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(d) \(\frac{c(a-b)}{a^2}\)

Given, α, β are the roots of the equation ax² + bx + c = 0. Then, 

α + β = – \(\frac{b}{a}\) , αβ = \(\frac{c}{a}\) 

Then, αβ² + α²β + αβ = αβ (α + β) + αβ = \(\big(\frac{c}{a}\big)\) \(-\frac{b}{a}\) + \(\frac{c}{a}\)

= - \(\frac{bc}{a^2}+\frac{c}{a}=\frac{-bc+ac}{a^2}=\)\(\frac{c(a-b)}{a^2}\)

Correct option is (B) x² – 4x + 1 = 0

Roots of quadratic equation are \(2+\sqrt{3}\) and \(2-\sqrt{3}.\) 

\(\therefore\) Required quadratic equation is

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

\(\Rightarrow((x-2)-\sqrt3)((x-2)+\sqrt3)=0\)

\(\Rightarrow(x-2)^2-(\sqrt3)^2=0\)     \((\because(a-b)(a+b)=a^2-b^2)\)

\(\Rightarrow x^2-4x+4-3=0\)       \((\because(a-b)^2=a^2-2ab+b^2)\)

\(\Rightarrow x^2-4x+1=0\)

Correct option is B) x² – 4x + 1 = 0

Correct option is (A) (+7,-2)

The given quadratic polynomial is

\(P(x)=x^2-5x-14\)

For zeros of polynomial, we have

P(x) = 0

\(\Rightarrow x^2-5x-14=0\)

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

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

\(\Rightarrow\) (x - 7)  (x+2) = 0

\(\Rightarrow\) x - 7 = 0 or x+2 = 0

\(\Rightarrow\) x = 7 or x = -2

Hence, the zeros of the quadratic polynomial are 7 & -2.

Correct option is A) (+7,-2)

Correct option is (C) 8

Given that 5 is the root of \(x^2-(K-1)x+10=0\)

\(\Rightarrow5^2-(K-1)5+10=0\)

\(\Rightarrow\) 5 (K - 1) = 25+10 = 35

\(\Rightarrow\) K - 1 \(=\frac{35}5\) = 7

\(\Rightarrow\) K = 7+1 = 8

Correct option is C) 8

2x² + x+ 4 = 0

⇒ 4x² +2x + 8 = 0

(Multiplying both sides by 2)

⇒ 4x² +2x = - 8

⇒ (2x)² + 2 \(\times\) 2x \(\times\) 1/2 + (1/2)2 = - 8 + (1/2)²

[Adding (1/2)² on both sides]

⇒ (2x + 1/2)² = - 8 + 1/4 = - 31/4 < 0

But, (2x + 1/2)² cannot be negative for any real value of x

So, there is no real value of x satisfying the given equation. 

Hence, the given equation has no real roots.

Correct option is (D) \(x^2-3x+2=0\)

Given that \(\alpha\;and\;\beta\) are the roots of the quadratic equation \(x^2-2x+1=0.\)

\(\Rightarrow(x-1)^2=0\)

\(\Rightarrow\) x = 1, 1

\(\therefore\alpha=1,\beta=1\)

\(\therefore\alpha+\beta=1+1=2\)

& \(\alpha\beta=1.1=1\)

\(\therefore\) Required quadratic equation is

\(x^2-\) (Sum of roots)x + Product of roots = 0

\(\Rightarrow x^2-(\alpha+\beta+\alpha\beta)x+(\alpha+\beta)\,\alpha\beta=0\)

\(\Rightarrow x^2-(2+1)x+2.1=0\)

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

Correct option is D) x² – 3x + 2 = 0

Correct option is (D) 2 – √3

Given quadratic equation is \(x^2-4x+1=0\)

Let the other root be x.

