**Exercise 8.6**

**Question 1:Â ****Determine the nature of the roots of the following quadratic equations.**

**Solutions:Â **

**Important Notes:Â **

- A quadratic equation is in the formÂ ax
^{2 }+bx +c =0 - To find the nature of roots, first find determinant “D”
- D = b
^{2}â€“ 4ac - If D > 0, equation has real and distinct roots
- If D < 0, equation has no real roots
- If D = 0, equation has 1 root

**(i) 2x ^{2} -3x + 5 =0**

Here,Â a= 2, b= -3, c= 5

D = b^{2} â€“ 4ac

= (-3)^{2} -4(2)(5)

= 9 â€“ 40

= -31<0

Here D<0 and so, the equation does not have any real roots.

**(ii) 2x ^{2} -6x + 3=0**

Here, a= 2, b= -6, c= 3

D = (-6)^{2} -4(2)(3)

= 36 â€“ 24

= 12<0

As D>0, the expression have real and distinct roots.

**(iii) For what value of k (4-k)x ^{2 }+ (2k+4)x +(8k+1) =0 is a perfect square.**

Here, a= 4-k, b= 2k+4, c= 8k+1

D = b^{2 â€“ }4ac

= (2k+4)^{2} â€“ 4(4-k)(8k+1)

= (4k^{2}+16+ 16k) -4(32k+4-8k^{2}-k)

= 4(k^{2} +8k^{2}+4k-31k+4-4)

=4(9k^{2}-27k)

D = 4(9k^{2}-27k)

As the given equation is a perfect square, D= 0

So,

4(9k^{2}-27k) = 0

â‡’ 9k^{2}-27k=0

â‡’Â k^{2 }-3k =0

â‡’Â K (k-3) =0

Now, for this equation to be true,

k = 0 orÂ k = 3

âˆ´ The value of k has to be either 0 or 3 for the equation to be a perfect square.

**(iv) Find the least positive value of k for which the equation x ^{2}+kx+4=0 has real roots.**

Here, a= 1, b= k, c= 4

Since the equation has real roots,Â D will be positive

D = b^{2Â }–Â 4ac â‰¥0

= k^{2} â€“ 16 â‰¥ 0

â‡’ k â‰¤ 4 ,k â‰¥ -4

âˆ´ The least positive value of kÂ will be 4 for which the given equation will have real roots.

**(v) Find the value of k for which the given quadratic equation has real roots and distinct roots.**

**Kx ^{2}+ 2x+1 =0**

Here, a= k, b= 2, c= 1

D = b^{2 }â€“ 4ac â‰¥0

= 4 -4k â‰¥ 0 = 4k â‰¤4

âˆ´ K â‰¤ 1

**(vi) Kx ^{2} +6x+1 = 0**

Here, a= k, b= 6, c= 1

D = b^{2 }â€“4ac â‰¥0

= 36 -4k â‰¥ 0

= 4k â‰¤36

âˆ´ K â‰¤ 9

**(vii) x ^{2 }â€“kx+9 =0**

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

It is given that the equation has real and distinct roots.

So, D = b^{2 }â€“ 4ac â‰¥0

= k^{2} â€“ 4(1)(9) â‰¥ 0

= k^{2 }-36 â‰¥ 0

â‡’Â kâ‰¥ -6 and k â‰¤6

âˆ´ For the equation to have real and distinct roots, the value of k has to be between -6 and 6.

**Question 2: Find the value of k.**

**(i) Kx ^{2}+4x+1=0**

Here,

a= k, b= 4, c= 1

As the equation has real and equal roots,

D = b^{2}-4ac = 0

= 4^{2}-4(k)(1)=0

= 16-4k=0

â‡’Â k = 4

**(ii) kx ^{2Â }âˆ’Â 2âˆš5xÂ +Â 4Â =Â 0**

Here, a= k , b= 2âˆš5Â ,Â c= 4

It is given that the equation has real and equal roots so,

D = 0

Or,Â b^{2}-4ac = 0

=Â âˆ’(2âˆš5)^{2Â }âˆ’Â 4Â Ã—Â kÂ Ã—Â 4Â =Â 0

= 20- 16k =0

â‡’Â kÂ =Â 5/4

**(iii) 3x ^{2}-5x+2k =0**

Here,

a= 3, b= -5, c= 2k

As the equation has real and equal roots, D = 0

Or, b^{2}-4ac = 0

= (-5)^{2}-4(3)(2k) = 0

= 25- 24k =0

K = 25/24

**(iv) 4x ^{2}+kx+9=0**

Given,

a= 4, b= k, c= 9

Here, D = 0 as the equation has real and equal roots.

