NCERT Exemplar: Quadratic Equations - 1

### Exercise 4.1

Choose the correct answer from the given four options in the following questions:
Q.1. Which of the following is a quadratic equation?
(a) x2 + 2x + 1 = (4 – x)2 + 3
(b) –2x2 = (5 – x)(2x-(2/5))
(c) (k + 1) x2 + (3/2) x = 7, where k = –1
(d) x3 – x2 = (x – 1)3

The standard form of a quadratic equation is given by,
ax2 + bx + c = 0, a ≠ 0
(A) Given, x2 + 2x + 1 = (4 – x)2 + 3
x2 + 2x + 1 = 16 – 8x + x2 + 3
10x – 18 = 0
which is not a quadratic equation.
(B) Given, -2x2 = (5 – x) (2x – 2/5)
-2x2 = 10x – 2x2 – 2 +2/5x
52x – 10 = 0
which is not a quadratic equation.
(C) Given, (k + 1) x2 + 3/2 x  = 7, where k = -1
(-1 + 1) x2 + 3/2 x = 7
3x – 14 = 0
which is not a quadratic equation.
(D) Given, x3 – x2 = (x – 1)3
x3 – x2 = x3 – 3x2 + 3x – 1
2x2 – 3x + 1 = 0

Q.2. Which of the following is not a quadratic equation?
(a) 2(x – 1)2 = 4x2 – 2x + 1
(b) 2x – x2 = x2 + 5
(c) (2x + 3)2 + x2 = 3x2 − 5x
(d) (x2 + 2x)2 = x4 + 3 + 4x3

A quadratic equation is represented by the form,
ax2 + bx + c = 0, a ≠ 0
(a) Given, 2(x – 1)2 = 4x2 – 2x + 1
2(x2 – 2x + 1) = 4x2 – 2x + 1
2x2 + 2x – 1 = 0
(b) Given, 2x – x2 = x2 + 5
2x2 – 2x + 5 = 0
(c) Given, (√2x + √3)2  = 3x2 – 5x
2x2 + 2√6x + 3  = 3x2 – 5x
x2 – (5 + 2√6)x – 3 = 0
(d) Given, (x2 + 2x)2 = x4 + 3 + 4x2
x4 + 4x3 + 4x2 = x4 + 3 + 4x2
4x3 – 3 = 0
which is a cubic equation and not a quadratic equation.

Q.3. Which of the following equations has 2 as a root?
(a) x2 – 4x + 5 = 0
(b) x2 + 3x – 12 = 0
(c) 2x2 – 7x + 6 = 0
(d) 3x2 – 6x – 2 = 0

If 2 is a root then substituting the value 2 in place of x should satisfy the equation.
(a) Given,
x2 – 4x + 5 = 0
(2)2 – 4(2) + 5 = 1 ≠ 0
So, x = 2 is not a root of x2 – 4x + 5 = 0
(b) Given, x2 + 3x – 12 = 0
(2)2 + 3(2) – 12 = -2 ≠ 0
So, x = 2 is not a root of x2 + 3x – 12 = 0
(c) Given, 2x2 – 7x + 6 = 0
2(2)2 – 7(2) + 6 = 0
Here, x = 2 is a root of 2x2 – 7x + 6 = 0
(d) Given, 3x2 – 6x – 2 = 0
3(2)2 – 6(2) – 2 = -2 ≠ 0
So, x = 2 is not a root of 3x2 – 6x – 2 = 0

Q.4. If ½ is a root of the equation x2 + kx – 5/4 = 0, then the value of k is
(a) 2
(b) – 2
(c) ¼
(d) ½

If ½ is a root of the equation
x2 + kx – 5/4 = 0 then, substituting the value of ½ in place of x should give us the value of k.
Given, x2 + kx – 5/4 = 0 where, x = ½
(½)2 + k (½) – (5/4) = 0
(k/2) = (5/4) – ¼
k = 1

Q.5. Which of the following equations has the sum of its roots as 3?
(a) 2x2 – 3x + 6 = 0
(b) –x2 + 3x – 3 = 0