\(\therefore\) Sum of roots \(=\frac{-(-4)}1=4\)

\(\Rightarrow\) \(x+2+\sqrt{3}=4\)

\(\Rightarrow\) \(x=4-2-\sqrt3\)

\(\Rightarrow\) x = \(2-\sqrt{3}\)

Hence, the other root is \(2-\sqrt{3}.\)

Correct option is D) 2 – √3

Correct option is B) p² > 4q

Correct option is (B) 8

Given that \(\alpha,\beta\) are roots of \(x^2-10x+9=0\)

\(\therefore\) Sum of roots \(=\frac{-(-10)}1=10\)

\(\Rightarrow\) \(\alpha+\beta=10\)      ______________(1)

And product of roots \(=\frac91=9\)

\(\Rightarrow\) \(\alpha\beta=9\)            ______________(2)

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

\(=10^2-4\times9\)

= 100 - 36

= 64 \(=8^2\)

\(\Rightarrow\) \(|\alpha-\beta|=8\)

Correct option is B) 8

given 4x² + 3x + 7 = 0 

We know sum of the roots = \(\frac{-3}{4}\) 

Product of the roots = \(\frac{7}{4}\) 

As given that α and β are roots then,

α + β =\(\frac{-3}{4}\) 

αβ =  \(\frac{7}{4}\) 

given \(\frac{1}{α }+ \frac{1}{β}\) 

= \(\frac{α+β}{αβ}\) 

= \(-\frac{3}{\frac{4}{\frac{7}{4}}}\) 

= \(\frac{-3}{7}\) 

(b) p² - 4q = 1. 

Given, x² – px + q = 0 

Let α, β be the roots of the given equation. Then, 

α + β = – \(\frac{(-p)}{1}\) = p             ...(i),                αβ = \(\frac{q}{1}\) = q                        ...(ii) 

Also, α - β = 1 (given)                                     ...(iii) 

∴ From (i) and (iii), 2α = p + 1 ⇒ α = \(\frac{p+1}{2}\)

∴ From (i) and (iii), 2β = p – 1 ⇒ b = \(\frac{p-1}{2}\) 

Substituting these values of a and b in (ii), we have \(\big(\frac{p+1}{2}\big)\)\(\big(\frac{p-1}{2}\big)\) = q

⇒ \(\frac{p^2-1}{4}\) = q ⇒ p² - 1 = 4q ⇒ p² - 4q = 1. 

2(a² + b²) x² + 2 (a + b) x + 1 = 0

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

a = 2 (a² + b²), b = 2(a + b), c = 1

Discriminant:

D = b² – 4ac

=[2(a + b)]² – 4. 2 (a² + b²).1

= 4a² + 4b² + 8ab – 8a² – 8b²

= – 4a² – 4b² + 8ab

= – 4(a² + b² – 2ab)

= – 4(a – b)² < 0

Hence the equation has no real roots.

Difference of the roots = \(\frac{\sqrt{D}}{|a|}\)

1 = \(\frac{\sqrt{b^2-4ac}}{|a|}\) 

1 =\(\frac{\sqrt{(-p)^2}-4(1)(q)}{|1|}\)

1 = p² – 4q = 1 

p² – 4q + 4q² – 4q² = 1 

p² + 4q² = 1 + 2(2)(q) + (2q)² 

p² + 4q² = (1 + 2q)²

4x² + px + 3 = 0

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

a = 4, b = p, c = 3

Find discriminant:

D = b² – 4ac

= p² – 4 x 4 x 3

= p² - 48

Since roots are real and equal (given)

Put D = 0

p² – 48 = 0

p² = 48 = (±4√3)²

p = ± 4√3

Hence p= 4√3 or p = -4√3

x² + px – q² = 0

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

a = 1, b = p, c = -q²

Using discriminant formula:

D = b² – 4ac

= (p)² – 4 x 1 x (-q²)

= p² + 4 q² > 0

Hence roots are real for all real values of p and q.

