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**Exercise 4.1**

**Choose the correct answer from the given four options in the following questions:****1. Which of the following is a quadratic equation?****(a) x ^{2} + 2x + 1 = (4 – x)^{2} + 3 **

The standard form of a quadratic equation is given by,

ax

(A) Given, x

x

10x – 18 = 0

which is not a quadratic equation.

(B) Given, -2x

-2x

52x – 10 = 0

which is not a quadratic equation.

(C) Given, (k + 1) x

(-1 + 1) x

3x – 14 = 0

which is not a quadratic equation.

(D) Given, x

x

2x

which represents a quadratic equation.

**Ans: (d)****Explanation:**

A quadratic equation is represented by the form,

ax^{2} + bx + c = 0, a ≠ 0

(A) Given, 2(x – 1)^{2} = 4x^{2} – 2x + 1

2(x^{2} – 2x + 1) = 4x^{2} – 2x + 1

2x^{2} + 2x – 1 = 0

which is a quadratic equation.

(B) Given, 2x – x^{2} = x^{2} + 5

2x^{2} – 2x + 5 = 0

which is a quadratic equation.

(C) Given, (√2x + √3)^{2} = 3x^{2} – 5x

2x^{2} + 2√6x + 3 = 3x^{2} – 5x

x^{2} – (5 + 2√6)x – 3 = 0

which is a quadratic equation.

(D) Given, (x^{2} + 2x)^{2} = x^{4} + 3 + 4x^{2}

x^{4} + 4x^{3} + 4x^{2} = x^{4} + 3 + 4x^{2}

4x^{3} – 3 = 0

which is a cubic equation and not a quadratic equation.**3. Which of the following equations has 2 as a root?****(a) x ^{2} – 4x + 5 = 0 **

**Ans: (c)****Explanation:**

If 2 is a root then substituting the value 2 in place of x should satisfy the equation.

(A) Given,

x^{2} – 4x + 5 = 0

(2)^{2} – 4(2) + 5 = 1 ≠ 0

So, x = 2 is not a root of x^{2} – 4x + 5 = 0

(B) Given, x^{2} + 3x – 12 = 0

(2)^{2} + 3(2) – 12 = -2 ≠ 0

So, x = 2 is not a root of x^{2} + 3x – 12 = 0

(C) Given, 2x^{2} – 7x + 6 = 0

2(2)^{2} – 7(2) + 6 = 0

Here, x = 2 is a root of 2x^{2} – 7x + 6 = 0

(D) Given, 3x^{2} – 6x – 2 = 0

3(2)^{2} – 6(2) – 2 = -2 ≠ 0

So, x = 2 is not a root of 3x^{2} – 6x – 2 = 0**4. If ½ is a root of the equation x ^{2} + kx – 5/4 = 0, then the value of k is**

If ½ is a root of the equation

x

Given, x

(½)

(k/2) = (5/4) – ¼

k = 2

**(c) √2x ^{2 }– 3/√2x + 1 = 0 **

Explanation:

The sum of the roots of a quadratic equation ax

Coefficient of x / coefficient of x

(A) Given, 2x

Sum of the roots = – b/a = -(-3/2) = 3/2

(B) Given, -x

Sum of the roots = – b/a = -(3/-1) = 3

(C) Given, √2x

2x

Sum of the roots = – b/a = -(-3/2) = 3/2

(D) Given, 3x

Sum of the roots = – b/a = -(-3/3) = 1

**Exercise 4.2**

**1. State whether the following quadratic equations have two distinct real roots. Justify your answer.**

**(i) x ^{2} – 3x + 4 = 0**

**(ii) 2x ^{2} + x – 1 = 0**

**(iii) 2x ^{2} – 6x + 9/2 = 0**

**(iv) 3x ^{2} – 4x + 1 = 0**

**(v) (x + 4) ^{2} – 8x = 0**

**(vi) (x – **√**2) ^{2} – 2(x + 1) = 0**

**(vii) **√**2 x ^{2} –(3/√2)x + 1/√2 = 0**

**(viii) x (1 – x) – 2 = 0**

**(ix) (x – 1) (x + 2) + 2 = 0**

**(x) (x + 1) (x – 2) + x = 0****Solution:**

(i)

The equation x^{2} – 3x + 4 = 0 has no real roots.

D = b^{2} – 4ac

= (-3)^{2} – 4(1)(4)

= 9 – 16 < 0

Hence, the roots are imaginary.

(ii)

The equation 2x^{2} + x – 1 = 0 has two real and distinct roots.

D = b^{2} – 4ac

= 1^{2} – 4(2) (-1)

= 1 + 8 > 0

Hence, the roots are real and distinct.

(iii)

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

D = b^{2} – 4ac

= (-6)^{2} – 4(2) (9/2)

= 36 – 36 = 0

Hence, the roots are real and equal.

(iv)

The equation 3x^{2} – 4x + 1 = 0 has two real and distinct roots.

D = b^{2} – 4ac

= (-4)^{2} – 4(3)(1)

= 16 – 12 > 0

Hence, the roots are real and distinct.

(v)

The equation (x + 4)^{2} – 8x = 0 has no real roots.

Simplifying the above equation,

x^{2} + 8x + 16 – 8x = 0

x^{2} + 16 = 0

D = b^{2} – 4ac**=** (0) – 4(1) (16) < 0

Hence, the roots are imaginary.

(vi)

The equation (x – √2)^{2 }– √2(x+1)=0 has two distinct and real roots.

