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

Determine the nature of roots for the following quadratic
equations

(i) *x*^{2} − *x* − 20 = 0

(ii) 9*x*^{2} − 24*x* + 16 = 0

(iii) 2*x*^{2} − 2*x* + 9 = 0

(i) *x*^{2}** **−

Here, *a *=* *1,* b *= −1* *,* c *= −20

Now, Δ = *b*^{2} − 4*ac*

Δ = (− 1)^{2} −
4(1)(−20) = 81

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

(ii) 9*x*^{2} − 24*x* + 16 = 0

Here, *a *=* *9* *,* b *= −24* *,* c *=*
*16

Now, Δ = *b*^{2} − 4*ac* = (−24)^{2} − 4
(9 )(16)=0

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

(iii) 2*x*^{2} − 2*x* + 9 = 0

Here, *a *=* *2* *,* b *= −2* *,* c *=*
*9

Now, Δ = *b*^{2}
− 4*ac* = (−2 )^{2} − 4(2 )(9 ) = −68

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

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

*kx *^{2}* *−* *(8*k *+* *4)* *+* *81* *=* *0*
*has real and equal roots?

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

(*k* + 9)*x*^{2} + (*k* + 1)*x* + 1 =
0 has no real roots?

(i)* kx*^{2}** **−

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

That is, *b*^{2} − 4*ac* = 0

Here, *a* = *k* , *b* = −(8*k* + 4) , *c*
= 81

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

64*k*^{2} + 64*k* + 16 − 324*k* = 0

64*k*^{2} − 260*k* + 16 = 0

dividing by 4 we get 16*k*^{2} − 65*k* + 4 = 0

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

(ii) (*k* + 9)*x*^{2} + (*k* + 1)*x* +
1 = 0

Since the equation has no real roots, Δ < 0

That is, *b*^{2} − 4*ac* < 0

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

That is, *(k + 1) ^{2} − 4(k + 9)(1) < 0*

*k *^{2}* *+* *2*k + 1 *– 4*k *−* *36* *<* *0

*k *^{2}* *−* *2*k *−* *35* *<* *0

(*k* + 5)(*k* − 7) < 0

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

Prove that the equation** ***x*^{2}** **(*p*^{2}** **+** ***q*** **^{2}** **)** **+** **2*x*(*pr*** **+** ***qs*)** **+** ***r*^{2}** **+** ***s*^{2}** **=** **0** **has no real** **roots. If *ps* = *qr*
, then show that the roots are real and equal.

The given quadratic equation is,* **x*^{2}** **(

Here,

*a *=* p*^{2}* *+*q*^{2}* *,* b *=*
*2(*pr *+*qs*)* *,* c *=* r*^{2 }+ *s*^{2}

Now, Δ = *b*^{2} − 4*ac* = [2( *pr* + *qs*)]^{2} − 4( *p*^{2}
+ *q*^{2} )(* r* ^{2 }+ *s*^{2} )

= 4 [*p*^{2}*r*^{2}
+ 2*pqrs* + *q*^{2}*s*^{2} − q^{2}r^{2}
– *p ^{2}s^{2}* –

= 4 [–*p*^{2}*s*^{2}
+ 2*pqrs* − q^{2}r^{2}] = −4 [(*ps* − *qr*)^{2}]
<0 ...(1)

since, Δ = *b*^{2} − 4*ac*
< 0 , the roots are not real.

If *ps* = *qr* then Δ = − 4[*ps*
– *qr*]^{2} = - 4[*qr − qr*]^{2} = 0 (using (1))

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

Tags : Example Solved Problem | Mathematics Algebra Example Solved Problem | Mathematics Algebra

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