Quadratic Equations Class 10 MCQs RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-5. We Provide Step by Step Answer of Exe-MCQs on Quadratic Equations as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-9.
| Board | ICSE |
| Subject | Maths |
| Class | 10th |
| Chapter-5 | Quadratic Equations |
| Writer/Book | RS Aggarwal |
| Topics | Solution of MCQs |
| Academic Session | 2024-2025 |
How to Solve MCQs on Quadratic Equations
Page- 62,63
Multiple Choice Questions :
Que-1: The degree of a quadratic equation is :
(a) 1 (b) 2 (c) 3 (d) none of these
Solution- (b) 2
Reason: If any equation having the highest power equal to 2 then that equation is known as the quadratic equation. So the degree of the quadratic equation is 2.
Que-2: The root of the equation px²+qx+r=0, where p ≠ 0, are given by :
(a) x = [-p±√(q²-4pr)]/2p (b) x = [-q±√(q²-2pr)]/4p (c) x = [-q±√(q²-4pr)]/2p (d) x = [-q±√(q²-4pr)]/2q
Solution- (c) x = [-q±√(q²-4pr)]/2p
Reason: px²+qx+r = 0
Compare it with :
ax²+bx+c = 0
x = [-b±√(b²-4ac)]/2a
So, px²+qx+r = 0 is
x = [-q±√(q²-4pr)]/2p.
Que-3: The discriminant of the quadratic equation ax²+bx+c=0, a ≠ 0 is given by :
(a) b²-2ac (b) b²-ac (c) b²-4ac (d) none of these
Solution- (c) b²-4ac
Reason: D = b²-4ac
Que-4: For real roots of the quadratic equations, the discriminant must be :
(a) grater than or equal to zero (b) greater than zero (c) less than or equal to zero (d) less than zero
Solution- (a) grater than or equal to zero
Reason: D > 0 means two real, distinct roots.
Que-5: If the roots of the quadratic equation, ax²+bx+c=0, a ≠ 0 are real and equal, then each root is equal to :
(a) -a/2b (b) -b/2a (c) -2a/b (d) -c/2a
Solution- (b) -b/2a
Reason: The roots of a quadratic function are given by x = [−b±√(b²−4ac)]/2a
D = b²−4ac is called the discriminant
As D = 0
x = −b/2a is the only root.
Que-6: If the discriminant of the quadratic equation, ax²+bx+c=0, a ≠ 0 is greater than zero and a perfect square and a,b,c are rational, then the roots are :
(a) rational and equal (b) irrational and unequal (c) irrational and equal (d) rational and unequal
Solution- (d) rational and unequal
Reason: D = 0
rational and unequal
Que-7: If the discriminant of the quadratic equation, ax²+bx+c=0, is greater than zero and a perfect square a and b are irrational, then the roots are :
(a) irrational and unequal (b) irrational and equal (c) rational and unequal (d) rational and equal
Solution- (a) irrational and unequal
Reason: D < 0 , irrational and unequal.
Que-8: Which of the following is quadratic equation ?
(a) x²-2√x+7=0 (b) 2x²-5x=(x-1)² (c) x-(1/x)=2x² (d) x²+(1/x²)=2
Solution- (b) 2x²-5x=(x-1)²
Reason: Simplify all of them to determine if it can be changed in the form of ax2 + bx + c = 0,
Que-9: Which of the following is quadratic equation ?
(a) x²+1=(2-x)²+3 (b) 2x²+3=(5+x)(2x-3) (c) x³-x²=(x-1)³ (d) none of these
Solution- (c) x³-x²=(x-1)³
Reason: x³ – x² = (x – 1)³
By using algebraic identity,
(a – b)³ = a³ – b³ – 3a²b + 3ab²
x³ – x² = x³ – 1 – 3x² + 3x
By grouping,
x³ – x³ – x² + 3x² – 3x + 1 = 0
2x² – 3x + 1 = 0
The degree of the equation is 2.
Therefore, x³ – x² = (x – 1)³ is a quadratic equation.
Que-10: Which of the following is not a quadratic equation ?
(a) 3x-x²=x²+5 (b) (x+2)²=2(x²-5) (c) (√2x+3)²=2x²+6 (d) (x-1)²=3x²+x-2
Solution- (c) (√2x+3)²=2x²+6
Reason: (√2x + √3)² + x² = 3x² – 5x
By using algebraic identity,
(a + b)² = a² + 2ab + b²
2x² + 3 + 2√6x + x² = 3x² – 5x
By grouping,
3x² – 3x² + 2√6x + 5x + 3 = 0
5x + 2√6x + 3 = 0
The degree of the equation is 1
Therefore, (√2x + √3)² + x² = 3x² – 5x is not a quadratic equation.