(c) √2x– 3/√2x + 1 = 0
(d) 3x2 – 3x + 3 = 0

The sum of the roots of a quadratic equation ax2 + bx + c = 0, a ≠ 0 is given by,
Coefficient of x / coefficient of x2 = – (b/a)
(a) Given, 2x2 – 3x + 6 = 0
Sum of the roots = – b/a = -(-3/2) = 3/2
(b) Given, -x2 + 3x – 3 = 0
Sum of the roots = – b/a = -(3/-1) = 3
(c) Given, √2x– 3/√2x+1=0
2x2 – 3x + √2 = 0
Sum of the roots = – b/a = -(-3/2) = 3/2
(d) Given, 3x2 – 3x + 3 = 0
Sum of the roots = – b/a = -(-3/3) = 1

Q.6. Values of k for which the quadratic equation 2x2 – kx + k = 0 has equal roots is
(a) 0 only
(b) 4
(c) 8 only
(d) 0, 8

Given, the equation is 2x² - kx + k = 0
We have to find the values of k for which the equation has equal roots.
A quadratic equation ax² + bx + c = 0 has equal roots when the discriminant of the equation is zero.
Discriminant = b² - 4ac
Here, a = 2, b = -k and c = k
(-k)² - 4(2)(k) = 0
k² - 8k = 0
k(k - 8) = 0
k = 0
k - 8 = 0
k = 8
Therefore, the roots of the equation are 0 and 8.

Q.7. Which constant must be added and subtracted to solve the quadratic equation   by the method of completing the square?
(a) 1/8
(b) 1/64
(c) 1/4
(d) 9/64

Given, the quadratic equation is 9x² + (3/4)x - √2 = 0
We have to find the constant to be added and subtracted to solve the quadratic equation by the method of completing the square.
Let y = 3x
Now, (3x)² + (3x)/4 - √2 = 0
y² + (1/4)y - √2 = 0 --------------------- (1)
By using algebraic identity,
(a + b)² = a² + 2ab + b² --------------------- (2)
Comparing (1) and (2),
a² = 1
a = 1
2ab = 1/4
2(1)b = 1/4
b = 1/8
b² = (1/8)² = 1/64
So, y² + (1/4)y - √2 + 1/64 - 1/64 = 0
On rearranging,
y² + (1/4)y + 1/64 - √2 - 1/64 = 0
(y + 1/8)² = √2 + 1/64
Now, (3x + 1/8)² = √2 + 1/64
Therefore, the constant to be added and subtracted is 1/64.

Q.8. The quadratic equation 2x2 – √5x + 1 = 0 has
(a) two distinct real roots
(b) two equal real roots
(c) no real roots
(d) more than 2 real roots

Given, the quadratic equation is 2x² - √5x + 1 = 0

We have to find the nature of the roots of the equation.

Discriminant = b² - 4ac

Here, a = 2, b = -√5 and c = 1

b² = (-√5)² = 5

4ac = 4(2)(1) = 8

b² - 4ac = 5 - 8 = -3

b² - 4ac < 0

A quadratic equation ax² + bx + c = 0 has no real roots when the discriminant of the equation is less than zero.

Therefore, the equation has no real roots.

Q.9. Which of the following equations has two distinct real roots?
(a) 2x2 -3√2x + 9/4 = 0

(b) x2 + x – 5 = 0

(c) x2 + 3x + 2√2 = 0
(d) 5x2 – 3x + 1 = 0

We have to find the equation which has two distinct real roots.
A quadratic equation ax² + bx + c = 0 has 2 distinct real roots when the discriminant of the equation is greater than zero.
Discriminant = b² - 4ac
From the options,
(a) 2x² - 3√2x + 9/4 = 0
Here, a = 2, b = -3√2 and c = 9/4
b² - 4ac = (-3√2)² - 4(2)(9/4)
= 18 - 18
= 0
Therefore, the equation has equal roots.
(b) x² + x - 5 = 0
Here, a = 1, b = 1 and c = -5
b² - 4ac = (1)² - 4(1)(-5)
= 1 + 20
= 21 > 0
Therefore, the equation has 2 distinct real roots.
(c) x² + 3x + 2√2 = 0
Here, a = 1, b = 3 and c = 2√2
b² - 4ac = (3)² - 4(1)(2√2)
= 9 - 8√2 < 0
Therefore, the equation has no real roots.