â‡’ b^{2}-4ac=0

=Â k^{2}-4(4)(9)=0

= k^{2 }-144=0

â‡’Â k = 12

**(v) 2kx ^{2}-40x+25=0**

Given,

a= 2k, b= -40, c= 25

Here D = 0 since the given equation has real and equal roots

âˆ´ b^{2}-4ac=0

= (-40)^{2}-4(2k)(25)=0

= 1600 – 200k=0

â‡’ k = 8

**(vi) 9x ^{2}-24x+k = 0**

Here,

a= 9, b= -24, c= k

As the equation has real and equal roots,

D = b^{2}-4ac=0

Or, (-24)^{2}-4(9)(k)=0

= 576-36k = 0

â‡’ k = 16

**(vii) 4x ^{2}-3kx+1 =0**

Here,

a= 4, b= -3k, c= 1

Since the given equation has real and equal roots, D = 0

Or, b^{2}-4ac=0

= (-3k)^{2}-4(4)(1)=0

= 9k^{2}-16=0

K = 4/3

**(viii) x ^{2}-2(5+2k)x+3(7+10k) = 0**

From the equation,

a= 1, b=+2(52k) , c= 3(7+10k)

Here the nature of the roots of the equation are real and equal roots.

So, D = 0

Or, b^{2}-4ac=0

â‡’ (+2(52k))^{2}-4(1)( 3(7+10k))=0

â‡’Â 4(5+2k)^{2 }-12(7+10k)=0

â‡’Â 25+4k^{2}+20k-21-30k=0

= 4k^{2}-10k+4=0

â‡’Â 2k^{2}-5k+2=0

â‡’Â 2k^{2}-4k-k+2=0

â‡’ 2k(k-2)-1(k-2) =0

â‡’Â (k-2)(2k-1) = 0

âˆ´ K = 2 Or k = Â½

**(ix) (3k+1)x ^{2}+2(k+1)x+k =0**

Here a= 3k+1, b=+2(k+1) , c= (k)

It is given that the nature of the roots of the equation are real and equal. So, D = 0

Or, b^{2}-4ac=0

= [2(k+1)]^{2} -4(3k+1)(k) =0

= (k+1)^{2}-k(3k+1)=0

= -2k^{2}+k+1 =0

Multiplying the equation with -1,

2k^{2Â }– k – 1 = 0

The value of k can be obtained by

K = \(\frac{1+\sqrt{9}}{4}\) = 1

Or , \(k=\frac{1-\sqrt{9}}{4}\)Â = \(\frac{-1}{2}\)

The value of k are 1 and \(\frac{-1}{2}\)

**(x) Kx ^{2}+kx+1 = -4x^{2}-x**

This equation has to be made in the form of a quadratic equation i.e.Â ax^{2}+bx+c=0.

So, bring all the x components to LHS,

Now the equation becomes

x^{2}(4+k)+x(k+1)+1 =0

Here a=4+k,b=+k+1 , c= 1

Now, D = 0 as the nature of the roots of the equation are real and equal

or, b^{2}-4ac=0

= (k+1)^{2}-4(4+k)(1) =0

= k^{2Â }– 2k-10 =0

It can be seen that this equation also in the form ax^{2}+bx+c=0

So, solve the equation as a quadratic.

Here, a=1 , b= -2 , c = -15

\(k=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

Or, k = 5 and -3

**(xi) (k+1)x ^{2}+2(k+3)x+k+8 =0**

Here,

a=k+1,b=2(k+3), c= k+8

b^{2}-4ac =0

= [2(k+30]^{2}-4(k+1)(k+8)=0

= 4(k+3)^{2}-4(k+1)(k+8) =0

Take out 4 as common from the LHS and divide the same on the RHS

= (k+3)^{2}-(k+1)(k+8) =0

â‡’ k^{2}+9+6k â€“ ( k^{2}+9k+18) =0

â‡’Â 9+6k-9k-8 =0

â‡’= -3k+1 =0

â‡’Â 3k = 1

k = 1/3

**(xii) x ^{2}-2kx+7k-12=0**

Here a=1,b=-2k, c= 7k-12

Given the nature of the roots of the equation are real and equal .