Que-11: The roots of the quadratic equation 2x²-x-6=0 are :
(a) -2,3/2 (b) 2,-3/2 (c) -2,-3/2 (d) 2,3/2
Solution- (b) 2,-3/2
Reason: 2x²-x−6 = 0
⇒ 2x²+3x−4x−6 = 0
⇒ x(2x+3)-2(2x+3) = 0
⇒ (x-2)(2x+3) = 0
⇒ x-2 = 0 or 2x+3 = 0
⇒ x = 2 or x = -3/2
∴ The roots of the given equation are 2, -3/2.
Que-12: Which of the following quadratic equations has 2 and 3 as its roots ?
(a) x²-5x+6=0 (b) x²+5x+6=0 (c) x²-5x-6=0 (d) x²+5x-6=0
Solution- (a) x²-5x+6=0
Reason: x²-5x+6 = 0
x²-2x-3x+6 = 0
x(x-2)-3(x-2) = 0
(x-2)(x-3) = 0
x = 2, 3.
Que-13: Which of the following is a root of the quadratic equation, 3x²+ 13x + 14 = 0 ?
(a) -1/3 (b) -3/2 (c) -5/3 (d) -7/3
Solution- (d) -7/3
Reason: 3x²+13x+14 = 0
3(-7/3)²+13(-7/3)+14 = 0
3(49/9)+13(-7/3)+14 = 0
(49/3)-(91/3)+14 = 0
(49-91+42)/3 = 0
91-91 = 0
0 = 0
So, -7/3 is a root of 3x²+13x+14 = 0
Que-14: If x = -1/2 is the solution of the quadratic equation3x²+2kx-3=0, then the value of k is :
(a) -3/4 (b) -5/4 (c) -9/4 (d) -4/5
Solution- (c) -9/4
Reason: 3x²+2kx-3 = 0
x = -1/2
3(-1/2)²+2(-1/2)k-3 = 0
3(1/4)-k = 3
-k = 3-(3/4)
-k = (12-3)/4
-k = 9/4
k = -9/4.
Que-15: If ax²+bx+c=0 has equal roots, then c = ?
(a) b/2a (b) b²/4a (c) -b/2a (d) -b²/4a
Solution- (b) b²/4a
Reason: For the equation to have equal roots, the discriminant must be equal to zero.
D = b²−4ac = 0
⇒ b² = 4ac
⇒ c = b²/4a
Que-16: If the equation, x²-ax+1=0 has two distinct and real roots, then :
(a) |a| ≥ 2 (b) |a| ≤ 2 (c) |a| > 2 (d) |a| < 2
Solution- (c) |a| > 2
Reason: x²-ax+1=0
d = b2 – 4ac > 0
Here a = 1, b = a, c = 1
So,
a2 – 4(1) (1) >0
a2 – 4 > 0
a2 >4
|a| > 2.
Que-17: The positive value of k for which the equations x²+kx+64=0 and x²- 8x + k = 0 will both have real roots, is :
(a) 16 (b) 8 (c) 12 (d) 4
Solution- (a) 16
Reason: x² + kx + 64 = 0
Here a = 1, b = k and c = 64
To have equal roots
b² – 4ac = 0
Therefore, (k)² – 4 (1) 64 = 0
k² – 256 = 0
k² = 256
k = 16
k = 16 is the positive value of k.
Que-18: If 3 is a root of quadratic equation x²-px+3=0, then p is not equal to :
(a) 4 (b) 3 (c) 5 (d) 2
Solution- (a) 3
Reason: x²-px+3 = 0
x = 3
(3)²-3p+3= 0
9-3p+3 = 0
12 = 3p
p = 4
So, 3 is not equal to p.
Que-19: If the roots of the quadratic equation 2x²+8x+k=0 are equal, then the value of k is
(a) 2 (b) 8 (c) 4 (d) none of these
Solution- (b) 8
Reason: 2x²+8x+k = 0
D = 0
D = b²-4ac
0 = (8)² – 4x2xk
0 = 64 – 8k
8k = 64
k = 64/8
k = 8.
Que-20: The solution set for the quadratic equation 2x²-x+(1/8)=0 is :
(a) {1/4,1/4} (b) {-1/4,1/4} (c) {-1/2,1/4} (d) {4,4}
Solution- (a) {1/4,1/4}
Reason: 2x²-x+(1/8) = 0
The given equation is 2x²−x+18. This can be written as,
1/8(16x²−8x+1) = 0
1/8(16x²−4x−4x+1) = 0
1/8(4x(4x−1)−1(4x−1)) = 0
1/8(4x−1)² = 0
(4x−1)² = 0
4x−1 = 0
x = 1/4
So, both the roots of the given equation is {1/4, 1/4}.