(d) 5x² - 3x + 1 = 0
Here, a = 5, b = -3 and c = 1
b² - 4ac = (-3)² - 4(5)(1)
= 9 - 20
= -11 < 0
Therefore, the equation has no real roots.

Q.10. Which of the following equations has no real roots?

(a) x– 4x + 3√2 = 0
(b) x2 + 4x – 3√2 = 0

(c) x2 – 4x – 3√2 = 0
(d) 3x2 + 4√3x + 4 = 0

We have to find an equation that has no real roots.

A quadratic equation ax² + bx + c = 0 has no real roots when the discriminant of the equation is less than zero.
Discriminant = b² - 4ac
From the options,
(a) x² - 4x + 3√2 = 0
Here, a = 1, b = -4 and c = 3√2
b² - 4ac = (-4)² - 4(1)(3√2)
= 16 - 12√2
= 16 - 12(1.414)
= 16 - 16.968
= -0.968 < 0
Therefore, the equation has no real roots.
(b) x² + 4x - 3√2 = 0
Here, a = 1, b = 4 and c = -3√2
b² - 4ac = (4)² - 4(1)(-3√2)
= 16 + 12√2 > 0
Therefore, the equation has 2 distinct real roots.
(c) x² - 4x - 3√2 = 0
Here, a = 1, b = -4 and c = -3√2
b² - 4ac = (-4)² - 4(1)(-3√2)
= 16 + 12√2 > 0
Therefore, the equation has 2 distinct real roots.
(d) 3x² + 4√3x + 4 = 0
Here, a = 3, b = 4√3 and c = 4
b² - 4ac = (4√3)² - 4(3)(4)
= 48 - 48
= 0

Q.11. (x2 + 1)2 – x2 = 0 has
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real root.

Given, the equation is (x² + 1)² - x² = 0
We have to find the nature of the roots of the equation.
By using algebraic identity,
(a + b)² = a² + 2ab + b²
(x² + 1)² = x⁴ + 2x² + 1
So, x⁴ + 2x² + 1 - x² = 0
By grouping,
x⁴ + 2x² - x² + 1 = 0
x⁴ + x² + 1 = 0
Let y = x²
So, y² + y + 1 = 0
Here, a = 1, b = 1 and c = 1
Discriminant = b² - 4ac = (1)² - 4(1)(1)
= 1 - 4

= -3 < 0
A quadratic equation ax² + bx + c = 0 has no real roots when the discriminant of the equation is less than zero.
Therefore, the equation has no real roots.
Hence, the correct answer is option [c].

### Exercise 4.2

(i) x2 – 3x + 4 = 0

(ii) 2x2 + x – 1 = 0

(iii) 2x2 – 6x + 9/2 = 0

(iv) 3x2 – 4x + 1 = 0

(v) (x + 4)2 – 8x = 0

(vi) (x – 2)2 – 2(x + 1) = 0

(vii) 2 x2 –(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

(i) The equation x2 – 3x + 4 = 0 has no real roots.
D = b2 – 4ac
= (-3)2 – 4(1)(4)
= 9 – 16 < 0
Hence, the roots are imaginary.
(ii) The equation 2x2 + x – 1 = 0 has two real and distinct roots.
D = b2 – 4ac
= 12 – 4(2) (-1)
= 1 + 8 > 0
Hence, the roots are real and distinct.
(iii) 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.
(iv) 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.
(v) The equation (x + 4)2 – 8x = 0 has no real roots.
Simplifying the above equation,
x2 + 8x + 16 – 8x = 0
x2 + 16 = 0
D = b2 – 4ac
= (0) – 4(1) (16) < 0
Hence, the roots are imaginary.
(vi) The equation (x – √2)– √2(x+1)=0 has two distinct and real roots.
Simplifying the above equation,
x2 – 2√2x + 2 – √2x – √2 = 0
x2 – √2(2+1)x + (2 – √2) = 0
x2 – 3√2x + (2 – √2) = 0
D = b2 – 4ac
= (– 3√2)2 – 4(1)(2 – √2)
= 18 – 8 + 4√2 > 0
Hence, the roots are real and distinct.
(vii) 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.
(viii) The equation x (1 – x) – 2 = 0 has no real roots.
Simplifying the above equation,
x2 – x + 2 = 0
D = b2 – 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,
x2 – x + 2x – 2 + 2 = 0
x2 + x = 0
D = b2 – 4ac
= 12 – 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,
x2 + x – 2x – 2 + x = 0
x2 – 2 = 0
D = b2 – 4ac
= (0)2 – 4(1) (-2)
= 0 + 8 > 0
Hence, the roots are real and distinct.

