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d = b²– 4ac

d = (–2√2)² – 4 (√3) (–2√3)

d = 8 + 24

d = 32

Correct option is (B) -1, -7

Given quadratic equation is

\(x^2+8x+7=0\)

\(\Rightarrow x^2+7x+x+7=0\)

\(\Rightarrow\) x (x+7) + 1 (x+7) = 0

\(\Rightarrow\) (x+7) (x+1) = 0

\(\Rightarrow\) x+7 = 0 or x+1 = 0

\(\Rightarrow\) x = -7 or x = -1

Hence, roots of given quadratic equation are x = -1 & x = -7.

Correct option is B) -1, -7

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

D = b² – 4ac

If D ≥ 0, roots are real

ax² + bx + c = 0,

⇒ b² ≥ ac ------ (1)

\(b\text{x}^2-2\sqrt{acx}+b=0\)

⇒ 4ac – 4b² ≥ 0 

⇒ b² ≤ ac ----- (2)

For both (1) and (2) to be true

⇒ b² = ac

(i) The given equation is 2x² - 8x + 5 = 0

This is of the form ax² + bx + c = 0, where a = -8 and c = 5

∴ Discriminant, D = b² - 4ac = (-8)² - 4 x 2 x 5 = 64 - 40 = 24 > 0

Hence, the given equation has real and unequal roots.

(ii) The given equation is 3x² - 2√6x + 2 = 0

This is of the form ax² + bx + c, where a = 3, b = -2√6 and c = 2.

∴ Discriminant, D = b² - 4ac = (-2√6)² - 4 x 3 x 2 = 24 - 24 = 0

Hence, the given equation has real and equal roots.

(iii) The given equation is 5x² - 4x + 1 = 0

This is of the form ax² + bx + c = 0, where a = 5,b = -4 and c = 1

∴ Discriminant, D = b² - 4ac = (-4)² - 4 x 5 x 1 = 16 - 20 = -4 < 0

Hence, the given equation has no real roots.

It is given that the roots of the equations ax² + 2bx + c = 0 are real.

∴ D₁ = (2b)² - 4 x a x c ≥ 0

⇒ 4(b² - ac) ≥ 0

⇒ b² - ac ≥ 0.........(i)

Also, the roots of the equation bx² - 2\(\sqrt{acx}\) + b = 0 are equal

∴ D₁ = (-2\(\sqrt{ac}\))² - 4 x b x b ≥ 0

⇒ 4(ac - b²) ≥ 0

⇒ -4(b² - ac) ≥ 0

⇒ b² - ac ≥ 0........(2)

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if

b² - ac = 0

⇒ b² = ac

(1 + m²) x² + 2mcx + c² – a² = 0

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

a = (1 + m²), b = 2mc and c = c² – a²

Since roots are equal, so D = 0

(2mc)² – 4.(1 + m²)(c² – a²) = 0

4 m²c² – 4c² + 4a² – 4 m²c² + 4 m²a² = 0

a² + m²a² = c²

or c² = a² (1 + m²)

Hence Proved

(i) The given equation is

5x (x - 2) + 6 = 0

⇒ 5x² - 10x + 6 = 0

This is of the form ax² + bx + c = 0,where a = 5,b = -10 and c = 6

∴ Discriminant, D = b² - 4ac = (-10)² - 4 x 5 x 6 = 100 - 120 = - 20 <0

Hence, the given equation has real and equal roots.

(ii) The given equation is 12x² - 4\(\sqrt{15}\)x + 5 = 0

This is of the form ax² + bx + c = 0,where a = 12,b = -4\(\sqrt{15}\) and c = 5

∴ Discriminant, D = b² - 4ac = (- 4\(\sqrt{15}\))² - 4 x 12 x 5 = 240 - 240 = 0

Hence, the given equation has real and equal roots.