D= b^{2}-4ac =0

= (2k)^{2}-4(1)(7k-12) =0

= 4k^{2}-28k+48 =0

= k^{2}-7k+12 =0

The value of k can be obtained by

\(k=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

Here a = 1 , b = -7k , c = 12

By calculating the value of k is \(\frac{7\pm \sqrt{1}}{2}\)Â = 4 , 3

The value of k for the given equation is 4 and 3 respectively.

**(xiii) (k+1)x ^{2}-2(3k+1)x+8k+1 =0**

Here a=k+1,b=-2(k+1), c= 8k+1

D= b^{2}-4ac =0

= (-2(k+1))^{2}-4(k+1)(8k+1) =0

= 4(3k+1)^{2}-4(k+1)(8k+1) =0

Take out 4 as common from LHS and divide the same on RHS

= (3k+1)^{2}-(k+1)(8k+1) =0

= 9k^{2}+6k+1 â€“( 8k^{2}+9k+1) =0

= 9k^{2}+6k+1 â€“ 8k^{2}-9k-1 =0

= k^{2}-3k =0

= k(k-3) =0

Either k =0 Or, k =3

**(xiv) 5x ^{2}-4x+2 +k(4x^{2}-2x+1) =0**

The given equation can be written as x^{2}(5+4k)-x(4+2k)+2-k =0

Here a = 5 + 4k, b = -(4 + 2k), c= 2 – k

As the nature of the roots of the equation are real and equal, D = 0

Or, b^{2}-4ac =0

= [-(4+2k)]^{2}-4(5+4k)(2-k)=0

= 16+4k^{2}+16- 4(10-5k+8k-4k^{2}] =0

= 16+4k^{2}+16- 40+20k-32k+16k^{2} =0

= 20k^{2 }-4k-24 =0

Take out 4 as common from LHS and divide the same on RHS

= 5k^{2}-k-6 =0

The value of k can be obtained by equation

\(k=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

Here, a = 5 , b= -1 , c =-6

\(k=\frac{1\pm \sqrt{-1^{2}-4(5)(-6)}}{2(5)}\)

\(k = \frac{6}{5} and -1\)

The value of k for the given equation are \(k = \frac{6}{5} and -1\) respectively.

**(xv) (4-k)x ^{2}+(2k+4)x+(8k+1) =0**

Here a=4-k, b=(2k+4), c= 8k+1

D = b^{2}-4ac =0

= (2k+4)^{2}-4(4-k)(8k+1)=0

= 4k^{2}+16k+16 -4(-8k^{2}+32k+4-k)=0

= 4k^{2}+16k+16 + 32k^{2}-124k-16 =0

â‡’Â 36k^{2}-108k =0

Take out 4 as common from LHS and divide the same on RHS

= 9k^{2}-27k =0

= 9k(k-3) =0

Either 9k =0

K =0

Or, K = 3

**(xvi) (2k+1)x ^{2}+2(k+3)x+(k+5)Â = 0**

Here a=2k+1, b=2(k+3), c= k+5

D= b^{2}-4ac =0

= [2(k+3)]^{2}-4(2k+1)(k+5) =0

Take out 4 as common from LHS and divide the same on the RHS

= [(k+3)]^{2}-(2k+1)(k+5) =0

= K^{2}+9+6k-(2k^{2}+11k+5) =0

= -k^{2}– 5k+4 =0

= k^{2}+5k-4 =0

The value of k can be obtained by \(k = \frac{6}{5} and -1\) respectively.

Here a= 1 , b = 5 , c= -4

Now \(k=\frac{-5\pm \sqrt{5^{2}-4(1)(-4)}}{2\times 1}\)

K = \(k=\frac{-5\pm \sqrt{41}}{2}\)

The value of k for the given equation is \(k=\frac{-5\pm \sqrt{41}}{2}\)

**(xvii) 4x ^{2}-2(k+1)x+(k+4) =0**

Here a=4, b=-2(k+1), c= k+4

D= b^{2}-4ac =0

= [-2(k+1)]^{2}-4(4)(k+4)=0

Take out 4 as common from LHS and divide the same on RHS

= (k+1)^{2}-4(k+4)=0

= k^{2}+1+2k-4k-16 =0

= k^{2}-2k-15 =0

The value of k can be obtained by \(k = \frac{6}{5} and -1\) respectively.