Q.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 of a quadratic equation  have the same sign , then the quadratic equation has no real roots.

(i) False. For example, a quadratic equation x2 – 9 = 0 has two distinct roots – 3 and 3.
(ii) False. For example, equation x2 + 4 = 0 has no real root.
(iii) False. For example, a quadratic equation x2 – 4+ 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 ax2+ bx + c = 0, as a and c have opposite sign, ac < 0
⟹ Discriminant = b2 – 4ac > 0.
(vi) True, because in this case discriminant is always negative.
For example, in ax2+ bx + c = 0, as b = 0, and a and c have same sign then ac > 0
⟹ Discriminant = b2 – 4ac = – 4 a c < 0

No, a quadratic equation with integral coefficients may or may not have integral roots.
Justification
Consider the following equation,
8x2 – 2x – 1 = 0
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.

Q.4. Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

Yes, for this Discriminant should not be a perfect square.
Consider the quadratic equation 2x2 +x−4=0, which has rational coefficients.
The roots of the given quadratic equation are  which are irrational.

Q.5. Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals? Why?

Yes, consider the quadratic equation with all distinct irrationals coefficients i.e. √ 3x2−7√3x +12√3 = 0. The roots of this quadratic equation are 3 and 4, which are rationals.

Q.6. Is 0.2 a root of the equation x2 – 0.4 = 0? Justify.

Given, the quadratic equation is x² - 0.4 = 0
We have to determine if 0.2 is a root of the equation.
Put x = 0.2 in the equation,
x² - 0.4 = (0.2)² - 0.4
= 0.04 - 0.4
= -0.36
Therefore, 0.2 is not a root of the equation.

Q.7. If b = 0, c < 0, is it true that the roots of x2 + bx + c = 0 are numerically equal and opposite in sign? Justify.

Given, the equation is x² + bx + c = 0.
b = 0 and c < 0
We have to determine if the roots of the equation are numerically equal and opposite in sign.
x = [-b ± √b² - 4ac]/2a
Here, a = 1, b = 0
Since c < 0 take c as -c.
x = [-0 ± √0 - 4(1)(-c)]/2(1)
x = ± √4c/2
x = ± c
Therefore, the roots are c and -c.

### Exercise 4.3

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

(i) 2 x2 – 3x – 5 = 0

(ii) 5x2 + 13x + 8 = 0

(iii) –3x2 + 5x + 12 = 0

(iv) –x2 + 7x – 10 = 0

(v) x2 + 22x – 6 = 0

(vi) x2 – 35x + 10 = 0

(vii) (½)x2– √11x + 1 = 0

ax2 + bx + c = 0, a ≠ 0 is given by,

(i) 2 x2 – 3x – 5 = 0

(ii) 5x2 + 13x + 8 = 0

(iv) –x2 + 7x – 10 = 0

(v) x2 + 2 √2x – 6 = 0

(vi) x2 – 3 √5x + 10 = 0

(vii) (½)x2– √11x + 1 = 0

Q.2. Find the roots of the following quadratic equations by the factorisation method:
(i) 2x2 + 5/3x - 2 = 0
(ii) 2/5x2 - x - 3/5 = 0
(iii) 3√2 x2 – 5x – √2  = 0
(iv) 3x2 + 5√5x - 10 = 0
(v) 21x2
-2x + 1/21 = 0

