Quadratic Equations MCQ

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Any equation that can be rearranged in standard form, or in other words, the equation of the second degree, is referred to as a quadratic equation in algebra. It will, therefore, have at least one squared phrase. In the equation ax2 + bx + c = 0, for example, x is an unknown number, whereas a, b, and c are known numbers or numerical coefficients, with a 0. When a quadratic polynomial is equated with zero, it is transformed into a quadratic equation. The values of the variables that satisfy a given quadratic equation are known as the roots.

The video below explains this:

Quadratic Equations Detailed Video Explanation:

Ques 1. In the quadratic equation 5x2 – 4x + 5 = 0, which one of them has the following roots?

Real and Unequal
Real and Equal
Non-real and Equal
Not real

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Ans. (d) Not real

Explanation: Calculate b2 – 4ac in order to find out the nature.

So, b2 – 4ac will be

= 42 – 4 x 5 x 5

= 16 – 100

= -84 < 0

Ques 2. When a natural number is multiplied by 12, it becomes 160 times its reciprocal. Calculate the number.

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Ans. (a) 8

Explanation: Assume the number as x

Then according to the given question,

x + 12 = 160/x

⇒ x2 + 12x – 160 = 0

⇒ x2 + 20x – 8x – 160 = 0

⇒ (x + 20) (x – 8) = 0

⇒ x = -20, 8

We only consider positive values because the number is natural.

Ques 3. 300 is the result of two consecutive integral multiples of 5. Find out the numbers.

10,15
15, 20
30, 35
25, 30

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Ans. (b) 15, 20

Explanation: Let 5n and 5(n + 1) be the consecutive integral multiples, with n being a positive integer.

In answer to the question:

5n × 5(n + 1) = 300

⇒ n2 + n – 12 = 0

⇒ (n – 3) (n + 4) = 0

⇒ n = 3 and n = – 4

n = – 4 will be eliminated because n is a positive natural number.

As a result, the numbers 15 and 20 are used.

Ques 4. Rohini could have gotten 10 additional marks out of a possible 30 on her math test, which would have been the square of her actual score 9 times. How many marks did she receive on the exam?

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Ans. (c) 15

Explanation: Assume her actual marks will be x

As 9 (x + 10) = x2

⇒x2 – 9x – 90 = 0

⇒x2 – 15x + 6x – 90 = 0

⇒x(x – 15) + 6 (x – 15) = 0

⇒(x + 6) (x – 15) = 0

So, x = – 6 or x =15

As x is assumed to be as the marks obtained, x ≠ – 6. Hence, x will be = 15.

Ques 5. A right triangle's altitude is 7 cm less than its base. The other two sides of the triangle are equal to: If the hypotenuse is 13 cm, the other two sides are equal to:

Base=12cm and Altitude=5cm
Base=12cm and Altitude=10cm
Base=10cm and Altitude=5cm
Base=14cm and Altitude=10cm??

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Ans. (a) Base=12cm and Altitude=5cm

Explanation: Let base will be as x cm.

So, Altitude = (x – 7) cm

As we know that in a right triangle,

Base2 + Altitude2 = Hypotenuse2 (Pythagoras theorem)

∴ x2 + (x – 7)2 = 132

On solving the above equation, we will get-

⇒ x = 12 or x = – 5

As the side of the triangle cannot be negative.

So, base = 12cm and altitude = 12 - 7 = 5cm

Also read: Nature of Roots of Quadratic Equation

Ques 6. If one of the roots of the equation 4x2-2x+k-4=0 is reciprocal to the other, then k will have the value as:

-4
-8
4
8

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Ans. (d) 8

Explanation: α x 1/α = (k-4)/4

k-4 = 4

k = 8

Ques 7. For a quadratic equation, the maximum number of roots is equal to

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Ans. (c) 2

Explanation: Because the degree of a quadratic equation is 2, the maximum number of roots for a quadratic equation is equal to 2.

Ques 8. In the following quadratic equation, 2x2 – √5x + 1 = 0 has

No more than two real roots
No real roots
Two distinct real roots
Two equal real roots

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Ans. (b) No real roots

Explanation: Given equation 2x2 – √5x + 1 = 0

When compared to a quadratic equation in its usual form,

a = 2, b = -√5, c = 1

Now, b2 – 4ac = (-√5)2 – 4(2)(1)

= 5 – 8

= -3 < 0

Hence, the given equation will be having no real roots.

Ques 9. The equation with the sum of its roots equal to 3 will be

√2x2 – 3/√2x + 1 = 0
3x2 – 3x + 3 = 0
–x2 + 3x – 3 = 0
2x2 – 3x + 6 = 0

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Ans. (c) –x2 + 3x – 3 = 0

Explanation: The sum of the roots of the quadratic equation ax2 + bx + c = 0, a 0 is,

x coefficient / x2 coefficient = –(b/a)

Let's have a look at the options.

√2x2 – 3/√2x + 1=0

2x2 – 3x + √2 = 0

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

3x2 – 3x + 3 = 0

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

-x2 + 3x – 3 = 0

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

2x2 – 3x + 6 = 0

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

Ques 10. A train travels 360 kilometres at a consistent speed. It would have taken 1 hour less to travel the same distance if the pace had been increased by 5 km/h. Determine the train's speed.

30 km/hr
40 km/hr
50 km/hr
60 km/hr

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Answer: (b) 40 km/hr

Explanation: Consider x km/hr to be as the speed of train.

Time required to cover 360 km = 360/x hr.

According to the given question,

⇒ (x + 5)(360-1/x) = 360

⇒ 360 – x + 1800-5/x = 360

⇒ x2 + 5x + 10x – 1800 = 0

⇒ x(x + 45) -40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45

The negative value is not considered for calculating speed so the answer will be 40km/hr.

Read More:

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Quadratic Equations MCQ

Quadratic Equations Detailed Video Explanation:

CBSE X Related Questions

1.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not

3.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

5.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Similar Mathematics Concepts

Comments

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Quadratic Equations MCQ

Quadratic Equations Detailed Video Explanation:

CBSE X Related Questions

1. If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos2 A – sin2 A or not

3. Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

5. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not

3.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

5.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.