Que-21: The solution set for the quadratic equation 2x²=288, is :
(a) {12,12} (b) {-12,-12} (c) {-12,18} (d) {-12,12}
Solution- (d) {-12,12}
Reason: 2x² = 288
x² = 288/2
x² = 144
x = √144
x = 12
Solution set is {-12,12}
Que-22: The solution set for the quadratic equation 3x²-18x=0, is :
(a) {6,6} (b) {0,-6} (c) {0,6} (d) none of these
Solution- (c) {0,6}
Reason: 3x²-18x = 0
3x(x-6) = 0
3x = 0 or x-6 = 0
x = 0 or x = 6
x = {0,6}
Que-23: The values of k for which the quadratic equation, 9x²-3kx+k=0, has equal roots, are :
(a) 0,1 (b) 0,2 (c) 2,4 (d) 0,4
Solution- (d) 0,4
Reason: We know that D = 0 then the equation has equal and real roots
D = b²−4ac = 0
a = 9, b = −3k, c = k
Substitute the values we get,
(−3k)² − (4×9×k) = 0
⇒ 9k²−36k = 0
⇒ 9k² = 36k
⇒ 9k(k−4) = 0
∴k = 0 or k = 4
Que-24: If the roots of the quadratic equation, px(x-2)+6=0 are equal, then the value of p is :
(a) 0 (b) 4 (c) 6 (d) none of these
Solution- (c) 6
Reason: px(x-2)+6 = 0
px²-2px+6 = 0
D = 0
D = b²-4ac
(-2p)² – (4xpx6) = 0
4p²-24p = 0
4p(p-6) = 0
4p = 0 and (p-6) = 0
p = 0 and p = 6
∴ p ≠ 0
p = 6.
Que-25: If the quadratic equation, px²-2√5px+15=0 has two equal roots, then the value of p is :
(a) 0 (b) 3 (c) 6 (d) both 0 and 3
Solution- (b) 3
Reason: Given quadratic equation is, px² -2√5 px + 15 = 0
Compare px² -2√5 px + 15 = 0 with ax² + bx + c = 0
a = p , b = -2√5p , c = 15
We know that , If the roots of the quadratic equation are equal , then it’s discriminant (D) equals to zero.
discriminant = 0
b² – 4ac = 0
( -2√5 p )² – 4× p × 15 = 0
20p² – 60p = 0
20p ( p – 3 ) = 0
p – 3 = 0
p = 3.
Que-26: If 1 is a root of the quadratic equation, ky²+ky+3=0, then the value of k is :
(a) -2/3 (b) -1/3 (c) -1/2 (d) -3/2
Solution- (d) -3/2
Reason: ky²+ky+3 = 0
y = 1
k(1)²+k(1)+3 = 0
k+k+3 = 0
2k = -3
k = -3/2.
Que-27: If the equation x²+5kx+16=0 has no real roots, then :
(a) k > 8/5 (b) k < -8/5 (c) -8/5 < k < 8/5 (d) none of these
Solution- (c) -8/5 < k < 8/5
Reason: x²+5kx+16 = 0
D = b²-4ac < 0
(5k)² – (4x1x16) < 0
25k²-64 < 0
25k² < 64
k² < 64/25
k < √(64/25)
k < 8/5
-8/5 < k < 8/5.
Que-28: The discriminant of the equation, 3x²-2x+(1/3)=0 is :
(a) 0 (b) 1 (c) 2 (d) 4
Solution- (a) 0
Reason: 3x²-2x+(1/3) = 0
D = b²-4ac
D = (2)² – (4x3x1/3)
D = 4-4
D = 0.
Que-29: What is the nature of the roots of the equation, 2x²-6x+3=0 ?
(a) rational and unequal (b) irrational and unequal (c) real and equal (d) imaginary and unequal
Solution- (b) irrational and unequal
Reason: 2x2 – 6x + 3 = 0
a = 2, b = – 6, c = 3
b2 – 4ac = (- 6)2 – 4(2)(3)
= 36 – 24
= 12
b2 – 4ac > 0
So, irrational and unequal.
Que-30: The nature of the roots of the equation, 3x²-4√3x+4=0 is :
(a) real and equal (b) irrational and unequal (c) rational and unequal (d) imaginary and unequal
Solution- (a) real and equal
Reason: 3x2 – 4√3 x + 4 = 0
a = 3, b = – 4√3, c = 4
b2 – 4ac = (- 4√3)2 – 4(3)(4)
= 16 × 3 – 4 × 4 × 3
= 48 – 48
= 0
b2 – 4ac = 0
Hence the equation has two equal real roots.
