Quadratic Equations- 1 - Free MCQ Practice Test with solutions, SSC CGL

• Given, x² + 9 = 6x
• ⇒ x² - 6x + 9 = 0
• ⇒ x² -3x -3x + 9 = 0
• ⇒ x ( x - 3) - 3 ( x - 3) = 0 
• Taking (x-3) common from the equation , we get
• ⇒ (x - 3) (x - 3) = 0
• ⇒ x = 3

• Let 3 consecutive terms A.P is a –d, a , a + d. and the sum is 51
• so, (a –d) + a + (a + d) = 51
• ⇒ 3a –d + d = 51
• ⇒ 3a = 51
• ⇒ a = 17
• The product of first and third terms is 273
• So  it stand for ( a –d) (a + d) = 273
• ⇒ a² –d² = 273
• ⇒ 172 –d ² = 273
• ⇒ 289 –d ² = 273
• ⇒  d ² = 289 –273
• ⇒ d ² = 16
• ⇒ d = 4
• Hence the 3rd terms ( a+d )= 17 + 4 =  21

• Using the quadratic formula:
• (3 ± √25) /4
• The two roots are:
• 8 /4 = 2
• -2 /4 = -¹⁄₂
• Since only x = 2 is an integer, the number of integral solutions is: 1

• Given the equation 3x² + tx + s = 0 with roots 2 and 3:
• Sum of roots: 2 + 3 = 5 ⇒ -t /3 = 5 ⇒ t = -15
• Product of roots: 2 × 3 = 6 ⇒ s /3 = 6 ⇒ s = 18
• s + t = 18 + (-15) = 3
• Answer: 3

• Step 1: Question statement and Inferences
• Given equation is x³−7x²+12x=0
• and p, q, r are the roots of this equation.
• We need to find the value of the expression  
• . For this we need to find the values of p, q and r by solving the given equation.
• Step 2 & 3: Finding required values and calculating the final answer
• x³−7x²+12x=0
• ⇒x(x²−7x+12)=0 ....(I)
• Note that x²−7x+12
• is a standard form quadratic equation. It can be factorized as follows:
• x²−4x−3x+12=(x−3)(x−4)
• Therefore (I) can be rewritten as:
• x(x-3)(x-4) = 0
• By putting each factor equal to zero, we can find the roots of (I):
• x = 0(Root 1)
• x – 3 = 0
• ⇒ x = 3 (Root 2)
• x – 4 = 0
• ⇒ x = 4 (Root 3)
• Therefore possible values for p, q, r: {0, 3, 4}
• Since it is given that p < q < r, we have:
• p = 0, q = 3 and r =4
• Therefore
• =
• When a number is raised to the power zero, we get 1.
• So,=1
• ⇒=1
• Answer: Option (C)

• Step 1: Set up the equation based on the description
• 
• Step 2 & 3: Finding required values and calculating the final answer
• This is a quadratic equation, so x has either one or two possible answers (Every quadratic equation has two roots but variable x may have some constraints imposed on it, which make one of the roots an impossible value to have for x)
• The constraint on x in the given equation is that x cannot be equal to zero (since x is in the denominator).
• To find out how many values of x are possible, we set up the equation to find the roots.
• First, multiply both sides by x to get rid of the fraction:
• 2=x²+x
• Next, set the equation equal to zero:
• x²+x−2=0
• Finally, factor out the expressions:
• (x+2)(x−1)=0
• x has two possible values: -2 and 1,
• Answer: Option (C)

Step 1: Question statement and Inferences

We are given two quadratic equations in x. We will solve both of them and look for a common value of x

Step 2 & 3: Finding required values and calculating the final answer

Approach:

1. We observe that both the given quadratic equations are already in factored form.
2. By putting each factor of an equation equal to zero, we can find the roots of an equation. So, we will be able to find the two roots of each equation.
3. The common root of both equations will be our answer

Implementing the approach:

Finding the roots of the first equation:

(x - 3)(2x - 5)

Either

x - 3 = 0

• x = 3  . . . Root 1A

Or

2x - 5 = 0

• x = 5/2 . . . Root1B

Finding the roots of the second equation:

Either

• x = 5/2 . . . Root 2A

Or

x - 4 = 0

• x = 4 . . . Root 2B

We observe that x=5/2 is the common root is a common root of both equations.