Here a= 1 , b = -2 , c= -15

K = \(k=\frac{2\pm \sqrt{69}}{2}\)

The value of k for the given equation is \(k=\frac{2\pm \sqrt{69}}{2}\)

**Question 3:Â ****In the following, determine the set of values of k for which the given quadratic equation has real roots:**

**Solution:**

**(i) 2x ^{2}+3x+k = 0**

From the equation,

a= 2 , b= 3 , c =k

Here, the given quadratic equation has equal and real roots

So, D = b^{2}-4ac â‰¥ 0

= 9 â€“ 4(2)(k) â‰¥ 0

= 9-8k â‰¥ 0

= Â kÂ â‰¤Â 9/8

The value of k does not exceed kÂ â‰¤ (9/8)Â toÂ have a real root.

**(ii) 2x ^{2}+kx+3 = 0**

From the equation,

a= 2 , b= k , c =3

Here,Â D = b^{2}-4ac â‰¥ 0

= k^{2}-4(2)(3) â‰¥ 0

= k^{2}-24 â‰¥ 0

= k^{2} â‰¥ 24

kÂ â‰¥Â âˆš24

kÂ â‰¥Â âˆš24

The value of k should not exceedÂ kÂ â‰¥Â âˆš24Â order to obtain real roots.

**(iii)2x ^{2}-5x-k =0**

Here, a= 2 , b= -5 , c =-k

AS the given quadratic equation has equal and real roots,

D = b^{2}-4ac â‰¥ 0

= 25 -4(2)(-k) â‰¥ 0

= 25-8k â‰¥ 0

= kÂ â‰¤Â 25/8

The value ofÂ k should not exceedÂ kÂ â‰¤Â 25/8

**(iv) Kx ^{2}+6x+1 =0**

Here,

a= k , b= 6 , c =1

Also,

D = b^{2}-4ac â‰¥ 0 (as the quadratic equation has equal and real roots)

= 36 -4(k)(1) â‰¥ 0

= 36-4k â‰¥ 0

= k â‰¤ 9

The value of k for the given equation is k â‰¤ 9

**(v) x ^{2}-kx+9 =0**

Here,

a= 1 , b= -k , c =9

Also, D = b^{2}-4ac â‰¥ 0 (as the given equation has equal and real roots)

= k^{2}-4(1)(-9) â‰¥ 0

= k^{2}-36 â‰¥ 0

= k^{2} â‰¥ 36

= kÂ â‰¥Â âˆš36

â‡’ K â‰¥ 6 or k â‰¤ -6

**Question 4:Â ****Determine the nature of the roots of the following quadratic equations.**

**Solution:**

**(i) 2x ^{2} -3x + 5 =0**

Here,

a= 2, b= -3, c= 5

D = b^{2} â€“ 4ac

= (-3)^{2} -4(2)(5)

= 9 â€“ 40

= -31 < 0

As D < 0, the equation does not having any real roots.

**(ii) 2x ^{2} -6x + 3=0**

Here,

a= 2, b= -6, c= 3

D = b^{2} â€“ 4ac

= (-6)^{2} -4(2)(3)

= 36 â€“ 24

= 12<0

As D>0, the equation does having any real and distinct roots.

**(iii) For what value of k (4-k)x ^{2 }+ (2k+4)x +(8k+1) =0 is a perfect square**

Here, a= 4-k, b= 2k+4, c= 8k+1

D = b^{2 }â€“ 4ac

= (2k+4)^{2} â€“ 4(4-k)(8k+1)

= (4k^{2}+16+ 16k) -4(32k+4-8k^{2}-k)

= 4(k^{2} +8k^{2}+4k-31k+4-4)

=4(9k^{2}-27k)

D = 4(9k^{2}-27k)