(iii) The given equation is x² - x + 2 = 0

This is of the form ax² + bx + c = 0,where a = 1,b = -1 and c = 2

∴ Discriminant, D = b² - 4ac = (- 1)² - 4 x 1 x 2 = 1 - 8 = -7 < 0

Hence, the given equation has no real roots.

a = 1, b = – k, c = 9

For real and distinct roots, then D > 0

(-k)² – 36 > 0

k > 6 

or k < -6

Given:

2x² + px + 8 = 0

Here

a = 2, b = p and c = 8

Discriminant D is given by:

D = (b² - 4ac)

= p² - 4 x 2 x 8

= (p² - 64)

If D ≥ 0, the roots of the equation will be real

⇒ (p² - 64) ≥ 0

⇒ (p + 8) (p - 8) ≥ 0

⇒ p ≥ 8 and p ≤ - 8

Thus, the roots of the equation are real for p ≥ 8 and  p ≤ - 8.

Given,

2x² + x + k = 0

It’s of the form of ax² + bx + c = 0

Where, a = 2, b = 1, c = k

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

D = 1² – 4(2)(k) ≥ 0

⇒ 1 – 8k ≥ 0

⇒ k ≤ \(\frac{1}{8}\)

The value of k should not exceed \(\frac{1}{8}\) to have real roots.

Given,

kx² + 6x + 1 = 0

It’s of the form of ax² + bx + c = 0

Where, a =k, b = 6, c = 1

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

D = (6)² – 4(1)(k) > 0

⇒ 36 – 4k > 0

⇒ 4k < 36

⇒ k < 9

The value of k should be lesser than 9 to have real and distinct roots.

Given:

9x² + 8kx + 16 = 0

Here

a = 9,b = 8k and c = 16

It is given that the roots of the equation are real and equal; therefore, we have:

D = 0

⇒ (b² - 4ac) = 0

⇒ (8k)² - 4 x 9 x 16 = 0

⇒ 64k² - 576 = 0

⇒ 64k² = 576

⇒ k² = 9

⇒ k = ≠3

∴ k = 3 or k = -3

Given

12abx² - (9a² - 8b²)x - 6ab = 0

⇒ 12abx² - 9a²x + 8b²x - 6ab = 0

⇒ 3ax (4bx - 3a) + 2b (4bx - 3a) = 0

⇒ (3ax + 2b) (4bx - 3a) = 0

⇒ 3ax + 2b = 0 or 4bx - 3a = 0

⇒ x = -2b/3a or x = 3a/4b

Hence, the roots of the equation are -2b/3a and 3a/4b.

The equation 3x² – 4x + 1 = 0 has two real and distinct roots.

D = b² – 4ac

= (-4)² – 4(3)(1)

= 16 – 12 > 0

Hence, the roots are real and distinct.

The equation 2x² – 6x + (9/2) = 0 has real and equal roots.

D = b² – 4ac

= (-6)² – 4(2) (9/2)

= 36 – 36 = 0

Hence, the roots are real and equal.

The equation 2x² – 6x + (9/2) = 0 has real and equal roots.

D = b² – 4ac

= (-6)² – 4(2) (9/2)

= 36 – 36 = 0

Hence, the roots are real and equal.

b²x {a²x + 1} – 1 {a²x + 1} = 0

(b²x – 1) (a²x + 1) = 0

b²x – 1 = 0 a²x + 1 = 0

x = 1/b², x = -1/a²

Therefore, the roots of the equation are 1/b²,  -1/a²​​​​​​​.

Given

4x² - 2(a² + b²)x + a²b² = 0

⇒ 4x² - 2a²x - 2b²x + a²b² = 0

⇒ 2x(2x - a²) - b²(2x - a²) = 0

⇒ (2x - b²) (2x - a²) = 0

⇒ 2x - b² = 0 or 2x - a² = 0

⇒ x = b²/2 or x = a²/2

Hence, the roots of the equation are b²/2 and a²/2

The equation √2x² – 3x/√2 + ½ = 0 has two real and distinct roots.