(α – 12) x² + 2(α – 12) x + 2 = 0

Roots of given equation are equal ( given)

So, D = 0

4(α – 12) (α – 14) = 0

α – 14 = 0 {(α – 12) ≠ 0}

α = 14

Hence the value of α is 14

Given: 3 is a root of equation x² – x + k = 0

Substitute the value of x = 3

(3)² – (3) + k = 0

9 – 3 + k = 0

k = -6

Now, x² + k (2x + k + 2) + p = 0

x² + (-6)(2x – 6 + 2) + p = 0

x² – 12x + 36 – 12 + p = 0

x² – 12x + (24 + p) = 0

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

a = 1, b = -12, c = 24 + p

Find Discriminant:

D = b² – 4ac

= (-12)² – 4 x 1 x (24 + p)

= 144 – 96 – 4p = 48 – 4p

Since roots are real and equal, put D = 0

48 – 4p = 0

4p = 48

p = 12

The value of p is 12.

2x² + px + 8 = 0

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

a = 2, b = p, c = 8

Find D:

D = b² – 4ac

= p² – 4 x 2 x 8

= p² – 64

Since roots are real, so D ≥ 0

p² – 64 ≥ 0

p² ≥ 64

≥ (±8)²

Either p ≥ 8 or p ≤ -8

Given: -5 is a root of the quadratic equation 2x² + px – 15 = 0

Substitute the value of x = -5

2(-5)² + p(-5) – 15 = 0

50 – 5p – 15 = 0

35 – 5p = 0

p = 7

Again,

In quadratic equation p(x² + x) + k = 0

7 (x² + x) + k = 0 (put value of p = 7)

7x² + 7x + k = 0

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

a = 7, b = 7, c = k

Find Discriminant:

D = b² – 4ac

= (7)² – 4 x 7 x k

= 49 – 28k

Since roots are real and equal, put D = 0

49 – 28k = 0

28k = 49

k = 7 / 4

The value of k is 7/4.

Let the natural number = x

When the number increased by 12 = x + 12

Reciprocal of the number = 1/x

According to the question, we have,

x + 12 = 160 times of reciprocal of x

x + 12 = 160/ x

x( x + 12 ) = 160

x² + 12x – 160 = 0

x² + 20x – 8x – 160 = 0

x( x + 20) – 8( x + 20)= 0

(x + 20) (x – 8) = 0

x + 20 = 0 or x – 8 = 0

x = – 20 or x = 8

Since, natural numbers cannot be negative.

The required number = x = 8

Let the two parts be x and y.

From the given information,

x + y = 20 ⟹y = 20 − x

3x² = (20 − x) + 10

3x² = 30 − x

3x² + x − 30 = 0

3x² − 9x + 10x − 30 = 0

3x(x − 3) + 10(x - 3) = 0

(x - 3) (3x + 10) = 0

x = 3, -10/3

Since, x cannot be equal to, -10/3 so, x=3

Thus, one part is 3 and other part is 20 − 3 = 17.

Correct answer is (c) c = a 

Let the roots of the equation (ax² + bx + c = 0) be a and 1/a.

∴ Product of the roots = a x 1/a = 1

⇒ c/a = 1

⇒ c = a

(b) 3, 3

\(\frac13\) is a root of the equation 3x² - 10x + p = 0

⇒ 3 \(\big(\frac13\big)^2\) - 10 x \(\frac13\) + p = 0

⇒ \(\frac13\) - \(\frac{10}{3}\) + p = 0 ⇒ \(-\frac93\) + p = 0 ⇒ p = 3

∴ The equation becomes 3x² - 10x + 3 = 0

⇒ 3x² - 9x - x + 3 = 0

⇒ 3x (x - 3) - 1 (x - 3) = 0 ⇒ (3x - 1)(x - 3) = 0

⇒ x = \(\frac13\), 3

∴ p = 3 and the other root = 3.

Let α and β be the roots of the equation 3x² - 10x + k = 0

∴ α = 1/β 

(Given)

⇒  αβ = 1

(Product of the roots = c/α)

⇒ k = 3

Hence, the value of k is 3.