• The value of x that satisfies both equations is x=5/2

Answer: Option (C)

Step 1: Question statement and Inferences

Given that x is not a positive integer and we need to find the value of x from the given equation.

Step 2 & 3: Finding required values and calculating the final answer

Given equation is

Squaring on both sides

​Squaring on both sides

• 8x  =x²
• x² -8x=0
• x(x -8)=0

 Putting each factor to zero, we get:

x = 0 or x = 8

However, since it is already given that x is not a positive integer, x≠8.

Therefore x = 0.

 Answer: Option (A)

Steps 1 & 2: Understand Question and Draw Inferences

We are given that x is a positive integer

• x > 0

Step 3: Analyze Statement 1

Writing the information given in statement 1 in an equation form:

 x=(√x)/2

Squaring both sides:

x²=x/4

x²−x/4=0  (a standard form Quadratic equation)

x(x−1/4)=0

Now that the equation is in factorized form, we can find the roots.

Either x = 0

Or,

x–1/4=0

x=1/4

Thus, two values of x can satisfy the given equation.

However, we are given that x > 0

• 0 cannot be a value of x.

This means that  x=1/4

Thus, a unique value of x is obtained from the information given in Statement 1

SUFFICIENT

Step 4: Analyze Statement 2

We are given a quadratic equation in the standard form:

4x² + 7x – 2 = 0                              

We will factorize this equation to find its roots:

4x² + 8x – x – 2 = 0

• 4x(x + 2) -1 (x+2) = 0
• (4x-1)(x+2) = 0

Now that the given equation has been factorized, we can find the roots by putting each factor equal to zero.

Either

4x – 1 = 0

 x=1/4

Or

x + 2 = 0

      x = -2

Given: x > 0

Therefore, x cannot have the value -2.

Thus, x=1/4

A unique value of x is obtained from the information given in Statement 2.

SUFFICIENT

Step 5: Analyze Both Statements Together (if needed)

We get a unique answer in steps 3 and4, so this step is not required

Answer: Option (D)

Step 1: Question statement and Inferences

It is given that

 x+y=−10 ....(i)

xy=16  ....(ii)

We need to find which of the given statements must be true.

Step 2 & 3: Finding required values and calculating the final answer

Squaring (i) on both sides, we get:

(x+y)²=100

x²+y²+2xy=100

....(iii)

We know that

(x−y)²=x²+y²−2xy

(x−y)²=x²+y²+2xy−4xy

(x−y)²=(x+y)²−4xy

Using (ii) and (iii) we get:

(x−y)²=100−64

(x−y)²=36

⇒(x−y)=±6

Therefore (I) is not a must be true statement.

Now let us look at (II).

(iii) – 2*(ii) gives us:

(x²+y²+2xy)−2xy=100−32

⇒x²+y²=68

Therefore statement II is true.

We already observed above that

(x−y)²=36

We know that

36=35+136=7∗5+1

Therefore

(x−y)²

 leaves a remainder of 1 when divided by 7.

Answer: Option (E)

Step 1: Question statement and Inferences

We are given an equation to calculate the distance travelled (d) by a free falling body within a time (t).

...(I)

It is given that on earth, a ball dropped from a height of 16 feet takes 1 second to reach the ground.

Therefore from (I), we have

16=

⇒g=32

Thus, we have found the value of acceleration due to gravity for earth.

Let the acceleration due to gravity of the other planet be G_P

It is given that G_P=7g

⇒G_P=7*32……. (II)

Step 2 & 3: Finding required values and calculating the final answer

Once again, using (I) we can write the equation for the free-fall of the ball on the other planet:

2800=

However, from (II), we know that

G_P=7*32

Using this in the above equation, we get,

•  
• t²= 25
• t² – 25 = 0
• (t -5)(t + 5) = 0
• t = 5 or t = -5

However, t denotes time. Therefore, it cannot have a negative value.