As the given equation is a perfect square,

D = 0

4(9k^{2}-27k) = 0

â‡’ 9k^{2}-27k=0

â‡’ K^{2 }-3k =0

â‡’ K (k-3) =0

âˆ´ k = 0 Or,Â k =3

**(iv) Find the least positive value of k for which the equation x ^{2}+kx+4=0 has real roots.**

Here,

a= 1, b= k, c= 4

D = b^{2 }â€“ 4ac â‰¥0

= k^{2} â€“ 16 â‰¥ 0

= kâ‰¤ 4Â Or, kâ‰¥-4

**(v) Find the value of k for which the given quadratic equation has real roots and distinct roots.**

**Kx ^{2}+ 2x+1 =0**

Here,

a= k, b= 2, c= 1

D = b^{2 }â€“4ac â‰¥0

= 4 -4k â‰¥ 0

= 4k â‰¤4

K â‰¤ 1

For the quadratic equation to have real and equal roots, k â‰¤ 1.

**(vi) Kx ^{2} +6x+1 = 0**

Here,

a= k, b= 6, c= 1

D = b^{2 }â€“ 4ac â‰¥0

= 36 -4k â‰¥ 0

= 4k â‰¤36

= k â‰¤ 9

For the quadratic equation to have real and equal roots, kÂ â‰¤ 9.

**(vii) x ^{2 }â€“kx+9 =0**

Here,

a= 1, b= -k, c= 9

As the equation is having real and distinct roots,

D = b^{2 }â€“ 4ac â‰¥0

= k^{2} â€“ 4(1)(9) â‰¥ 0

= k^{2 }-36 â‰¥ 0

= k â‰¥ -6 and k â‰¤6

The value of k lies between -6 and 6 for the equation to have real and distinct roots.

**Question 5:Â ****Find the values of k for which the given quadratic equation has real and distinct roots.**

**Solution:**

**(i) Kx ^{2}+2x+1 =0**

Here, a= k, b= 2 , c =1

D = b^{2}-4ac â‰¥ 0

= 4-4(1)(k) â‰¥ 0

= 4k â‰¤ 4

= k â‰¤ 1

**(ii) Kx ^{2}+6x+1 =0**

Here,

a= k, b= 6 , c =1

D = b^{2}-4ac â‰¥ 0

= 36-4(1)(k) â‰¥ 0

= 4k â‰¤ 36

= k â‰¤ 9

**Question 6:Â ****For what value of k , (4-k)x ^{2}+(2k+4)x+(8k+1)=0 , is a perfect square.**

**Solution:**

Here,

a= 4-k , b= 2k+4 , c =8k+1

D = b^{2}-4ac

= (2k+4)^{2}-4(4-k)(8k+1)

= 4k^{2 }+16+4k-4(32+4-8k^{2}-k)

= 4(k^{2 }+4+k-32-4+8k^{2}+k)

=4(9k^{2}-27k)

It is given that the given equation is a perfect square

âˆ´ D =0

= 4(9k^{2}-27k) =0

= (9k^{2}-27k) =0

= 3k (k-3) =0

â‡’ k = 0 Or, k =3

**Question 7:Â ****If the roots of the equation (b-c)x ^{2} +(c-a)x+(a-b)=0 are equal , then prove that 2b = a + c.**

**Solution:**

Here,

a= (b-c), b= (c-a), c= (a-b)

It is given that the equation has real and equal roots.

So, D = b^{2 }-2ac =0

= (c-a)^{2 }-4(b-c)(a-b) =0

= c^{2}+a^{2 }-2ac â€“ 4(ab-b^{2}-ac+cb) =0

= c^{2}+a^{2 }-2ac â€“ 4ab+4b^{2}+4ac-4cb =0

= c^{2}+a^{2 }+2ac â€“ 4ab+4b^{2}-4cb =0

= (a+c)^{2 }-4ab +4b^{2 }-4cb =0

= (c+a-2b )^{2} = 0

= (c+a-2b ) = 0

= c+a = 2b

âˆ´ It is proved.

**Question 8:Â ****If the roots of the equation (a ^{2 }+b^{2}) x^{2}â€“ 2(ac+bd)x+(c^{2}+ d^{2})= 0 are equal. Prove that a**

**Ã·b = c**

**Ã· d.**

**Solution:**

Here,

a= (a^{2 }+b^{2}) ,b= â€“ 2(ac+bd) , c=(c^{2}+ d^{2}) .