Correct option is (D) c = a

Let \(\alpha\;and\;\frac1\alpha\) be roots of equation \(ax^2+bx+c=0.\)

\(\therefore\) Product of roots \(=\frac ca\)

\(\Rightarrow\) \(\alpha\times\frac1\alpha\) \(=\frac ca\)

\(\Rightarrow\) 1 \(=\frac ca\)

\(\Rightarrow\) a = c

Hence, roots of equation \(ax^2+bx+c=0\) will be reciprocal if c = a.

Correct option is D) c = a

Correct option is (B) x + 1/x = 5/2

Let x be a number.

Then its reciprocal is \(\frac1x.\)

Given that sum of number and its reciprocal is \(\frac{5}{2}.\)

\(\therefore\) \(x+\frac{1}{x}=\frac{5}{2}\) which represent the given situation.

Correct option is B) x + 1/x = 5/2

x² + y² = a².

Differentiating given equation w.r.t. x, we get

2x + 2y(dy/dx) = 0

⇒  x + y(dy/dx) = 0.

The speed of the train 36km/hr

Let her actual marks be x 

Therefore, 9 (x +10) = x² 

i.e., x² – 9x – 90 = 0 

i.e., x² – 15x + 6x – 90 = 0 

i.e., x(x – 15) + 6(x –15) = 0 

i.e., (x + 6) (x –15) = 0 

Therefore, x = – 6 or x = 15 

Since x is the marks obtained, x ≠ – 6. Therefore, x = 15. 

So, Ajita got 15 marks in her mathematics test.

Let its original average speed be x km/h. Therefore,

63/x + 72/(x + 6) = 3

7/x + 8/(x + 6) = 3/9 = 1/3

(7(x + 6) + 8x)/(x (x + 6)) = 1/3

i.e., 21 (x + 6) + 24x = x (x + 6) 

i.e., 21x + 126 + 24x = x² + 6x 

i.e., x² – 39x – 126 = 0 

i.e., (x + 3) (x – 42) = 0 

i.e., x = – 3 or x = 42 

Since x is the average speed of the train, x cannot be negative. 

Therefore, x = 42. 

So, the original average speed of the train is 42 km/h.

Let the breadth be ‘a’ m 

Given, rectangular mango grove whose length is twice its breadth. 

Length = 2a 

Area = 800 m² 

⇒ 2a × a = 800 

⇒ a² = 400 

⇒ a = 20 m 

Thus, length = 40m

Let breadth be X cm 

length = 2X cms 

area = 800 

2X (X) = 800

X(X) = 400 

X² = 400

X = 20

Yes it is possible to design a rectangular mangrove having breadth = 20 cm and length = 40 cm.

Let the total number of camels be x.

Number of camels in forest = x/4 .

Number of camels gone to mountains = 2√x

Remaining camels on bank of the river = 15

Total camels =  x/4 + 2√x + 15

x = x/4 + 2√x + 15

4x = x + 8√x + 60

3x – 60 = 8√x

Squaring both the sides,

9x² + 3600 – 360x = 64x

9x² – 424x + 3600 = 0

On simplifying further,

9x² – 100x – 324x + 3600 = 0

x(9x – 100) – 36(9x – 100) = 0

(x – 36)(9x – 100) = 0

x = 100/9 or X = 36

The number of camels is 36. (As whole values are only considered)

Given,

9x² – 3x – 2 = 0.

⇒ 9x² – 3x – 2 = 0.

⇒ 9x² – 6x + 3x – 2 = 0

⇒ 3x (3x – 2) + 1(3x – 2) = 0

⇒ (3x – 2)(3x + 1) = 0

Now, either 3x – 2 = 0 ⇒ x = \(\frac{2}{3}\)

Or, 3x + 1= 0 ⇒ x = \(\frac{-1}{3}\)

Thus, the roots of the given quadratic equation are x = \(\frac{2}{3}\) and x = \(\frac{-1}{3}\) respectively.