Rejecting the negative value, we get t = 5

Thus, on the other planet, it will take a ball 5 seconds to reach the ground from a height of 2800 feet under conditions of free fall.

Answer: Option (C)

Steps 1 & 2: Understand Question and Draw Inferences

Given equation is

x²+bx+c=0 …………….. (I)

Let us suppose that the roots of this equation are p and q.

Given that b and c are constants, we need to find the value of c.

We know that for a quadratic equation

ax²+bx+c=0

Sum of roots = −b/a

Product of roots = c/a

In the given equation, a = 1

Therefore,

Sum of roots = -b

Product of roots = c

Step 3: Analyze Statement 1

It is given that the sum of roots of (I) is zero.

So, we have:

-b = p + q = 0

• b = 0 ………….. (II)

We cannot find the value of c from this information.

Statement 1 alone is not sufficient to arrive at a unique answer.

Step 4: Analyze Statement 2

Now let us look at statement 2.

Writing it in equation form, we have:

p²+q²=18

We know that

p²+q²=(p+q)²−2pq

• p²+q²=(−b)²−2c

• (−b)²−2c=18
• b²−2c=18 ………….. (III)

But we need the value of b here to find the value of c.

Statement 2 alone is not sufficient to arrive at a unique answer.

Step 5: Analyze Both Statements Together (if needed)

Now, let us look at both the statements together.

From statement 1, we have

b = 0

From statement 2, we have

b² - 2c = 18

Combining both of these, we get:

-2c = 18

• c = -9

Answer: Option (C)

Given equation is

And, we need to find the value of x, given that x is an integer.

Step 2 & 3: Finding required values and calculating the final answer

Let us rewriting the last term of (I) to get x-1 in the denominator:

Let us now bring the terms with the same denominator together:

• (x + 1)² = {2(x – 1)}²
• [(x+1)²-{2(x-1)}² ]=0 ………….. (II)

• Observe that this is of the form  a² – b² = (a+b)(a-b)

  So let us say:

  a = x + 1             

  b = 2(x – 1) = 2x - 2

  Therefore:

  a + b = x + 1 + 2x – 2 = 3x -1

  a – b = x + 1 - 2x + 2 = -x + 3

  Therefore we can rewrite (II) as:

  (3x-1)(3-x)=0

• Putting each factor to zero:

Since it is given that x is an integer,

Thus, x = 3 is the required answer.

Answer: Option (E)

Steps 1 & 2: Understand Question and Draw Inferences

Given equation is

 ax²+bx+c=0

It is also given that a, b and c are constants and that a≠0.

We need to find the value of b.

Step 3: Analyze Statement 1

It is given that 3 and 4 are the roots of the equation.

We know that for a quadratic equation
ax²+bx+c=0

Sum of roots = −b/a

Product of roots = c/a

Therefore, we have:

−ba  = 3 + 4 = 7………..(I)

c/ a  = 3 * 4 = 12 …………(II)

Note that these are two linear equations in three variables. So we cannot solve exclusively for b.

Statement 1 alone is not sufficient to arrive at a unique answer.

Step 4: Analyze Statement 2

It is given that the product of roots is 12.

Therefore, we have:

c/a  = 12 …….(III)

Notice that (III) is a single linear equation in two variables. Moreover, it doesn’t provide any information about b.

So statement 2 alone is not sufficient to arrive at a unique answer.

Step 5: Analyze Both Statements Together (if needed)

Now let us look at both the statements together.

Statement 1 gives us (I) and (II)

Statement 2 gives us (III).

However, notice that (III) is essentially same as (II).

In other words, we are simply left with (I) and (II) (essentially the same situation as in statement 1).

Therefore statement 1 and statement 2 combined together are not sufficient to arrive at a unique answer.

Answer: Option (E)

In the given equation,
a = (1 + t²)
b = 2 t c
c = (c² – a²)

⇒ b² – 4 a c = 0

⇒ 4 t² c² – 4 (1 + t²) (c² – a²) = 0

⇒ t² c² – t² c² + a² – c² + a² t² = 0

⇒ a² (1 + t²) = c²

∴ a² (1 + t²) = c² is true.