D = b² – 4ac

= (- 3/√2)² – 4(√2) (½)

= (9/2) – 2√2 > 0

Hence, the roots are real and distinct.

x² – 6x = –4

Add the coefficient of (x/2)² to both sides

x² – 6x + (3)² = –4 + (3)²

(x – 3)² = –4 + 9

x – 3 = √5

x = ±√5 + 3

The given Q.E. is x² + 5 = -6x 

⇒ x² + 6x = -5 

⇒ (x)² + 2.(x).3 = -5 

Now L.H.S. is of the form a² + 2ab where b = 3. 

Adding b² = 3² on both sides we get 

x² + 2(x)(3) + 3² = -5 + 3² 

(x + 3)² = -5 + 9 = 4 

∴ x + 3 = 74 = ± 2 

⇒ x = +2 – 3 or – 2 – 3 

\(\therefore\) x= -1 or -5 are the roots of the given Q.E.

The given equations is x² + x + 2 = 0

Comparing it with ax² +bx + c = 0.we get

a = 1, b = 1  and c = 2

∴ Discriminant d = b² - 4ac = 1² - 4 x 1 x 2 = 1 - 8 = - 7 < 0

Hence, the given equation has no real roots (or real roots does not exist).

Correct option is (C) 1/√2, 1/√2

Given quadratic equation is

\(2x^2-2\sqrt{2}x+1=0\)

\(\Rightarrow2x^2-\sqrt{2}x-\sqrt{2}x+1=0\)

\(\Rightarrow\sqrt{2}x(\sqrt{2}x-1)-1(\sqrt{2}x-1)=0\)

\(\Rightarrow(\sqrt{2}x-1)(\sqrt{2}x-1)=0\)

\(\Rightarrow\sqrt{2}x-1=0\) \(or\;\sqrt{2}x-1=0\)

\(\Rightarrow\) \(x=\frac1{\sqrt2}\) or \(x=\frac1{\sqrt2}\)

Hence, the roots of given quadratic equation are \(\frac{1}{\sqrt{2}}\;and\;\frac{1}{\sqrt{2}}.\)

Correct option is C) 1/√2 , 1/√2

x² + (1 – 2i)x – 2i = 0 

Given x² + (1 – 2i)x – 2i = 0 

⇒ x² + x – 2ix – 2i = 0 

⇒ x(x + 1) – 2i(x + 1) = 0 

⇒ (x + 1)(x – 2i) = 0 

⇒ x + 1 = 0 or x – 2i = 0 

∴ x = –1 or 2i 

Thus, the roots of the given equation are –1 and 2i.

x² + 10ix – 21 = 0 

Given x² + 10ix – 21 = 0 

⇒ x² + 10ix – 21 × 1 = 0 

We have i² = –1 

⇒ 1 = –i² 

By substituting 1 = –i² in the above equation, we get 

x² + 10ix – 21(–i²) = 0 

⇒ x² + 10ix + 21i² = 0 

⇒ x² + 3ix + 7ix + 21i² = 0 

⇒ x(x + 3i) + 7i(x + 3i) = 0 

⇒ (x + 3i)(x + 7i) = 0

⇒ x + 3i = 0 or x + 7i = 0 

∴ x = –3i or –7i 

Thus, the roots of the given equation are –3i and –7i.

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(25x(x+1)=-4\)

⇒ 25x² + 25x + 4 = 0 

⇒ 25x² + 20x + 5x + 4 = 0 

⇒ 5x(5x + 4) + (5x + 4) = 0 

⇒ (5x + 1)(5x + 4) = 0 

⇒ x = -1/5, -4/5

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(16x-\frac{10}{x}=27\)

⇒ 16x² – 27x – 10 = 0 

⇒ 16x² – 32x + 5x – 10 = 0 

⇒ 16x(x – 2) + 5(x – 2) = 0 

⇒ (16x + 5)(x – 2) = 0 

⇒ x = -5/16, 2

6x² – √2x – 2 

= 6x² – 3√2 x + 2√2 x – 2 

= 3x (2x – √2 ) + √2 (2x – √2 ) 

= (3x + √2 ) (2x – √2) 

Now, 6x² – √2x – 2 = 0 gives (3x +√2 ) (2x –√2 ) = 0, i.e., 3x +√2 = 0 or 2x – √2 = 0 

So, the roots are −√2/3 and √2/2.

