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# class-ten-math quadratic equation NCERT solutions

## Exercise 4.3 NCERT solutions

3. Find the roots of the following equations :

⇒ x2 – 3x – 1 = 0
This is of the form ax2 + bx + c = 0.
Where a = 1, b = – 3, c = –1
Now, b2 – 4ac = (–3)2 – 4 × 1 × (–1) = 9 + 4 = 13

⇒ – 11 × 30 = 11(x2 – 3x – 28)

⇒ –30 = x2 – 3x – 28

⇒ x2 – 3x – 28 + 30 = 0

⇒ x2 – 3x + 2 = 0

This is of the form ax2 + bx + c = 0.

Where, a = 1, b = –3, c = 2

Now, b2 – 4ac = (–3)2 – 4 × 1 × 2 = 9 – 8 = 1

So, roots are 3+1/2 and 3−1/2

or 4/2 and 2/2 or 2 and 1

Hence, 2 and 1 are the roots of a given quadratic equation

⇒  6x + 6 = x2 + 2x – 15

x2 + 2x – 15 – 6x – 6 = 0

x2 – 4x – 21 = 0,

which is a quadratic equation in x. This is of the form ax2 + bx + c = 0

Where, a = 1, b = – 4, c = –21

Now, b2 – 4ac = (– 4)2 – 4 × 1 × (–21)

= 16 + 84 = 100

So, roots are 4 +10/ 2  and 4 −10/ 2  or 7 and –3

Since Age cannot be negative.

∴ x = 7

Hence, Rehman’s present age = 7 years.

5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Sol. Let Shefali got x marks in Mathematics.

∴ Shefali’s marks in English = 30 – x

As per condition :

Shefali’s marks in Mathematics = x + 2

Shefali’s marks in English = 27 – x

∴ Required product = (x + 2) (27 – x)

= 27x – x2 + 54 – 2x

= –x2 + 25x + 54

According to the given problem,

–x2 + 25x + 54 = 210

⇒ –x2 + 25x + 54 – 210 = 0

⇒ –x2 + 25x – 156 = 0

⇒ x2 – 25x + 156 = 0

⇒ x2 – 12x – 13x + 156 = 0

⇒ (x – 12) (x – 13) = 0 ⇒ x = 12 or x = 13

Case I : When x = 13, then

Shefali’s marks in Mathematics = 13

Shefali’s marks in English = 30 – 13 = 17

Case II : When x = 12, then

Shefali’s marks in Mathematics = 12

Shefali’s marks in English = 30 – 12 = 18

Hence, Shefali’s marks in the two subjects are 13 and 17 or 12 and 18.

6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Sol. Let shorter side of the rectangular field = AD = x m

Longer side of the rectangular field = AB

= (x + 30)m

Diagonal of rectangular field = DB = (x + 60) m

Using Pythagoras Theorem in ΔDAB,

we have, (DB)2 = (AD)2 + (AB)2

⇒ (x + 60)2 = (x)2 + (x + 30)2

⇒ x2 + 3600 + 120x = x2 + x2 + 900 + 60

x2 + 3600 + 120x – 2x2 – 900 – 60x = 0

⇒ –x2 + 60x + 2700 = 0

x2 – 60x – 2700 = 0

Using the quadratic formula, we get

Since a side cannot be negative, so we will take the positive value.

So, x = 90

Hence, shorter side of rectangular field = 90 m

∴ Longer side of the rectangular field

= (90 + 30) m = 120 m

7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Sol. Let larger number = x. Smaller number = y

As per condition :

x2 – y2 = 180 …(i) (∵   x > y)

Also, y2 = 8x …(ii)

From eqn. (i) and (ii), we have x2 – 8x = 180

⇒ x2 – 8x – 180 = 0

Using the quadratic formula, we get

or 36/2 and −20/2 or 18 and –10.

When, x = –10, then from eqn. (ii),

y2 = 8(–10)

= – 80, which is impossible.

When, x = 18, then from eqn. (ii),

y2 = 8(18) = 144

⇒ y = ± 144 ⇒ y = ±12

Hence, the required numbers are 18 and 12 or 18 and –12.

8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Sol. Let the speed of the train = x km/hour

Distance covered by the train = 360 km

Increased speed of the train = (x + 5) km/hour
∴ Time is taken by the train with increased speed

As per condition :

⇒ 1800 = x2 + 5x

⇒ x2 + 5x – 1800 = 0

Using the quadratic formula, we get

Since, speed cannot be negative. We will take the positive value.

So, x = 40.

Hence, speed of train = 40 km/hour.

9. Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Sol. Let the time taken by larger tap = x hours

Time taken by smaller tap = (x + 10) hours

A portion of a tank filled in one hour by the larger tap = 1/x

A portion of a tank filled in one hour by the smaller tap = 1/ x +10

∴   Portion of tank filled in one hour by both taps = 1/x + 1/ x +10 …(i)

But, two taps together can fill the tank in 9 38 hours, i.e., in 75/8 hours.

Now, a portion of a tank filled by two taps together in one hour = 8/75 …(ii)

From eqn. (i) and eqn. (ii), we have

⇒ 75(2x + 10) = 8 (x2 + 10x)

⇒ 150x + 750 = 8x2 + 80x

⇒ 8x2 + 80x – 150x – 750 = 0

⇒ 8x2 – 70x – 750 = 0

⇒ 4x2 – 35x – 375 = 0

Using the quadratic formula, we get

Since x is the time, it cannot be negative. Therefore, x = 15.

⇒ 1452 = xx 2 11+

⇒ x2 + 11x – 1452 = 0

Using the quadratic formula, we get

So, roots of the equation are −11+77/2and −11−77/2 or 66/2 and – 88/2 or 33 and – 44.

is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Sol. Let length of side of one square = x m

∴ Area of square = x2 m2

Perimeter of square = 4x m

Let length of side of another square = y m

∴ Area of that square = y2 m2

Perimeter of that square = 4y m

As per condition,

x2 + y2 = 468 …(i)

And 4x – 4y = 24

⇒ 4(x – y) = 24 ⇒ x – y = 6 …(ii)

Using the quadratic formula, we get

⇒ y = 12 and – 18
As, length cannot be negative. We will take the positive value.
So, y = 12
From (ii), x = 6 + 12 = 18
Hence, the sides of the two squares are 12 m and 18 m.

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