Given equation is 5x² – 3x – 2 = 0.

⇒ 5x² – 3x – 2 = 0.

⇒ 5x² – 5x + 2x – 2 = 0

⇒ 5x(x – 1) + 2(x – 1) = 0

⇒ (5x + 2)(x – 1) = 0

Now, either 5x + 2 = 0 ⇒x = \(\frac{-2}{5}\)

Or, x -1= 0 ⇒x = 1

Thus, the roots of the given quadratic equation are 1 and x = \(\frac{-2}{5}\) respectively.

Given,

16x – \(\frac{10}{x}\) = 27

On multiplying x on both the sides we have,

⇒ 16x² – 10 = 27x

⇒ 16x² – 27x – 10 = 0

⇒ 16x² – 32x + 5x – 10 = 0

⇒ 16x(x – 2) +5(x – 2) = 0

⇒ (16x + 5) (x – 2) = 0

Now, either 16x + 5 = 0 ⇒ x = \(\frac{-5}{16}\)

Or, x – 2 = 0 ⇒ x = 2

Thus, the roots of the given quadratic equation are x =\(\frac{-5}{16}\) and x = 2 respectively.

Given equation is 48x² – 13x – 1 = 0.

⇒ 48x² – 13x – 1 = 0.

⇒ 48x² – 16x + 3x – 1 = 0.

⇒ 16x(3x – 1) + 1(3x – 1) = 0

⇒ (16x + 1)(3x – 1) = 0

Either 16x + 1 = 0 ⇒ x = \(\frac{-1}{16}\)

Or, 3x – 1=0 ⇒ x = \(\frac{1}{3}\)

Thus, the roots of the given quadratic equation are x = \(\frac{-1}{16}\) and x = \(\frac{1}{3}\) respectively.

Given equation is 25x(x + 1) = -4

25x(x + 1) = -4

⇒ 25x² + 25x + 4 = 0

⇒ 25x² + 20x + 5x + 4 = 0

⇒ 5x (5x + 4) + 1(5x + 4) = 0

⇒ (5x + 4)(5x + 1) = 0

Now, either 5x + 4 = 0 therefore x = \(\frac{– 4}{5}\)

Or, 5x + 1 = 0 therefore x = \(\frac{– 1}{5}\)

Thus, the roots of the given quadratic equation are x = \(\frac{– 4}{5}\) and x = \(\frac{– 1}{5}\) respectively.

Given equation is 6x² + 11x + 3 = 0.

⇒ 6x² + 9x + 2x + 3 = 0

⇒ 3x (2x + 3) + 1(2x + 3) = 0

⇒ (2x +3) (3x + 1) = 0

Now, either 2x + 3 = 0 ⇒ x = \(\frac{-3}{2}\)

Or, 3x + 1= 0 ⇒ x = \(\frac{-1}{3}\)

Thus, the roots of the given quadratic equation are x = \(\frac{-3}{2}\) and x = \(\frac{-1}{3}\) respectively.

Given equation is 3x² = -11x – 10

⇒ 3x² + 11x + 10 = 0

⇒ 3x² + 6x + 5x + 10 = 0

⇒ 3x(x + 2) + 5(x + 2) = 0

⇒ (3x + 2)(x + 2) = 0

Now, either 3x + 2 = 0 ⇒ x = \(\frac{-2}{3}\)

Or, x + 2 = 0 ⇒ x = -2

Thus, the roots of the given quadratic equation are x = \(\frac{-2}{3}\) and -2 respectively.