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

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

⇒ √3x² – 3√2x + √2x – 2√3 = 0 

⇒ √3x² – √3×√3×√2x + √2x – 2×√3×√3 = 0

⇒ √3x² – √3×√6x + √2x – √6×√3 = 0 

⇒ √3x(x - √6) + √2(x - √6) = 0 

⇒ (√3x + √2)(x - √6) = 0 

⇒ x = √6, -√(2/3)

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

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

⇒ √2x² – 4x + x – 2√2 = 0 

⇒ √2x(x – 2√2) + (x – 2√2) = 0 

⇒ (√2x + 1)(x – 2√2) = 0 

⇒ x = -1/√2, 2√2

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized

\(\frac{2}{x + 1}+\frac{3}{2(x-2)}=\frac{23}{5x};x\neq0,-1,2\)

⇒ 5x(4x – 8 + 3x + 3) = 46(x + 1)(x – 2) 

⇒ 35x² – 25x = 46x² – 92 – 46x 

⇒ 11x² - 19x - 92 = 0 

⇒ 11x² - 44x + 23x – 92 = 0 

⇒ 11x(x – 4) + 23(x – 4) = 0 

⇒ (11x + 23)(x – 4) = 0 

⇒ x = 4, - 23/11

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(4\sqrt{3}x^{2}+5x-2\sqrt{3}=0\)

⇒ 4√3x² + 8x – 3x – 2√3 = 0 

⇒ 4x(√3x + 2) – √3(√3x + 2) = 0 

⇒ (4x - √3)(√3x + 2) = 0 

⇒ x = √3/4, - 2/√3

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(3(\frac{7x+1}{5x-3})-4(\frac{5x-3}{7x+1})=11;x\neq\frac{3}{5},-\frac{1}{7}\)

⇒ 3(7x + 1)² – 4(5x – 3)² 

= 11(5x – 3)(7x + 1) 

⇒ 3(49x² + 1 + 14x) – 4(25x² + 9 – 30x) 

= 11(35x² – 3 – 16x) 

⇒ 338x² – 338x = 0 

⇒ x(x – 1) = 0 

⇒ x = 0, 1

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

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

⇒ 5x² – 5x + 2x – 2 = 0

⇒ 5x(x – 1) + 2(x – 1) = 0 

⇒ (5x + 2)(x – 1) = 0 

⇒ x = -2/5, 1

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(6x^{2}+11x+3=0\)

⇒ 6x² + 9x + 2x + 3 = 0 

⇒ 3x(2x + 3) + (2x + 3) = 0 

⇒ (3x + 1)(2x + 3) = 0 

⇒ x = -1/3, - 3/2

6x² – 17ix – 12 = 0

Given 6x² – 17ix – 12 = 0

⇒ 6x² – 17ix – 12x1 = 0

We have i² = –1 

⇒ 1 = –i² 

By substituting 1 = –i² in the above equation, we get 

6x² – 17ix – 12(–i²) = 0 

⇒ 6x² – 17ix + 12i² = 0 

⇒ 6x² – 9ix – 8ix + 12i² = 0 

⇒ 3x(2x – 3i) – 4i(2x – 3i) = 0 

⇒ (2x – 3i)(3x – 4i) = 0 

⇒ 2x – 3i = 0 or 3x – 4i = 0 

⇒ 2x = 3i or 3x = 4i

x = \(\frac{3}{2}i,or\frac{4}{3}i\) 

Thus, the roots of the given equation are

\(\frac{3}{2}i,or\frac{4}{3}i\)

