1 in the given quadratic equation we get

(3(1) + 1)x² + 2(1 + 1)x + 1 = 0

4x² + 4x + 1 = 0

⇒ (2x + 1)² = 0

Thus, 2x = -1 ⇒ x = -1/2 is the root of the given quadratic equation.

The given equation (k +1)x² – 2(3k +1)x + 8k + 1 = 0 is in the form of ax² + bx + c = 0

Where a = (k +1), b = -2(3k + 1), c = 8k + 1

For the equation to have real and equal roots, the condition is

D = b² – 4ac = 0

⇒ (-2(3k + 1))² – 4(k +1)(8k + 1) = 0

⇒ 4(3k +1)² – 4(k + 1)(8k + 1) = 0

⇒ (3k + 1)² – (k + 1)(8k + 1) = 0

⇒ 9k² + 6k + 1 – (8k² + 9k + 1) = 0

⇒ 9k² + 6k + 1 – 8k² – 9k – 1 = 0

⇒ k² – 3k = 0

⇒ k(k – 3) = 0

Either k = 0 Or, k – 3 = 0 ⇒ k = 3,

So, the value of k can either be 0 or 3

For a quadratic equation, ax² + bx + c = 0, 

D = b² – 4ac 

If D = 0, roots are equal

\((3k+1)\text{x}^2+2(k+1)\text{x}+1=0\)

⇒ D = 4(k + 1)² – 4(3k + 1) = 0 

⇒ k² + 2k + 1 – 3k – 1 = 0 

⇒ k(k – 1) = 0 

⇒ k = 0, 1 

When k = 0, 

Eq. – x² + 2x + 1 = 0 

⇒ (x + 1)² = 0 

⇒ x = -1 

When k = 1, 

Eq. – 4x² + 4x + 1 = 0 

⇒ (2x + 1)² = 0 

⇒ x = -1/2

(3k + 1) x² + 2(k + 1) x + 1 = 0

Compare given equation with the general form of quadratic equation, which is ax² + bx + c = 0

a = (3k + 1), b = 2(k + 1), c = 1

Find discriminant:

D = b² – 4 ac

= (2(k + 1))² – 4(3k + 1) x 1

= 4k² + 4 + 8k – 12k – 4

= 4k (k – 1)

Since roots are real and equal (given)

Put D = 0

4k (k – 1) = 0

Either, k = 0 or k – 1 = 0

k = 0, k = 1

Correct option is (A) b² – 4ac

The discriminant of the quadratic equation \(ax^2+bx+c=0\) is \(D=b^2-4ac.\)

Correct option is A) b² – 4ac

Correct option is (B) \(b^2-2ac=a^2\)

Given quadratic equation is

\(ax^2+bx+c=0\)

Given that \(sin\;\alpha\) and \(cos\;\alpha\) are roots of the equation

\(ax^2+bx+c=0\)

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

\(\Rightarrow\) \(sin\;\alpha+cos\;\alpha\) \(=\frac{-b}a\)   ______________(1)

& Product of roots \(=\frac ca\)

\(\Rightarrow\) \(sin\;\alpha\;cos\;\alpha\) \(=\frac ca\)       ______________(2)

\(\because\) \(sin^2\alpha+cos^2\alpha=1\)

\(\Rightarrow\) \((sin\;\alpha+cos\;\alpha)^2\) \(-2\;sin\;\alpha\;cos\;\alpha=1\)

\(\Rightarrow(\frac{-b}a)^2-\frac{2c}a=1\)

\(\Rightarrow\) \(\frac{b^2-2ac}{a^2}=1\)

\(\Rightarrow\) \(b^2-2ac=a^2\)

Correct option is B) b² – 2ac = a²

Correct option is (C) a² – 4b = 1

Let \(\alpha\;and\;\beta\) are roots of quadratic equation \(x^2-ax+b=0.\)

\(\therefore\) Sum of roots \(=\frac{-(-a)}1=a\)

\(\Rightarrow\) \(\alpha+\beta=a\)      ______________(1)

& Product of roots \(=\frac b1=b\)

\(\Rightarrow\) \(\alpha\beta=b\)          ______________(2)

Now, \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\)

\(\Rightarrow\) \((\alpha-\beta)^2=a^2-4b\)    ______________(3)

Given that difference of the roots of given equation is 1.

\(\therefore\) \(\alpha-\beta=1\)

Then from (3), we have

\(a^2-4b=1^2=1\)

\(\Rightarrow\) \(a^2-4b=1\)

Correct option is C) a² – 4b = 1 

Correct option is (A) b² = 4ac

If roots of quadratic equation \(ax^2+bx+c=0\) are equal, then \(b^2-4ac=0\) or \(b^2= 4ac.\)

Correct option is A) b² = 4ac

Correct option is (D) are real and equal

Given equation is \(ax^2+bx+c=0\)

If discriminant = 0

Then roots are real and equal.

Correct option is D) are real and equal

Correct option is (A) a + b = -1

Let \(\alpha\) be common root of given quadratic equations.

\(\therefore\alpha^2+a\alpha+b=0\)        ______________(1)

& \(\alpha^2+b\alpha+a=0\)        ______________(2)

Subtract equation (2) from (1), we get

\((\alpha^2+a\alpha+b)-(\alpha^2+b\alpha+a)=0\)

\(\Rightarrow(a-b)\alpha+b-a=0\)

\(\Rightarrow(a-b)\alpha-(a-b)=0\)

\(\Rightarrow\alpha=\frac{a-b}{a-b}=1\)

Hence, x = 1 is a common root of given quadratic equations.

Put x = 1 in given equation, we get

\(1^2+a.1+b=0\)

\(\Rightarrow\) 1+a+b = 0

\(\Rightarrow\) a + b = -1

Correct option is A) a + b = -1