## Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.6

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

(i) 2x

^{2}Â -Â 3x + 5Â = 0

**Solution**

(ii) 2x

^{2}Â - 6x + 3Â = 0**Solution**

^{2}

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

As we know that D = b

^{2}Â â€“ 4ac

Putting the value of a = 2, b = -6 and c = 3

= (-6)2 - 4 Ã— 2 Ã— 3

= 36 â€“ 24

= 12

Since, D > 0

Therefore, root of the given are real and distinct .

(iii) 3/5Â x

^{2}Â - 2/3Â x + 1Â = 0**Solution**

(iv) 3x

^{2}Â - 4âˆš3Â x + 4Â = 0**Solution**

(v) 3x

^{2}Â - 2âˆš6Â x + 2Â = 0**Solution**

**2. Find the values of k for which the roots are real and equal in each of the following equation:Â**

(i) kx

^{2}Â + 4x + 1Â = 0**Solution**

(ii) kx

^{2}Â - 2âˆš5Â x + 4Â = 0
Â

**Solution**
(iii) 3x

^{2}Â - 5x + 2kÂ = 0**Solution**

(iv) 4x

^{2}Â + kx + 9Â = 0**Solution**

(v) 2kx

^{2}Â - 40Â x + 25Â = 0**Solution**

(vi) 9x

^{2}Â - 24x + kÂ = 0**Solution**

(vii) 4x

^{2}Â - 3kx + 1Â = 0**Solution**

(viii) x

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

(ix) (3k+1)x

^{2}Â + 2(k+1)Â x + kÂ = 0**Solution**

(x) kx

^{2}Â + kx + 1Â = -4x^{2}Â - x**Solution**

The give question isÂ kx

^{2}Â + kx + 1Â = -4x

^{2}Â - x

â‡’ kx

^{2}Â + kx + 1+Â 4x

^{2}Â + x = 0

â‡’ kx

^{2Â }+Â 4x

^{2}Â + kx + x + 1Â = 0

â‡’ x

^{2}(k+4)Â + x(k + 1) + 1Â = 0

Comparing withÂ Â ax

a = (k+4), b = (k+1), c = 1^{2}Â + bx + c = 0Given that the nature of the roots of this equation are real and equal.

i.e. D =Â b

^{2}Â - 4ac = 0

(x) (k+1)x

^{2}Â + 20Â = (k+3)xÂ + (k+8) = 0Â
Solution

(xii) x

^{2}Â - 2kx + 7k - 12Â = 0**Solution**

(xiii) (k+1)x

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

(xiv) 5x

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

(xv) (4-k)x

^{2}Â + (2k + 4)x + (8kÂ + 1)Â = 0**Solution**

(xvi) (2k+1)x

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

(xvii) 4x

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

3.In the following determine the set of values of k foe which the green quadratic equation has

real roots:

(i) 2x

^{2}Â + 3x + kÂ = 0**Solution**

(xvii) 2x

^{2}Â +Â x + kÂ = 0**Solution**

(iii) 2x

^{2}Â - 5x - kÂ = 0**Solution**

(iv) kx

^{2}Â + 6x + 1Â = 0**Solution**

(v) 3x

^{2}Â + 2x + kÂ = 0**Solution**

**4. Find the least positive value of k for which the equation have real and equal roots :Â**

**Solution**

(i) x

^{2}Â - 2(k+1)x + k^{2}Â = 0**Solution**

(i) The given quadric equation is x

Here ,^{2}Â -2 (k+1) x+k^{2}Â = 0, and roots are real and equal . Then find the value of k.a = 1, b = -2(k+1) and, c = k