Given as 21x² + 9x + 1 = 0

Now we shall apply discriminant rule,

Where, x = (-b ± √(b² – 4ac))/2a

Since, a = 21, b = 9, c = 1

Therefore,

x = (-9 ± √(9² – 4 (21)(1)))/ 2(21)

= (-9 ± √(81 - 84))/42

= (-9 ± √(-3))/42

= (-9 ± √3(-1))/42

Then we have i² = –1

On substituting –1 = i² in the above equation, we get

x = (-9 ± √3i²)/42

= (-9 ± √(√3i)²/42

= (-9 ± √3i)/42

= -9/42 ± √3i/42

= -3/14 ± √3i/42

∴ The roots of the given equation are -3/14 ± √3i/42

Given as 9x² + 4 = 0

9x² + 4 × 1 = 0

As we know, i² = –1 ⇒ 1 = –i²

On substituting 1 = –i² in the above equation, we get

Therefore,
9x² + 4(–i²) = 0

9x² – 4i² = 0

(3x)² – (2i)² = 0

[On using the formula, a² – b² = (a + b) (a – b)]

(3x + 2i) (3x – 2i) = 0

3x + 2i = 0 or 3x – 2i = 0

3x = –2i or 3x = 2i

x = -2i/3 or x = 2i/3

Thus, the roots of the given equation are 2i/3, -2i/3

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

(a + b)² = a² + 2ab + b²

^{\(x^2-4ax+4a^2-b^2 = 0\)}

⇒ x² – 2 × 2ax + 4a² = b² 

⇒ (x – 2a)² = b² 

⇒ x – 2a = ±b 

⇒ x = 2a + b, 2a – b

x² – 6x + 7= 0 , which has roots 3 + √2, 3– √2

Number of distinct line segments that can be formed out of n-points = \(\frac{n(n-1)}{2}\)

Given: No. of line segments 

\(\frac{n(n-1)}{2}\) = 10 

⇒ n² – n = 20 

⇒ n² – n – 20 = 0 

⇒ n² – 5n + 4n – 20 = 0 

⇒ n(n – 5) + 4(n – 5) = 0 

⇒ (n – 5) (n + 4) = 0 

⇒ n – 5 = 0 (or) n + 4 = 0 

⇒ n = 5 (or) -4 

∴ n = 5 [n – can’t be negative]

Let the digit in the units place = x 

Let the digit in the tens place = y 

∴ The number = 10y + x 

By interchanging the digits the number becomes 10x + y 

By problem (10x + y) – (10y + x) = 18 

⇒ 9x – 9y = 18 

⇒ 9(x – y) =18 

⇒ x – y = 18/9 = 2 

⇒ y = x – 2

(i.e.) digit in the tens place = x – 2 

digit in the units place = x 

Product of the digits = (x – 2) x 

By problem x² – 2x = 8

x² – 2x – 8 = 0 

⇒ x² – 4x + 2x – 8 = 0 

⇒ x(x – 4) + 2(x – 4) = 0 

⇒ (x – 4) (x + 2) = 0 

⇒ x – 4 = 0 (or) x + 2 = 0

⇒ x = 4 (or) x = -2 

∴ x = 4 [∵ x can’t be negative] 

∴ The number is 24.

Let the ones digit be ‘a’ and tens digit be ‘b’. 

Given, two-digit number is such that the product of its digits is 8. 

⇒ ab = 8 --- (1) 

Also, when 18 is subtracted from the number, the digits interchange their places 

⇒ 10b + a – 18 = 10a + b 

⇒ 9b – 9a = 18 ⇒ b – a = 2 

⇒ b = 2 + a 

Substituting in 1 

⇒ a × (2 + a) = 8 

⇒ a² + 2a – 8 = 0 

⇒ a² + 4a – 2a – 8 = 0 

⇒ a(a + 4) – 2(a + 4) = 0 

⇒ (a – 2)(a + 4) = 0 

⇒ a = 2 

Thus, b = 4 

Number is 42

Let’s consider the three consecutive numbers to be x, x + 1, x + 2 respectively. And x being the first integer of the sequence.