It is given that the equation has real and equal roots.

So, D = b^{2 }-4ac = 0

= [-2(ac+bd)]^{2} -4 (a^{2 }+b^{2})(c^{2 }+ d^{2}) =0

= (ac+bd)^{2} – (a^{2 }+b^{2})(c^{2 }+ d^{2}) =0

= a^{2}c^{2} + b^{2}d^{2 }+2abcd â€“ (a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2 }+ b^{2}d^{2}) =0

= 2abcd – a^{2}d^{2} – b^{2}c^{2 }Â =0

= abcd +abcd – a^{2}d^{2} – b^{2}c^{2 }Â =0

= ad(bc-ad) +bc(ad-bc) =0

= ad(bc-ad) -bc(bc-ad) =0

= (ad-bc)(bc-ad) =0

= ad â€“bc =0

= (aÃ·b) = (cÃ· d)

Hence, proved.

**Question 9:Â Â ****If the roots of the equation ax ^{2Â }+ 2bx + c = 0 andÂ bx^{2Â }âˆ’Â 2câˆš(ax)Â +Â bÂ =Â 0Â are simultaneously real , then prove that b^{2Â }– ac =0.**

**Solution:**

The given equations are ax^{2}+2bx +c=0 andÂ Â bx^{2}âˆ’2câˆš(ax)Â +Â bÂ =Â 0

These two equations are of the form ax^{2}+bx +c =0 .

Given that the roots of the two equations are real. Hence, D â‰¥ 0 that is b^{2}-4ac â‰¥ 0

Let us assume that ax^{2}+2bx +c=0 be equation (i)

And bx^{2}âˆ’2câˆš(ax)Â +Â bÂ =Â 0Â be (ii)

From equation (i) b^{2}-4ac â‰¥ 0

= 4 b^{2}-4ac â‰¥ 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (iii)

From equation (ii) b^{2}-4ac â‰¥ 0

=Â (2âˆš(ca))^{2}âˆ’4b^{2}â‰¥0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (iv)

Given, that the roots of equation (i) and (ii) are simultaneously real and hence equation (iii) = equation (iv).

= 4b^{2}-4ac = 4ac -4 b^{2}

= 8ac = 8b^{2}

= b^{2}-ac =0.

Thus, it is proved that b^{2}-ac =0.

**Â **

**Question 10:Â ****If p, q are the real roots and p ****â‰ q . Then show that the roots of the equation (p-q)x ^{2} +5(p+q)x -2(p-q) = 0 are real and equal.**

**Solution:**

The given equation is (p-q)x^{2} +5(p+q)x -2(p-q) =0

Given, p , q are real and p â‰ q.

Then, Discriminant (D) = b^{2 }â€“ac

= [5(p+q)]^{2} -4(p-q)(-2(p-q))

= 25(p+q)^{2} +(p-q)^{2}

We know that the square of any integer is always positive i.e. > zero.

As, D = b^{2 }â€“ac â‰¥ 0

As p , q are real and p â‰ q.

â‡’ 25(p+q)^{2} +(p-q)^{2 }Ëƒ 0 = D Ëƒ 0

âˆ´ The roots of this equation are real and unequal.

**Question 11:Â ****If the roots of the equation (c ^{2}-ab)x^{2 }-2(a^{2}-bc)x +b^{2}-ac =0 are equal , then prove that either a=0 or a**

**Â³+b**

**Â³+c**

**Â³ = 3abc .**

**Solution:**

Here,

a = (c^{2}-ab), b = -2(a^{2}-bc), c = b^{2}-ac.

It is given that the roots of the given question are equal.

So, D= 0 , b^{2}-4ac =0

= [-2(a^{2}-bc)]^{2} -4(c^{2}-ab)( b^{2}-ac) =0

= 4(a^{2}-bc)^{2} – 4(c^{2}-ab)( b^{2}-ac) =0

= 4a (aÂ³+bÂ³+cÂ³- 3abc) =0

Either 4a =0Â â‡’ a =0

Or, (aÂ³+bÂ³+cÂ³- 3abc) =0

= (aÂ³+bÂ³+cÂ³) =3abc

Hence proved.