^{2}

As we know that D = b

^{2}Â â€“ 4ac

Putting the value of a = 1, b = -2 (k+1) and, c =

= {-2(k+1)}

^{2}Â -4 Ã— 1 Ã— k

^{2}

= {4(k

^{2}+2k+1)} â€“ 4k

^{2}

= 4k

^{2}Â + 8k + 4 â€“ 4k

^{2}

= 8k + 4

The given equation will have real and equal roots, if D = 0

8k + 4 = 0

8k = -4

k = -4/8

= -1/2

Therefore, the value of k = -1/2

(ii)Â k

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

The given quadric equation k

Then find the value of k.^{2}x^{2}-2(2k-1)x+4 = 0 , and roots are real and equalÂHere,

a = k

^{2}, b = -2(2k-1) and , c = 4

As we know that D = b

^{2}Â â€“ 4ac

Putting the value of a = k

^{2}Â , b = -2(2k-1) and, c = 4

= {-2(2k-1)}

^{2}Â -4 Ã— k

^{2}Â Ã— 4

= {4(4k

^{2}Â â€“ 4k + 1)} â€“ 16k

^{2}

= 16k

^{2}Â â€“ 16k + 4 â€“ 16k

^{2}

= -16k + 4

The given equation will have real and equal roots , if D = 0

-16k + 4 = 0

16k = 4

k =4/16

= 1/4

Therefore , the value of k = 1/4

(iii) (k+1)x

^{2}Â - 2(k-1)x + 1Â = 0Â**Solution**

^{2}Â â€“ 2 (k-1)x+1 = 0, and roots are real and equal

Then find the value of k .

Here,

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

As we know that D = b

^{2}Â â€“ 4ac

Putting the value of a = k+1, b = -2(k-1) and , c = 1

= {-2(k-1)}

^{2}Â â€“ 4 Ã— (k+1) Ã— 1

= {4(k

^{2}Â â€“ 2k + 1)} â€“ 4k â€“ 4

= 4k

^{2}Â â€“ 8k + 4 â€“ 4k â€“ 4

= 4k

^{2}Â â€“ 12k = 0

The given equation will have real and equal roots, ifÂ D = 0

4k

^{2}Â â€“ 12k + 0 = 0

4k

^{2}Â â€“ 12k = 0

Now factorizing of the above equation

4k(k-3) = 0

k(k-3) = 0

So, either

(k-3) = 0

k = 0 or

k = 3

**5. Find the value of k for which the following equation have real roots**

(i) 2x

^{2}Â + kx + 3Â = 0**Solution**

(ii) kx(x-2) +6 = 0

**Solution**

(iii)Â x

^{2}Â - 4kx + kÂ = 0**Solution**

(iv) kx(x-2âˆš5)+ 10Â = 0

**Solution**

(v)Â kx(x-3)+ 9Â = 0

**Solution**

(vi) 4x

^{2}Â + kx + 3Â = 0**Solution**

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

(i)Â kx

^{2}Â + 2x + 1Â = 0**Solution**

(ii)Â kx

^{2}Â + 6x + 1Â = 0**Solution**

7. For what value of k, (4-k)x

^{2}Â + (2k+4)x+(8k+1) = 0, is a perfect square .Â

**Solution**

8. Find the least positive value of k for which the equation x

^{2}+kx+4 = 0 has real roots .**Solution**
9. Find the values of k for (3k + 1)x

^{2}Â +2(k+1) x+1 = 0 has equal roots . Also, find the roots .**Solution**

10. Find the values of p for which the quadratic equation (2p+1)x

^{2}Â â€“ (7p+2)x+(7p-3)=0 has equal roots . Also find these roots .

**Solution**

11. If -5 is a root of the quadratic equation 2x

^{2}Â + px -15 = 0 and the quadratic equation p(x

^{2}+x)+k = 0 has equal roots , find the value of k.

**Solution**

12. If 2 is a root of the quadratic equation 3x

^{2}Â + px â€“ 8 = 0 and the quadratic equation 4x

^{2}-2px + k = 0 has equal roots , find the value of k .

**Solution**

13. If 1 is a root of the quadratic equation is 3x

^{2}Â + ax â€“ 2 = 0 and the quadratic equation a(x

^{2}Â + 6x) â€“ b = 0 has equal roots, find the value of b .

**Solution**

The given quadratic equation is3x

^{2}Â + ax â€“ 2 = 0 and one root is 1.

Then , it satisfies the given equation .

3(1)

^{2}Â + a(1)-2=0

â‡’ 3 + a â€“ 2 = 0

â‡’ 1+ a = 0

â‡’ a = -1

The quadratic equation a(x

^{2}Â + 6x) â€“ b = 0 , has equal roots .