Que-31: The value of the discriminant of the equation 2x²-3x+5=0, is :
(a) 31 (b) √-31 (c) √31 (d) -31
Solution- (d) -31
Reason: 2x²-3x+5 = 0
D = b²-4ac
D = (-3)²-(4x2x5)
D = 9-40
D = -31.
Que-32: If the equation, ax²+2x+a=0 has two real and equal roots, then :
(a) a = 0,1 (b) a = 1,1 (c) a = 0,-1 (d) a = -1,1
Solution- (d) a = -1,1
Reason: The given quadratic equation is ax2 + 2x + a = 0 , and roots are distinct.
Discriminant (D) = b2 – 4ac
Here, a = a, b = 2, and c = a
Putting the value of a = a, b = 2, and c = a.
= (2)2 – 4 × a × a
= 4 – 4a2
The given equation will have real and distinct roots, if D > 0,
∴ 4 – 4a2 > 0
∴ 4(1 – a)2 > 0
∴ (1 – a)2 > 0
∴ 1 – a2 > 0
∴ 1 > a2
∴ a = ±1
a = {-1,1}
Que-33: The value of the discriminant of the equation, √3x²+10x+7√3=0 is :
(a) 4 (b) 16 (c) -16 (d) -12
Solution- (b) 16
Reason: D = b²-4ac
D = (10)²-4x√3×7√3
D = 100 – 84
D = 16.
Que-34: The value of the discriminant of the equation, x²-(√2+1)x+√2=0 is :
(a) 3+2√2 (b) 1-2√2 (c) 3-2√2 (d) 2-√2
Solution- (c) 3-2√2
Reason: x²-(√2+1)x+√2 = 0
D = b²-4ac
D = (√2+1)²-4x1x√2
D = (√2²+1²+2x√2×1) – 4√2
D = (2+1+2√2) – 4√2
D = 3+2√-4√2
D = 3-2√2.
Que-35: The roots of the equation, x+(1/x) = 3 are :
(a) (2±√5)/2 (b) (3±√5)/2 (c) (1±√3)/2 (d) none of these
Solution- (b) (3±√5)/2
Reason: x+(1/x) = 3
x²+1 = 3x
x²−3x+1 = 0
Now, by using quadratic formula,
x = [−b±√(b²−4ac)]/2a
Here a = 1, b = −3, c = 1
x = [3±√(−3)²−4×1×1]/2×1
x = [3±√(9−4)]/2
x = (3±√5)/2
Hence the roots are x = (3+√5)/2 or x = (3−√5)/2.
Que-36: The solution set for the quadratic equation 2x²+kx-k²=0 is :
(a) {k,k} (b) {-k,k} (c) {-k,k/2} (d) {-k/2,k}
Solution- (c) {-k,k/2}
Reason: 2x² + kx – k² = 0
2x² + 2kx – kx – k² = 0
2x(x+k) – k (x+k) = 0
(x+k) (2x-k) = 0
x = -k, k/2.
Que-37: The solution set for the equation, 25x(x+1)=-4, is :
(a) {1/5,4/5} (b) {-1/5,4/5} (c) {-4/5,-1/5} (d) {-4/5,1/5}
Solution- (c) {-4/5,-1/5}
Reason: 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
⇒ x = −1/5 or x = −4/5.
Que-38: The value of the discriminant of the equation, 3x²=-11x-10, is :
(a) 1 (b) 0 (c) 4 (d) 9
Solution- (a) 1
Reason: 3x² = -11x-10
3x²+11x+10 = 0
D = b²-4ac
D = (11)²-(4x10x3)
D = 121-120
D = 1.
Que-39: If -5 is a root of the quadratic equation 2x²+px-15=0 and the quadratic equation p(x²+x)+k=0 has equal roots, then the value of k is :
(a) 7/4 (b) 5/4 (c) 3/4 (d) 1/4
Solution- (a) 7/4
Reason: Given –5 is a root of the quadratic equation 2x2 + px – 15 = 0.
∴-5 satisfies the given equation.
∴ 2(5)2+ p(-5)-15 = 0
∴ 50 – 5p – 15= 0
∴ 35-5p = 0
∴5p = 35 ⇒ p = 7
Substituting p = 7 in p(x2 + x)+ k= 0,we get
7(x2+x)+k=0
∴7x2 + 7x + k = 0
The roots of the equation are equal.
∴ Discriminant b2 – 4ac = 0
Here, a=7, b=7, c=k
b2-4ac=0
∴ (7)2 – 4(7)(k)=0
∴ 49-28k= 0
∴28k = 49
k = 49/28
k = 7/4.
–: End of Quadratic Equations Class 10 MCQs RS Aggarwal Goyal Brothers ICSE Maths Solutions :–
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