Simplifying the above equation,

x^{2} – 2√2x + 2 – √2x – √2 = 0

x^{2} – √2(2+1)x + (2 – √2) = 0

x^{2} – 3√2x + (2 – √2) = 0

D = b^{2} – 4ac

= (– 3√2)^{2} – 4(1)(2 – √2)

= 18 – 8 + 4√2 > 0

Hence, the roots are real and distinct.

(vii)

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

D = b^{2} – 4ac

= (- 3/√2)^{2} – 4(√2) (½)

= (9/2) – 2√2 > 0

Hence, the roots are real and distinct.

(viii)

The equation x (1 – x) – 2 = 0 has no real roots.

Simplifying the above equation,

x^{2} – x + 2 = 0

D = b^{2} – 4ac**=** (-1)^{2} – 4(1)(2)

= 1 – 8 < 0

Hence, the roots are imaginary.

(ix)

The equation (x – 1) (x + 2) + 2 = 0 has two real and distinct roots.

Simplifying the above equation,

x^{2} – x + 2x – 2 + 2 = 0

x^{2} + x = 0

D = b^{2} – 4ac**=** 1^{2} – 4(1)(0)

= 1 – 0 > 0

Hence, the roots are real and distinct.

(x)

The equation (x + 1) (x – 2) + x = 0 has two real and distinct roots.

Simplifying the above equation,

x^{2} + x – 2x – 2 + x = 0

x^{2} – 2 = 0

D = b^{2} – 4ac

= (0)^{2} – 4(1) (-2)

= 0 + 8 > 0

Hence, the roots are real and distinct.**2. Write whether the following statements are true or false. Justify your answers.**

**(i) Every quadratic equation has exactly one root.**

**(ii) Every quadratic equation has at least one real root.**

**(iii) Every quadratic equation has at least two roots.**

**(iv) Every quadratic equations has at most two roots.**

**(v) If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.**

**(vi) If the coefficient of x2 and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.****Solution: **

(i) False. For example, a quadratic equation *x*^{2} – 9 = 0 has two distinct roots – 3 and 3.

(ii) False. For example, equation *x*^{2} + 4 = 0 has no real root.

(iii) False. For example, a quadratic equation *x*^{2} – 4*x *+ 4 = 0 has only one root which is 2.

(iv) True, because every quadratic polynomial has almost two roots.

(v) True, because in this case discriminant is always positive.

For example, in *ax*^{2}+ *bx* + *c* = 0, as *a* and *c* have opposite sign, *ac* < 0

⟹ Discriminant = *b*^{2} – 4*ac* > 0.

(vi) True, because in this case discriminant is always negative.

For example, in *ax*^{2}+ *bx* + *c* = 0, as *b* = 0, and *a* and *c* have same sign then *ac* > 0

⟹ Discriminant = *b*^{2} – 4*ac* = – 4 *a c* < 0**3. A quadratic equation with integral coefficient has integral roots. Justify your answer.Solution: **

No, a quadratic equation with integral coefficients may or may not have integral roots.

Justification

Consider the following equation,

8x

The roots of the given equation are ½ and – ¼ which are not integers.

Hence, a quadratic equation with integral coefficient might or might not have integral roots.

**Exercise 4.3**

**1. Find the roots of the quadratic equations by using the quadratic formula in each of the following:**

**(i) 2 x ^{2} – 3x – 5 = 0**

**(ii) 5x ^{2} + 13x + 8 = 0**

**(iii) –3x ^{2} + 5x + 12 = 0**

**(iv) –x ^{2} + 7x – 10 = 0**

**(v) x ^{2} + 2** √

**(vi) x ^{2} – 3** √

**(vii) (½)x ^{2}– √11x + 1 = 0**

The quadratic formula for finding the roots of quadratic equation

ax

**(v) x ^{2} + 2** √

**Exercise 4.4**

1. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.**Solution:**

Let the natural number = ‘x’.

According to the question,

We get the equation,

x² – 84 = 3(x + 8)

x² – 84 = 3x + 24

x² – 3x – 84 – 24 = 0

x² – 3x – 108 = 0

x² – 12x + 9x – 108 = 0

x(x – 12) + 9(x – 12) = 0

(x + 9) (x – 12)

⇒ x = -9 and x = 12

Since, natural numbers cannot be negative.

The number is 12.**2. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.****Solution:**

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

x^{2} + 12x – 160 = 0

x^{2} + 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**3. A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.****Solution:**

Let original speed of train = x km/h

We know,

Time = distance/speed

According to the question, we have,

Time taken by train = 360/x hour

And, Time taken by train its speed increase 5 km/h = 360/( x + 5)

It is given that,

Time taken by train in first – time taken by train in 2nd case = 48 min = 48/60 hour

360/x – 360/(x +5) = 48/60 = 4/5

360(1/x – 1/(x +5)) = 4/5

360 ×5/4 (5/(x² +5x)) =1

450 × 5 = x² + 5x

x² +5x -2250 = 0

x = (-5± √ (25+9000))/2

= (-5 ±√ (9025) )/2

= (-5 ± 95)/2

= -50, 45

But x ≠ -50 because speed cannot be negative

So, x = 45 km/h

Hence, original speed of train = 45 km/h**4. If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?****Solution:**

Let Zeba’s age = x

According to the question,

(x-5)²=11 + 5x

x² + 25 - 10x = 11 + 5x

x² - 15x + 14 = 0

x² - 14x - x + 14 = 0

x(x -14) -1(x - 14) = 0

x = 1 or x = 14

We have to neglect 1 as 5 years younger than 1 cannot happen.

Therefore, Zeba’s present age = 14 years.

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