Putting the value of a , we get

-1(x

^{2}Â + 6x) â€“ b = 0

â‡’ x

^{2}Â + 6x + b = 0

Here . A = 1, B = 6 and C = b

D = (6)

^{2}Â â€“ 4(1) (b)

= 36 â€“ 4b

The given equation will have real and equal roots , ifÂ Â D = 0

Thus, 36 â€“ 4b = 0

â‡’ 4b = 36

â‡’ b = 9

Therefore , the value of b is 9 .

14. Find the value of p for which the quadratic equationÂ (p+1)x

^{2}- 6(p+1) x + 3(p+9) = 0 , p â‰ -1 has equal roots . Hence, find the roots of the equation .

**Solution**Â

The given quadratic equation (p+1)x

^{2}- 6(p+1) x + 3(p+9) = 0 , has equal roots .
Here, a = p + 1, b = -6p -6 and c = 3p + 27.

As we know that D = b

^{2}Â â€“ 4ac
Putting the values of a = p + 1 , b = -6p â€“ 6 and c = 3p +27.

D = [-6(p+1)]

^{2}Â â€“ 4(p+1)[3(p+9)]
= 36(p

^{2}Â + 2p + 1) â€“ 12(p^{2}Â + 10p + 9)
= 36p

^{2}Â â€“ 12p^{2}Â + 72p â€“ 120p + 36 - 108
= 24p

^{2}Â â€“ 48 â€“ 72
The given equation will have real and equal roots, if D = 0

Thus , 24p

^{2}Â â€“ 48p â€“ 72 = 0
â‡’ p

^{2}Â â€“ 2p â€“ 3 = 0
â‡’ p

^{2}Â â€“ 3p + p â€“ 3 = 0
â‡’ p(p -3) + 1 (p-3) = 0

â‡’ (p+1)(p -3) = 0

â‡’ p+1 = 0 or p â€“ 3 = 0

â‡’ p = -1 or p = 3

Therefore , the value of p is -1,3.

It is given that p â‰ 1, thus p = 3 only .

Now the equation becomes

4x

^{2}Â -24x + 36 = 0
â‡’ x

^{2}Â â€“ 6x + 9 = 0
â‡’ x

^{2}Â â€“ 3x â€“ 3x + 9 = 0
â‡’ x(x-3)-3(x-3) = 0

â‡’ (x-3)

^{2}Â = 0
â‡’ x = 3 , 3

Hence , the root of the equation is 3.

15. Determine the nature of the roots of the following quadratic equations :

(i) (x-2a) (x â€“ 2b) = 4ab

(ii) 9a

^{2b}2x^{2}-24abcdx + 16c^{2}d^{2}Â = 0 , a â‰ 0, b â‰ 0
(iii) 2(a

^{2}+b^{2})x^{2}Â + 2(a+b)x+1 = 0
(iv) (b+c)x

^{2}Â â€“ (a+b+c) x + a = 0**Solution**

(i) The given quadratic equation is (x-2a)(x-2b) = 4ab

â‡’ x

^{2}Â â€“ 2(a+b)x + 4ab â€“ 4ab = 0

â‡’ x

^{2}Â â€“ 2(a+b) x = 0

Here, a = 1, b = -2(a+b) and, c = 0

As we know that D = b

^{2}Â â€“ 4ac

Putting the value of a = 1, b = -2(a+b) and , c = 0

= (-2(a+b))

^{2}-4Ã—1Ã—0

= 4(a

^{2}+2ab + b

^{2}) â€“ 0

Since , D > 0

Therefore , root of the given equation are real and distinct .

(ii) The given quadratic equation is 9a

^{2}b

^{2}x

^{2}-24abcdx + 16c

^{2}d

^{2}Â = 0

Here, a = 9a

^{2}b

^{2}d

^{2}

As we know that D = b

^{2}Â â€“ 4ac

Putting the value of a = 9a

^{2}b

^{2}, b = -24abcd and, c = 16c

^{2}d

^{2}

= (24abcd)

^{2}Â â€“ 4 Ã— 9a

^{2}b

^{2}Ã—16c

^{2}d

^{2}

= (576a

^{2}b

^{2}c

^{2}d

^{2}) â€“ 576a

^{2}b

^{2}c

^{2}d

^{2}

= 0

Since, D = 0Â

Therefore, root of the given equation are real and equal .