(i) Given,
2x2 + 5/3x - 2 = 0
⇒6x2 +5x−6=0
6x2 −4x+9x−6=0
2x(3x−2)+3(3x−2)=0
(3x−2)(2x+3)=0

(ii) Given, the quadratic equation is (2/5)x² - x - 3/5 = 0.
We have to find the roots of the equation by factorisation method.
The equation can be written as
2x² - 5x - 3 = 0
On factoring,
2x² - 6x + x - 3 = 0
2x(x - 3) + 1(x - 3) = 0
(2x + 1)(x - 3) = 0
Now, 2x + 1 = 0
2x = -1
x = -1/2
Also, x - 3 = 0
x = 3
Therefore, the roots of the equation are 3 and -1/2.

(iii) Given, the quadratic equation is 3√2x² - 5x - √2 = 0.
We have to find the roots of the equation by factorisation method.
On factoring,
3√2x² - 5x - √2 = 0
3√2x² - 6x + x - √2 = 0
3√2x(x - √2) + 1(x - √2) = 0
(3√2x + 1)(x - √2) = 0
Now, 3√2x + 1 = 0
3√2x = -1
x = -1/3√2
Also, x - √2 = 0
x = √2
Therefore, the roots of the equation are √2 and -1/3√2.

(iv) Given, the quadratic equation is 3x² + 5√5x - 10 = 0.
We have to find the roots of the equation by factorisation method.
On factoring,
3x² + 6√5x - √ 5x - 10 = 0
3x(x + 2√5) - √5(x + 2√5) = 0
(3x - √5)(x + 2√5) = 0
Now, 3x - √5 = 0
3x = √5
x = √5/3
Also, x + 2√5 = 0
x = -2√5
Therefore, the roots of the equation are -2√5 and √5/3.

(v) Given, the quadratic equation is 21x² - 2x + 1/21 = 0.

We have to find the roots of the equation by factorisation method.
The equation can be written as
441x² - 42x + 1 = 0
On factoring,
441x² - 21x - 21x + 1 = 0
21x(21x - 1) - (21x - 1) = 0
(21x - 1)(21x - 1) = 0
Now, 21x - 1 = 0
21x = 1
x = 1/21
Therefore, the roots of the equation are 1/21 and 1/21.

The document NCERT Exemplar: Quadratic Equations - 1 | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on NCERT Exemplar: Quadratic Equations - 1 - Mathematics (Maths) Class 10

 1. How to solve quadratic equations by factorization?
Ans. To solve a quadratic equation by factorization, follow these steps: 1. Write the equation in the form ax^2 + bx + c = 0. 2. Find two numbers that multiply to give ac and add to give b. 3. Rewrite the middle term using the two numbers found in step 2. 4. Factorize the quadratic expression. 5. Set each factor to zero and solve for the variable to find the solutions.
 2. What is the quadratic formula and how is it used to solve quadratic equations?
Ans. The quadratic formula is given by x = (-b ± √(b^2 - 4ac)) / 2a. To use the quadratic formula to solve a quadratic equation in the form ax^2 + bx + c = 0, simply substitute the values of a, b, and c into the formula and solve for x. The formula provides the roots of the quadratic equation.
 3. How many solutions can a quadratic equation have?
Ans. A quadratic equation can have either two real solutions, one real solution (in case the discriminant is zero), or two complex solutions (in case the discriminant is negative). The number of solutions is determined by the value of the discriminant b^2 - 4ac.
 4. Can a quadratic equation have no solutions?
Ans. Yes, a quadratic equation can have no real solutions if the discriminant b^2 - 4ac is negative, resulting in two complex solutions. In this case, the quadratic equation has no real roots.
 5. How can the graph of a quadratic equation help in solving it?
Ans. The graph of a quadratic equation can help in solving it by providing a visual representation of the equation. The x-intercepts of the graph represent the solutions of the quadratic equation. By analyzing the graph, one can determine the number of solutions and approximate their values. Additionally, the vertex of the parabola can provide information about the maximum or minimum value of the quadratic equation.

## Mathematics (Maths) Class 10

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