Correct option is (C) 2a = b + c

For equal roots, we have

Discriminant = 0

\(\Rightarrow(b-c)^2-4(a-b)(c-a)=0\)

\(\Rightarrow(b^2+c^2-2bc)\) \(-(4ac-4a^2-4bc+4ab)=0\)

\(\Rightarrow(b^2+c^2+4a^2+2bc-4ab-4ac)=0\)

\(\Rightarrow b^2+c^2+(-2a)^2+2bc\) \(+2\times b\times -2a+2\times c\times -2a=0\)

\(\Rightarrow(b+c-2a)^2=0\)     \((\because a^2+b^2+c^2+2ab+2bc+2ac=(a+b+c)^2)\)

\(\Rightarrow b+c-2a=0\)

\(\Rightarrow\) 2a = b + c

Correct option is C) 2a = b + c

Correct option is (B) real and equal

Given equation is

(x – a) (x – b) + (x – b)(x – c) + (x – c) (x – a) = 0

\(\Rightarrow x^2-(a+b)x+ab+x^2-(b+c)x\) \(+bc+x^2-(a+c)x+ac=0\)

\(\Rightarrow3x^2-((a+b)+(b+c)\) \(+(c+a))x+ab+bc+ac=0\)

\(\Rightarrow3x^2-2(a+b+c)x+ab+bc+ac=0\)

\(\Rightarrow3x^2-2\times3ax+a^2+a^2+a^2=0\)        \((\because a=b=c)\)

\(\Rightarrow3x^2-6ax+3a^2=0\)

\(\Rightarrow x^2-2ax+a^2=0\)

\(\Rightarrow(x-a)^2=0\)

\(\Rightarrow\) x - a = 0 or x - a = 0

\(\Rightarrow\) x = a or x = a

Hence, roots of given equation are real and equal.

Correct option is B) real and equal

Correct option is (D) 4

Given quadratic equation is

\(3x^2+ (2k + 1)\,x \,– (k + 5)= 0\)

\(\therefore\) Sum of roots \(=\frac{-(2k+1)}3\)

& Product of roots \(=\frac{-(k+5)}3\)

Given that sum of roots = Product of roots

\(\therefore\) \(\frac{-(2k+1)}3\) \(=\frac{-(k+5)}3\)

\(\Rightarrow-(2k+1)=-(k+5)\)

\(\Rightarrow2k+1=k+5\)

\(\Rightarrow2k-k=5-1\)

\(\Rightarrow k=4\)

Correct option is D) 4

Correct option is (C) 4

Let \(\alpha\;and\;\beta\) are roots of equation \(3x^2+(2k+1)\,x-(k+5)=0.\)

\(\therefore\) Sum of roots \(=\frac{-(2k+1)}3\)

\(\Rightarrow\) \(\alpha+\beta\) \(=\frac{-(2k+1)}3\)    _____________(1)

And product of roots \(=\frac{-(k+5)}3\)

\(\Rightarrow\) \(\alpha\beta\) \(=\frac{-(k+5)}3\)         _____________(2)

Given that sum of roots = Product of roots

\(\Rightarrow\) \(\frac{-(2k+1)}3\) \(=\frac{-(k+5)}3\)

\(\Rightarrow\) 2k+1 = k+5

\(\Rightarrow\) 2k - k = 5 - 1 = 4

\(\Rightarrow\) k = 4

Correct option is C) 4

Let the number of days in which B finishes the work be ‘b’. 

∴ Number of days in which A finishes the work = b – 10 

In 1 day, 

B finishes 1/b of the work 

A finishes 1/(b – 10) of the work 

Now, both A and B together can finish the work in 12 days

\(\Rightarrow \frac{1}{b}+\frac{1}{b\,-10}=\frac{1}{12}\)

⇒ 12(b – 10 + b) = b² – 10b 

⇒ 24b – 120 = b² – 10b 

⇒ b² - 34b + 120 = 0 

⇒ b² -30b – 4b + 120 = 0 

⇒ b(b – 30) – 4(b – 30) = 0 

⇒ b = 4, 30 b can’t be 4 as A takes 10 days less than B

Thus number of days in which B alone finishes the work is 30 days.

Let the slower pipe fill the reservoir in ‘a’ hours

Faster pipe fills it in ‘a – 10’ hours.

