### Class 10 Maths Objective Questions Chapter 4 Quadratic Equations

CBSE Class 10 Maths Chapter 4 – Quadratic Equations Objective Questions

There are many different ways to find out the unknown values of x. This chapter 4-Quadratic Equations of class 10 Maths deals with explanations of how to find the value of from a given equation when x is of unknown value and a, b, c are of known value. The theory found in the concept discussed is that an equation to be quadratic “a” should not be equal to 0. However, the equation is of the form ax2 + bx + c = 0. Also, the values of a, b, and c are always real numbers. Now, as per the latest exam pattern, more MCQs are expected to appear in the exam question paper. So, we have compiled some 20 objective questions from this chapter in this CBSE Class 10 Maths Chapter 4-Quadratic Equations Objective Questions, which the students can solve for practice.

These MCQs for CBSE Class 10 Maths Chapter 4- Quadratic Equations are prepared from the examination perspective. We have also given the list of topics, which are covered in these CBSE Class 10 Maths objective type of questions .

4.1 Introduction to Quadratic Equations (5 MCQs Listed From This Topic)

4.2 Solving QE by Factorisation (5 MCQs From The Topic)

4.3 Solving QE by completing square (5 MCQs Listed From The Topic)

4.4 Solving QE using quadratic formula (5 MCQs From This Topic)

Solution: The standard form of quadratic equation is ax 2 + bx + c = 0, a ≠ 0. So the degree of a quadratic equation is 2

Solution: For general quadratic equation ax 2 +bx+c=0.

Sum of the roots of the equation is 8

Answer: (B) x(x + 1) + 8 = (x + 2) (x – 2)

Solution: (a) (x−2) 2 + 1 = (2x – 3)

Solution: For a quadratic equation ax2+bx+c=0,

Thus, quadratic equation is x 2 −5x+5=0

Solution: Area of rectangle = length×breadth

Area of the rectangular field = x (2x+1) =3

We need to split the coefficient of x such that the sum

of the factors is 5 and their product is -14.

So we will find the coefficient as 7 and -2.

The sum of 7 and -2 is 5 and product is -14.

Now equate the factors to zero to find the roots.

So the roots of the equation are 2,-7

Solution: Comparing x 2 +5x+6 =0 to ax 2 + bx+ c = 0, we have a=1, b=5 and c=6

Now, we need to find two numbers whose product is 6 and whose sum is 5

Pairs of numbers whose product is 6

Of these pairs, the pair that gives the sum 5 is the third pair

Identifying the pair, we rewrite the given quadratic equation as

x 2 +5x+6= x 2 +2x+3x+6 = x(x+2) +3(x+2)

Given that the altitude of a right triangle is 7 cm less than its base

base 2 + altitude 2 = hypotenuse 2

Dividing with 2 on both sides the above equation simplifies to

Length cannot be negative so x cannot be equal to – 5

base x = 12cm; altitude = 12 – 7 = 5cm

Solution: Let the speed of the train be x km/hr.

Time taken to travel 210 km = 210/x hours

When the speed is increased by 5 km/h, the new speed is (x+5)

Time taken to travel 210 km with the new speed is 210 / (x+5) hours

The speed cannot be negative. Thus, the speed of the train is 30 km/hr

Solution: The roots of a quadratic equation do not change when it is multiplied by a constant non-zero real number. So when the equation x 2 +3x−18=0 is multiplied by 2, the roots still remain the same i.e. -6, 3.

Solution: Converting statement into an equation-

⇒ (5x+1) 2 =16 (Applying (a+b) 2 formula)

⇒5x + 1 = ± 4(Taking square root on both sides)

Dividing by the coefficient of x 2 , we get

Adding and subtracting the square of b/2=7/4, (half of coefficient of x)

We get, [x 2 −2 (7/4) x+ (7/4) 2 ] − (7/4) 2 +3/2=0

The equation after completing the square is:

Taking square root, (x−7/4) = (±5/4)

Solution: Multiplying the equation throughout by 5, we get 25x 2 –30x–10=0

(5x) 2 –[2× (5x) ×3] +3 2 –3 2 –10=0

Solution: If a natural number is x, the next natural number is greater than x by 1 and hence x+1. For eg. For 3, next natural number is 4. The product of the 2 numbers is x(x+1) =x 2 + x

Solution: The identity (a+b) 2 = (a 2 +2ab+b 2 ) represents a perfect square.

If we observe carefully we can see that x 2 +6x can be written in the form of (a 2 +2ab+b 2 ) by adding a constant.

To make x 2 +6x a perfect square, divide the co efficient of x by 2 and then add the square of the result to make this a perfect square.

We should add 9 to make x 2 +6x a perfect square.

Solution: Step 1:- For, x 2 +4x+c=0, value of discriminant D=4 2 –4c=16−4c

Step 2:- The roots of quadratic equation are real only when D ≥ 0

Answer: (D) 1, two distinct real roots

Solution: D = b 2 –4ac= (–5) 2 –4×3×2=1>0

Solution: Let x and y be the length and breadth of the rectangle respectively.

Area of the rectangle=length × breadth

The roots of the above quadratic equation will be

Considering positive value for breadth, we have y=14.20.

x=44.6 and y=14.20 (approximately).

Solution: Step 1:- For, x 2 +2(k+2) x+9k=0, value of discriminant D= [2(k+2)] 2 –4(9k) =4(K 2 +4−5k)

Step 2:- The roots of quadratic equation are real and equal only when D=0

Solution: Quadratic equation of the form ax 2 + bx + c = 0

The roots of the above quadratic equation will be

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