NEXT

# class-ten-math quadratic equation NCERT solutions

## Exercise 4.3 NCERT solutions

3. Find the roots of the following equations :

⇒ x^{2 }– 3x – 1 = 0

This is of the form ax^{2} + bx + c = 0.

Where a = 1, b = – 3, c = –1

Now, b2 – 4ac = (–3)^{2} – 4 × 1 × (–1) = 9 + 4 = 13

⇒ – 11 × 30 = 11(x^{2} – 3x – 28)

⇒ –30 = x^{2}
– 3x – 28

⇒ x^{2}
– 3x – 28 + 30 = 0

⇒ x^{2}
– 3x + 2 = 0

This is of the form ax^{2}
+ bx + c = 0.

Where, a = 1, b = –3, c = 2

Now, b^{2}
– 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

⇒ 6*x *+ 6 = *x*^{2} + 2*x *– 15

⇒ *x*^{2} + 2*x *– 15 –
6*x *– 6 = 0

⇒ *x*^{2} – 4*x *– 21 = 0,

which is a quadratic
equation in *x*. This is of the form *ax*^{2} + *bx *+ *c
*= 0

Where, *a *= 1,
*b *= – 4, *c *= –21

Now, *b*^{2}
– 4*ac *= (– 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 – x^{2} + 54 – 2x

= –x^{2} + 25x + 54

According to the given problem,

–x^{2} + 25x + 54 = 210

⇒ –x^{2} +
25x + 54 – 210 = 0

⇒ –x^{2} +
25x – 156 = 0

⇒ x^{2} –
25x + 156 = 0

⇒ x^{2} –
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}

⇒ x^{2} + 3600 + 120x = x^{2 }+ x^{2} + 900 + 60

⇒ *x*^{2} + 3600 + 120*x *– 2*x*^{2} – 900 – 60*x *=
0

⇒ –*x*^{2} + 60*x *+ 2700 = 0

⇒ *x*^{2} – 60*x *– 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 :

x^{2} – y^{2}
= 180 …(i) (∵ x >
y)

Also, y^{2}
= 8x …(ii)

From eqn. (i) and
(ii), we have x^{2} – 8x = 180

⇒ x^{2 }– 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),

y^{2} = 8(–10)

= – 80, which is impossible.

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

y^{2} = 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 = x^{2} + 5x

⇒ x^{2}
+ 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 (x^{2} + 10x)

⇒ 150x +
750 = 8x^{2} + 80x

⇒ 8x^{2}
+ 80x – 150x – 750 = 0

⇒ 8x^{2}
– 70x – 750 = 0

⇒ 4x^{2}
– 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+

⇒ x^{2} + 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 m^{2}. 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 = x^{2}
m^{2}

Perimeter of square = 4x m

Let length of side of another square = y m

∴ Area of that square = y^{2}
m^{2}

Perimeter of that square = 4y m

As per condition,

x^{2} + y^{2}
= 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.

NEXT