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Example Solved Problem | Mathematics Algebra - Nature of Roots of a Quadratic Equation | 10th Mathematics : Algebra

Chapter: 10th Mathematics : Algebra

Nature of Roots of a Quadratic Equation

Nature of Roots of a Quadratic Equation: Determine the nature of roots for the following quadratic equations : Example Solved Problem

Nature of Roots of a Quadratic Equation

The roots of the quadratic equation ax2 + bx +c = 0 , a ≠ 0 are found using the formula x Here, b2 - 4ac called as the discriminant (which is denoted 2a by Δ ) of the quadratic equation, decides the nature of roots as follows 


 

Example 3.41

Determine the nature of roots for the following quadratic equations 

(i) x2x − 20 = 0

(ii) 9x2 − 24x + 16 = 0

(iii) 2x2 − 2x + 9 = 0

Solution

(i) x2 x 20 = 0

Here, a = 1, b = −1 , c = −20

Now, Δ = b2 − 4ac

Δ  = (− 1)2 − 4(1)(−20) = 81

Here, Δ  = 81 > 0 . So, the equation will have real and unequal roots 

(ii) 9x2 − 24x + 16 = 0

Here, a = 9 , b = −24 , c = 16

Now, Δ = b2 − 4ac = (−24)2 − 4 (9 )(16)=0

Here, ∆ = 0 . So, the equation will have real and equal roots.

(iii) 2x2 − 2x + 9 = 0

Here, a = 2 , b = −2 , c = 9

Now, Δ  = b2 − 4ac = (−2 )2 − 4(2 )(9 ) = −68

Here, Δ = − 68 < 0 . So, the equation will have no real roots.

 

Example 3.42

(i) Find the values of ‘k’, for which the quadratic equation

kx 2 (8k + 4) + 81 = 0 has real and equal roots?

(ii) Find the values of ‘k’ such that quadratic equation

(k + 9)x2 + (k + 1)x + 1 = 0 has no real roots? 

Solution

(i) kx2 (8k + 4) + 81 = 0

Since the equation has real and equal roots, Δ = 0.

That is, b2 − 4ac = 0

Here, a = k , b = −(8k + 4) , c = 81

That is, [ −(8k + 4)]2 − 4(k )(81) = 0 

64k2 + 64k + 16 − 324k = 0

64k2 − 260k + 16 = 0

dividing by 4 we get 16k2 − 65k + 4 = 0

(16k − 1)(k − 4) = 0 then, k = 1/16 or k = 4

(ii) (k + 9)x2 + (k + 1)x + 1 = 0

Since the equation has no real roots, Δ < 0

That is, b2 − 4ac < 0

Here, a = k + 9 , b = k + 1 , c = 1

That is, (k + 1)2 − 4(k + 9)(1) < 0

k 2 + 2k + 1 – 4k 36 < 0

k 2 2k 35 < 0

(k + 5)(k − 7) < 0

Therefore, − 5 < k < 7 .     {If α < β and if (xα)(xβ) < 0 then, α < x < β }.

 

Example 3.43

Prove that the equation x2 (p2 + q 2 ) + 2x(pr + qs) + r2 + s2 = 0 has no real roots. If ps = qr , then show that the roots are real and equal.

Solution

The given quadratic equation is, x2 ( p2  + q2 ) + 2x ( pr + qs) + r2  + s2  = 0

Here,

a = p2 +q2 , b = 2(pr +qs) , c = r2 + s2

Now, Δ = b2 − 4ac  =  [2( pr + qs)]2 − 4( p2  + q2 )( r 2 + s2 )

= 4 [p2r2 + 2pqrs + q2s2 − q2r2p2s2q2r2q2s2 ]

= 4 [–p2s2 + 2pqrs − q2r2] = −4 [(psqr)2] <0           ...(1)

since, Δ = b2 − 4ac < 0 , the roots are not real.

If ps = qr then Δ = − 4[psqr]2 = - 4[qr − qr]2 = 0 (using (1))

Thus, Δ = 0 if ps = qr and so the roots will be real and equal.


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