(iii) The given quadratic equation is 2(a

Here , a = 2(a

As we know that D = b

Putting the value of a =Â 2(a

= (2(a+b))

= (4a

= 8ab â€“ 4a

Since, D < 0

Therefore , root of the given equation are not real.Â

(iv) The given quadratic equation isÂ (b+c)x

Here, a = (b+c), b = -(a+b+c) and, c = a

As we know that D = b

Putting the value of a = (b+c), b = -(a+b+c) and, c = a

= (-a(a+b+c))

= (a

= a

Since, D > 0

Therefore, root of the given equation are real and unequal .

(i) x

(ii) 2x

(iii) 4x

(iv) 2x

(i) The given quadratic equation x

Then find the value of k .

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

As we know that D = b

Putting the value of a = 1, b = -k and c = 9

= (-k)

= k

The given equation will have real roots , if D â‰¥ 0

k

k

k â‰¥ âˆš36 or k â‰¤ - âˆš36

k â‰¤ - 6 or k â‰¥ 6

Therefore, the value of k â‰¤ - 6 or k â‰¥ 6

(ii) The given quadratic equation is 2x

(iii) The given quadratic equation is 2(a

^{2}+b^{2})x^{2}Â + 2(a+b)x+1=0Here , a = 2(a

^{2}+b^{2}), b = 2(a+b) and, c = 1As we know that D = b

^{2}Â â€“ 4acPutting the value of a =Â 2(a

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

^{2}Â â€“ 4 Ã— 2(a^{2}Â + b^{2}) Ã— 1= (4a

^{2}Â + 4b^{2}Â + 8ab) â€“ 8a^{2}Â â€“ 8b^{2}= 8ab â€“ 4a

^{2}Â â€“ 4b^{2}Since, D < 0

Therefore , root of the given equation are not real.Â

^{}(iv) The given quadratic equation isÂ (b+c)x

^{2}Â â€“ (a+b+c)x+a = 0Here, a = (b+c), b = -(a+b+c) and, c = a

As we know that D = b

^{2}Â â€“ 4acPutting the value of a = (b+c), b = -(a+b+c) and, c = a

= (-a(a+b+c))

^{2}Â â€“ 4 Ã— (b+c) Ã— a= (a

^{2}Â + b^{2}Â + c^{2}Â + 2ab + 2bc + 2ca) â€“ 4ab â€“ 4ca= a

^{2}Â + b^{2}Â + c^{2}Â â€“ 2ab + 2bc â€“ 2caSince, D > 0

Therefore, root of the given equation are real and unequal .

^{}^{16.Â Determine the set of values of k for which the following quadratic equations have real roots:Â }(i) x

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

^{2}Â + kx + 2 = 0(iii) 4x

^{2}Â â€“ 3kx + 1 = 0(iv) 2x

^{2}Â + kx â€“ 4 = 0**Solution**(i) The given quadratic equation x

^{2}Â â€“ kx + 9 = 0, and roots are realThen find the value of k .

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

As we know that D = b

^{2}Â â€“ 4acPutting the value of a = 1, b = -k and c = 9

= (-k)

^{2}Â â€“ 4 Ã— 1 Ã— 9= k

^{2}Â â€“ 36The given equation will have real roots , if D â‰¥ 0

k

^{2}Â â€“ 36 â‰¥ 0k

^{2}Â â‰¥ 36k â‰¥ âˆš36 or k â‰¤ - âˆš36

k â‰¤ - 6 or k â‰¥ 6

Therefore, the value of k â‰¤ - 6 or k â‰¥ 6

(ii) The given quadratic equation is 2x

^{2}Â + kx + 2 = 0, and roots are real.^{}
Then find the value of k.