**Question 12:Â ****Show that the equation 2(a ^{2}+b^{2})x^{2}+2(a+b)x+1 =0 has no real roots , when a**

**â‰ b.**

**Solution:**

Here, a= 2(a^{2}+b^{2}) , b= 2(a+b) , c = +1.

Given, a â‰ b

D = b^{2 }-4ac

=[2(a+b)]^{2} -4 (2(a^{2}+b^{2}))(1)

= 4(a+b)^{2} -8(a^{2}+b^{2})

= 4(a^{2}+b^{2}+2ab) â€“8a^{2}-8b^{2}

= +2ab â€“4a^{2}-4b^{2}

According to the question a â‰ b, as the discriminant D has negative squares so the value of D will be less than zero.

âˆ´ D Ë‚ 0, when a â‰ b.

**Question 13:Â ****Prove that both of the roots of the equation (x-a)(x-b) +(x-c)(x-b)+(x-c)(x-a) =0 are real but they are equal only when a=b=c.**

**Solution:**

The equation has to be solved to get the form of a quadratic equation i.e.Â ax^{2}+bx+c = 0

So,

(x-a)(x-b) +(x-c)(x-b)+(x-c)(x-a) =0 â‡” 3x^{2}-2x(a+b+c)+(ab+bc+ca) =0

This equation is in the form of quadratic equation withÂ a =3 , b = 2(a+b+c) , c = (ab+bc+ca)

D =Â b^{2}-4ac

= [-2(a+b+c)]^{2} -4(3)(ab+bc+ca)

=4(a+b+c)^{2}-12(ab+bc+ca)

=4[(a+b+c)^{2 }-3(ab+bc+ca) ]

= 4[ a^{2}+b^{2}+c^{2 }â€“ab-bc-ca]

= 2[ 2a^{2}+2b^{2}+2c^{2 }â€“2ab-2bc-2ca]

= 2[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]

Here, clearly D â‰¥ 0 ,

If D = 0 then,

[(a-b)^{2}+(b-c)

^{2}+(c-a)

^{2}] =0

a â€“b = 0

b â€“ c = 0

c â€“ a = 0

Or, a = b = c = 0

Hence proved.

**Question 14:Â ****If a, b, c are real numbers such that ac ****â‰ 0, then, show that at least one of the equations ax ^{2 }+bx+c =0 and â€“ax^{2}+bx+c =0 has real roots.**

**Solution:**

The given equation are ax^{2}+ bx+c =0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (i)

And- ax^{2}+bx+c =0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)

Given, equations are in the form of ax^{2}+ bx+c =0 also given that a ,b, c are real numbers and ac â‰ 0.

D = b^{2 }-4ac

For equation (i) = b^{2 }-4ac â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(iii)

For equation (ii) = b^{2}-4(-a)(c)

= b^{2 }+4ac â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(iv)

As a , b , c are real and given that ac â‰ 0 , henceÂ b^{2 }-4ac Ëƒ 0 and b^{2 }+4ac Ëƒ 0

âˆ´ D Ëƒ 0

Hence proved.

**Question 15:Â ****If the equation (1+m ^{2})x+2mcx+(c^{2}-a^{2})=0 has real and equal roots , prove that c^{2}= a^{2}(1+m^{2}).**

**Solution:**

Here,

a = (1+m^{2}) , b= 2mc , c= +(c^{2}-a^{2})

It is given that the nature of the roots of this equation is equal and hence D=0 , b^{2}-4ac =0

= (2mc)^{2} -4(1+m^{2}) (c^{2}-a^{2}) =0

= 4m^{2}c^{2 }â€“ 4(c^{2}+m^{2}c^{2}-a^{2} â€“a^{2}m^{2}) =0

= 4(m^{2}c^{2}– c^{2}+m^{2}c^{2}+a^{2} +a^{2}m^{2}) =0

= m^{2}c^{2}– c^{2}+m^{2}c^{2}+a^{2} +a^{2}m^{2 }=0

Take m^{2Â }as common and rearrange the terms:

= a^{2} +a^{2}m^{2 }– c^{2 }=0

= a^{2} (1+ m^{2}) â€“ c^{2} =0

Or, c^{2 }= a^{2} (1+ m^{2})

Thus, it is proved that as D= 0 , then the roots are equal of c^{2 }= a^{2} (1+ m^{2}).