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Question -

State whether the following quadratic equations have two distinct real roots. Justify your answer.



Answer -

  1. x2 – 3x + 4 = 0
  2. 2x2 + x – 1 = 0
  3. 2x2 – 6x + 9/2 = 0
  4. 3x2 – 4x + 1 = 0
  5. (x + 4)2 – 8x = 0
  6. (x – 2)2 – 2(x + 1) = 0
  7. 2 x2 –(3/√2)x + 1/√2 = 0
  8. x (1 – x) – 2 = 0
  9. (x – 1) (x + 2) + 2 = 0
  10. (x + 1) (x – 2) + x = 0

Solution:

    Q i.             x2 –3x + 4 = 0

Solution:-

 (i)

The equation x2 – 3x + 4 = 0 hasno real roots.

D = b2 – 4ac

= (-3)2 – 4(1)(4)

= 9 – 16 < 0

Hence, the roots are imaginary.

 Q ii.    2x2 +x – 1 = 0

Solution:-

(ii)

The equation 2x2 + x – 1 = 0has two real and distinct roots.

D = b2 – 4ac

= 12 – 4(2) (-1)

= 1 + 8 > 0

Hence, the roots are real and distinct.

Q iii.    2x2 –6x + 9/2 = 0

Solution:-

(iii)

The equation 2x2 – 6x +(9/2) = 0 has real and equal roots.

D = b2 – 4ac

= (-6)2 – 4(2) (9/2)

= 36 – 36 = 0

Hence, the roots are real and equal.

Q iv.    3x2 – 4x + 1 = 0

Solution:-

 (iv)

The equation 3x2 – 4x + 1 = 0has two real and distinct roots.

D = b2 – 4ac

= (-4)2 – 4(3)(1)

= 16 – 12 > 0

Hence, the roots are real and distinct.

Q v.     (x + 4)2 – 8x = 0

Solution:-

 (v)

The equation (x + 4)2 – 8x = 0has no real roots.

Simplifying the above equation,

x2 + 8x + 16 – 8x = 0

x2 + 16 = 0

D = b2 – 4ac

= (0) – 4(1) (16)< 0

Hence, the roots are imaginary.

 

Q vi.    (x – 2)2 –2(x + 1) = 0

Solution:-

 (vi)

The equation (x – √2)– √2(x+1)=0 hastwo distinct and real roots.

Simplifying the above equation,

x2 – 2√2x + 2 – √2x – √2 = 0

x2 – √2(2+1)x + (2 – √2) = 0

x2 – 3√2x + (2 – √2) = 0

D = b2 – 4ac

= (– 3√2)2 – 4(1)(2 – √2)

= 18 – 8 + 4√2 > 0

Hence, the roots are real and distinct.

Q vii.   √2 x2 –(3/√2)x+ 1/√2 = 0

Solution:-

 (vii)

The equation √2x2 – 3x/√2 + ½= 0 has two real and distinct roots.

D = b2 – 4ac

= (- 3/√2)2 – 4(√2) (½)

= (9/2) – 2√2 > 0

Hence, the roots are real and distinct.

Q viii.  x (1 – x) – 2 = 0

Solution:-

 (viii)

The equation x (1 – x) – 2 = 0 has no realroots.

Simplifying the above equation,

x2 – x + 2 = 0

D = b2 – 4ac

= (-1)2 –4(1)(2)

= 1 – 8 < 0

Hence, the roots are imaginary.

Q ix.    (x – 1) (x + 2) + 2 = 0

Solution:-

 (ix)

The equation (x – 1) (x + 2) + 2 = 0 has tworeal and distinct roots.

Simplifying the above equation,

x2 – x + 2x – 2 + 2 = 0

x2 + x = 0

D = b2 – 4ac

= 12 –4(1)(0)

= 1 – 0 > 0

Hence, the roots are real and distinct.

Q x.     (x + 1) (x – 2) + x = 0

Solution:-

 (x)

The equation (x + 1) (x – 2) + x = 0 has tworeal and distinct roots.

Simplifying the above equation,

x2 + x – 2x – 2 + x = 0

x2 – 2 = 0

D = b2 – 4ac

= (0)2 – 4(1) (-2)

= 0 + 8 > 0

Hence, the roots are real and distinct.

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