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

As we know that D = b

Putting the value of a = 2 , b = k and , c = 2

= (k)

= k

The given equation will have real roots , if D â‰¥ 0

k

k

k â‰¥ âˆš16 or k â‰¤ - âˆš16

k â‰¤ -4 or k â‰¥ 4

Therefore , the value of k â‰¤ -4 or k â‰¥ 4

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

As we know that D = b

^{2}Â â€“ 4acPutting the value of a = 2 , b = k and , c = 2

= (k)

^{2}Â - 4Ã—2Ã—2= k

^{2}Â - 16The given equation will have real roots , if D â‰¥ 0

k

^{2}Â â€“ 16 â‰¥ 0k

^{2}Â â‰¥ 16k â‰¥ âˆš16 or k â‰¤ - âˆš16

k â‰¤ -4 or k â‰¥ 4

Therefore , the value of k â‰¤ -4 or k â‰¥ 4

(iv) The given quadratic equation is 2x

^{2}Â + kx â€“ 4 = 0, and roots are real
Then find the value of k .

Here, a = 2, b = k and c , = -4

As we know that D = b

^{2}Â â€“ 4ac
Putting the value of a = 2 , b = k and c = -4

= (k)

^{2}Â â€“ 4 Ã— 2 Ã— (-4)
= k

^{2}Â + 32
The given equation will have real roots , if D â‰¥ 0

k

^{2}Â + 32 â‰¥ 0
since left hand side is always positive . So k âˆˆ R

Therefore , the value of k âˆˆ R

17. If the roots of the equation (b-c) x

^{2}Â + (c-a)x+(a-b) = 0 are equal , then prove that 2b = a + c.**Solution**

18. If the roots of the equation (a

^{2}Â + b^{2}) x^{2}Â - 2(ac + bd) x + (c^{2}Â + d^{2}) = 0 are equal ,Â prove that a/b = c/d .Â**Solution**

19. If the roots of the equations ax

^{2}Â + 2bx + c = 0 and bx^{2}Â â€“ 2âˆšacx + b = 0 are simultaneously real , then prove that b^{2}Â = ac .**Solution**

20. If p ,q are real and p â‰ q, then show that the roots of the equation (p-q)x

^{2}Â +5(p+q) x â€“ 2(p-q) = 0 are real unequal .

Â Â Â Â Â Â Â

**Solution**

Â 21. If the roots of the equation (c

^{2}Â â€“ ab) x^{2}Â â€“ 2(a^{2}Â â€“ bc) x + b^{2}Â â€“ ac = 0 are equal , prove that either a = 0 or a^{3}Â + b^{3}Â + c^{3Â }= 3abc.Â Â**Solution**

22. Show that the equation 2(a

^{2}Â + b^{2}) x^{2}Â + 2(a+b) x + 1 = 0 has no real roots, when a â‰ b .Â Â**Solution**

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

Sol.Â

24. If a,b,c are real numbers such that ac â‰ 0, then show that at least one of the equations ax

^{2}Â + bx + c = 0 and - ax^{2}Â + bx + c = 0 has real roots .**Solution**

25. If the equation (1+m

^{2})x^{2}Â + 2mcx + (c^{2}Â â€“ a^{2}) = 0 has equal roots , prove that c^{2}Â = a^{2}(1+m^{2}).**Solution**

The given equation (1+m

^{2})x2 + 2mcx + (c

^{2}Â â€“ a

^{2}) = 0, has equal roots

Then prove that c

^{2}Â = (1+m^{2}).
Here,

a = (1+m

^{2}) , b = 2mc and ,Â c = (c^{2}Â â€“ a^{2})
As we know that D = b

^{2}Â â€“ 4ac
Putting the value of a = (1+m

^{2}) , b = 2mc and, c = (c^{2}Â â€“ a^{2})
D = b

^{2}Â â€“ 4ac
= {2mc}

^{2}Â â€“ 4 Ã— (1+m^{2}) Ã— (c^{2}Â â€“ a^{2})
= 4 (m

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

^{2}c^{2}Â â€“ 4c^{2}Â + 4a^{2}Â â€“ 4m^{2}c^{2}Â + 4m^{2}a^{2}
= 4a

^{2}Â + 4m^{2}a^{2}Â â€“ 4c^{2}
The given equation will have real roots , if D = 0

4a

^{2}Â + 4m^{2}a^{2}Â â€“ 4c^{2}Â = 0
4a

^{2}Â + 4m^{2}a^{2}Â = 4c^{2}
4a

^{2}Â (1+m^{2}) = 4c^{2}
a

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

^{2}Â = a^{2}Â (1+m^{2})