From the question, it’s understood that

x² + (x + 1)(x + 2) = 154

⇒ x² + x² + 3x + 2 = 154

⇒ 2x² + 3x – 152 = 0

Solving for x by factorization method, we have

⇒ 2x² + 19x – 16x – 152 = 0

⇒ x(2x + 19) – 8(2x – 19) = 0

⇒ (2x – 19)(x – 8) = 0

Now, either 2x – 19 = 0 ⇒ x = 19/2 (which is not an integer)

Or, x – 8 = 0 ⇒ x = 8

Hence, considering x = 8 the three consecutive integers are 8, 9 and 10.

Let the numbers be ‘a’ and ‘b’. 

Given, difference of squares of two numbers is 88. 

⇒ a² – b² = 88 

Also, the larger number is 5 less than twice the smaller number. 

⇒ a = 2b – 5 Thus, (2b – 5)² - b² = 88 

⇒ 4b² + 25 – 20b - b² = 88 

⇒ 3b² – 20b – 63 = 0 

⇒ 3b² – 27b + 7b – 63 = 0 

⇒ 3b(b – 9) + 7(b – 9) = 0 

⇒ (3b + 7)(b – 9) = 0 

⇒ b = 9 

Thus, a = 2 × 9 – 5 = 13

Let the consecutive odd positive integers be ‘a’ and a + 2 

⇒ a² + (a + 2)² = 970 

⇒ 2a² + 4a – 966 = 0 

⇒ a² + 2a – 483 = 0 

⇒ a² + 23a – 21a – 483 = 0 

⇒ a(a + 23) – 21(a + 23) = 0 

⇒ (a – 21)(a + 23) = 0 

Thus, a = 21 

Consecutive odd positive integers are 21, 23

Let the numbers be ‘a’ and ‘b’. 

Given, difference of the squares of two numbers is 180. 

⇒ a² – b² = 180 

Also, square of the smaller number is 8 times the larger. 

⇒ b² = 8a 

Thus, a² – 8a – 180 = 0

⇒ a² – 18a + 10a – 180 = 0 

⇒ a(a – 18) + 10(a – 18) = 0 

⇒ (a + 10)(a – 18) = 0 

⇒ a = -10, 18 

Thus, the other number is 324 – 180 = b² 

⇒ b = 12 

Numbers are 12, 18 or -12, 18

Let the larger number = x

Then the square of the smaller number = 8 times the larger number = 8x

and the square of the larger number = x²

According to the question,

  x² - 8x = 180

  =>  x² - 8x - 180 = 0

  =>  x² - 18x + 10x - 180 = 0

  =>  x(x - 18) + 10(x - 18) = 0

  =>  (x - 18) (x + 10)  = 0

  =>  x - 18 = 0  or x + 10 = 0

  =>  x = 18  or x = -10

Thus, the larger number = 18 or -10

Then, the square of the smaller number  = 8(18)  or     8(-10)

  = 144       or     -80

The square of a number can't be negative, so, the square of smaller number = 144

Hence, the smaller number = sqrt(144) = 12

The numbers are 12 and 18

Given: (x + l)² = 2(x – 3) 

⇒ x² + 2x + 1 = 2(x – 3) = 2x – 6

⇒ x² + 2x + l – 2x + 6 = 0 

⇒ x² + 7 = 0 is a Q.E.

Given: (x – 3) (2x + 1) = x(x + 5) 

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

⇒ 2x² + x – 6x – 3 = x² + 5x 

⇒ 2x² – 5x – 3 – x² – 5x = 0 

⇒ x² – 10x – 3 = 0 is a Q.E.

(or) 

Comparing the coefficients of x² on both sides. 

x . 2x and x . x 

⇒ 2x² and x² 

2x² ≠ x² 

Hence it’s a Q.E.