Practice Test: Quadratic Equations - Class 10 MCQ

Thus, it can be expressed as product of linear factors.

However, the condition "one year ago, the man was 8 times as old as his son" holds only when y=7, because one year ago, the man was 48 and the son was 6, and indeed 48 is 8 times 6.

Thus, the correct ages are:

• Son: 7 years
• Man: 49 years

Given equation is x²-5x+7=0
We have discriminant as b²-4ac=(-5)²-4*1*7= -3
And x = , Since we do not have any real number which is a root of a negative number, the roots are not real.

expand the given quadratic equation: x² + p(4x + p – 1) + 2 = 0

So, the quadratic equation becomes:

The discriminant is :

Let the two digit number be ab.

a×b=12

When 9 is added to the number

10(a+b)+9 = 10b+a

9a−9b+9=0

a−b+1=0

a−12a+1=0

a2+a−12=0

(a+4)(a−3)=0

a = -4 is not accepted. So, a = 3, b = 4

The number is 34 .

 Here a = 2, b = 1, c = -1
∴ D = b² – 4ac = (1)² – 4 × 2 × (-1) = 1 + 8 = 9 > 0
∴ Roots of the given equation are real and distinct.

Here a = 12, b = 4k, c = 3
Since the given equation has real and equal roots
∴ b² – 4ac = 0
⇒ (4k)² – 4 × 12 × 3 = 0
⇒ 16k² – 144 = 0
⇒ k² = 9
⇒ k = ±3

Since roots are equal
∴ D = 0 => b² – 4ac = 0
⇒ (c – a)² -4(b – c) (a – b) = 0
⇒ c² – b² – 2ac -4(ab -b² + bc) = 0 =>c + a-2b = 0 => c + a = 2b
⇒ c² + a² – 2ca – 4ab + 4b² + 4ac – 4bc = 0
⇒ c² + a² + 4b² + 2ca – 4ab – 4bc = 0
⇒ (c + a – 2b)² = 0
⇒ c + a – 2b = 0
⇒ c + a = 2b

Let one of the odd positive integer be x
then the other odd positive integer is x+2
their sum of squares = x² +(x+2)²
= x² + x² + 4x +4
= 2x² + 4x + 4
Given that their sum of squares = 290
⇒ 2x² +4x + 4 = 290
⇒ 2x² +4x = 290-4 = 286
⇒ 2x² + 4x -286 = 0
⇒ 2(x² + 2x - 143) = 0
⇒ x² + 2x - 143 = 0
⇒ x² + 13x - 11x -143 = 0
⇒ x(x+13) - 11(x+13) = 0
⇒ (x-11) = 0 , (x+13) = 0
Therfore , x = 11 or -13
We always take positive value of x
So , x = 11 and (x+2) = 11 + 2 = 13
Therefore , the odd positive integers are 11 and 13 .