Given, the two pipes will fill the reservoir together in 12 hours.

In 1 hour, part of reservoir filled = 1/12

\(\Rightarrow \frac{1}{a}+\frac{1}{a\,-\,10}=\frac{1}{12}\)

⇒ 12(a + a – 10) = a² – 10a 

⇒ 24a – 120 = a² – 10a 

⇒ a² - 34a + 120 = 0 

⇒ a² – 30a – 4a + 120 = 0 

⇒ a(a – 30) – 4(a – 30) = 0 

⇒ (a – 4)(a – 30) = 0 

Value of a can’t be 4 as (a – 10) will be negative 

Thus a = 30

Let the faster pipe fill the tank in ‘a’ min

Slower pipe fills it in ‘a + 5’ min.

Given, the pipes running together can fill a tank in \(11\frac{1}{9}\) = 100/9 minutes.

In 1 min, part of tank filled = 9/100

\(\Rightarrow \frac{1}{a}+\frac{1}{a\,+\,5}=\frac{9}{100}\)

⇒ 100(a + a + 5) = 9(a² + 5a) 

⇒ 200a + 500 = 9a² + 45a 

⇒ 9a² – 155a - 500 = 0 

⇒ 9a² – 180a + 25a - 500 = 0 

⇒ 9a(a – 20) + 25(a – 20) = 0 

⇒ (9a + 25)(a – 20) = 0 

⇒ a = 20 mins

Slower pipe will fill it in 25 min

Let the smaller diameter tap fill the reservoir in ‘a’ hours

Larger diameter tap fills it in ‘a – 10’ hours.

Given, two water taps together can fill a tank in \(9\frac{3}{8}=\) 75/8 hours.

In 1 hour, part of tank filled = 8/75

\(\Rightarrow \frac{1}{a}+\frac{1}{a\,-\,10}=\frac{8}{75}\)

⇒ 75(a + a – 10) = 8a² – 80a 

⇒ 150a – 750 = 8a² – 80a 

⇒ 8a² – 230a + 750 = 0 

⇒ 4a² – 115a + 375 = 0 

⇒ 4a² – 100a – 15a + 375 = 0 

⇒ 4a(a – 25) – 15(a – 25) = 0 

⇒ (4a – 15)(a – 25) = 0 

Value of a can’t be 15/4 as (a – 10) will be negative 

Thus a = 25 

Time taken by faster tap = 25 – 10 = 15 hours

(i) Since solution set is {3,5}

⟹ x = 3 Or x = 5

⟹ x – 3 = 0 Or x – 5 = 0

⟹ (x – 3) (x – 5) = 0

⟹ x² – 5x – 3x + 15 = 0

⟹ x² – 8x + 15 = 0 

Which is the required equation.

(ii) Since solution set is {−2, 3}

⟹ x = – 2 Or x = 3

⟹ x + 2 = 0 Or x – 3 = 0

⟹ (x + 2) (x – 3) = 0

⟹ x² – 3x + 2x − 6 = 0

⟹ x² – x − 6 = 0 

Which is the required equation.

(iii) Since solution set is {5,−4,}

⟹ x = 5 Or x = −4

⟹ x − 5 = 0 Or x + 4 = 0

⟹ (x − 5) (x + 4) = 0

⟹ x² – 5x + 4x − 20 = 0

⟹ x² – x − 20 = 0

Which is the required equation.

(iv) Since solution set is {−3, -2/5}

⟹ x = −3 Or x = −2/5

⟹ x + 3 = 0 Or 5x + 2 = 0

⟹ (x + 3) (5x + 2) = 0

⟹ 5x² + 2x + 15x + 6 = 0

⟹ 5x² + 17x + 6 = 0

Which is the required equation.

Correct option is (A) a, –a

Given equation is \(\frac{x}{a}=\frac{a}{x}\)

\(\Rightarrow x^2=a^2\)

\(\Rightarrow x=\pm a\)

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

Correct option is A) a, – a