# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.6 | Set 2

### Question 11. If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal-roots, find the value of k.

**Solution:**

2x² + px – 15 = 0

-5 is its one root

It will satisfy it

2(-5)

^{2}+p(-5)-15=02*25-5p-15=0

50-15-5p=0

-5p+35=0

-5p=-35

p=-35/-5=7

Now in equation

p(x

^{2}+x)+k=07(x

^{2}+x)+k=07x

^{2}+7x+k=0Here, a=7, b=7, c=k

D or b

^{2}-4ac = (7)^{2}-4*7*k=49-28k

Roots are real and equal

D or b

^{2}-4ac=049-28k=0⇒28k=49

k=49/28=7/4

### Question 12. If 2 is a root of the quadratic equation 3x² + px – 8 = 0 and the quadratic equation 4x² – 2px + k = 0 has equal roots, find the value of k.

**Solution:**

2 is a root of 3x

^{2}+px-8=0It will satisfy if

3(2)

^{2}+p*2-8=012+2p-8=0

4+2p=0

2p=-4

p=-4/2=-2

p=-2

Now in equation,

4x

^{2}-2px+k=04x

^{2}-2*(-2)x+k=04x

^{2}+4x+k=0Here, a=4, b=4, c=k

D=b

^{2}-4ac=(4)^{2}-4*4*k=16- 16k

Roots are real and equal

D or b

^{2}-4ac=016-16k=0

=> 16k = 16

k = 16

### Question 13. If 1 is a root of the quadratic equation 3x² + ax – 2 = 0 and the quadratic equation a(x² + 6x) – b=0 has equal roots, find the value of b.

**Solution:**

1 is one root of

3x² + ax – 2 = 0

3(1)

^{2}+a*1-2=03+a-2=0⇒a+1=0

a=-1

Now in equation a(x

^{2}+6x)-b=0-1(x

^{2}+6x)-b=0-x

^{2}-6x-b=0⇒x

^{2}+6x+b=0Here, a=1, b=6, c=b

D=b

^{2}-4ac=(6)^{2}-4*1*k=36-4k

Roots are equal

D or b

^{2}-4ac=036-4k=0

4k=-36

k=-36/4=-9

### Question 14. Find the value of p for which the quadratic equation (p + 1) x² – 6 (p + 1) x + 3 (p + q) = 0, p ≠ -1 has equal roots. Hence, find the roots of the equation.

**Solution:**

(p+1)x

^{2}-6(p+1)x+3(p+9)=0,p≠-1

Comparing with ax

^{2}+bx+c=0b

^{2}-4ac, c=p+1, b=-6(p+1), c=3(p+9)Now D=b

^{2}-4ac=[-6(p+1)]

^{2}-4*(p+1)*3(p+9)=36(p+9)

^{2}-12(p+1)(p+9)Roots are equal

D=0

⇒36(p+1)

^{2}-12(p+1)(p+9)=0⇒36(p+1)

^{2}=12(p+1)(p+9)⇒3(p+1)=p+9⇒3p+3=p+9

⇒3p-p=⇒9-3⇒2p=6

p=6/2=3

p=3

Hence, x=3,3

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

**(i) (x – 2a) (x – 2b) = 4ab**

**Solution:**

⇒x

^{2}-2bx-2ax+4ab-4ab=0Here, a=1,b=-2(a+b), c=0

Discriminant(D)=b

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

^{2}-4*1*0={-2(a+b)}

^{2}D>0

Roots are real and distinct

**(ii) 9a²b²x² – 24abcdx + 16c²d² = 0, a ≠ 0, b ≠ 0**

**Solution:**

Here a=9a

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

^{2}-4ac=(-24abcd)

^{2}-4*9a^{2}b^{2}*16c^{2}d^{2}=576a

^{2}b2c^{2}d^{2}-576a^{2}b^{2}c^{2}d^{2}=0

D=0

Roots are real and equal

**(iii) 2 (a² + b²) x² + 2 (a + b) x + 1 = 0**

**Solution:**

Here a=2(a

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

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

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

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

^{2}+4b^{2}+8ab-8a^{2}-8b^{2}=8ab-4a

^{2}-4b^{2}=-(4a

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

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

^{2}D<0

Root are not real

**(iv) (b + c) x² – (a + b + c) x + a = 0**

**Solution:**

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

D=b

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

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

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

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

^{2}D>0

Roots are real and distinct.

### Question 16. Determine the set of values of k for which the given following quadratic equation has real roots :

**(i) x² – kx + 9 = 0**

**Solution:**

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

D=b

^{2}-4ac=(-k)

^{2}-4*1*9=k

^{2}-36Roots are real

D≥⇒k2-36≥0

k

^{2}≥36⇒k^{2}(±6)^{2}k≥6 or k≤-6

**(ii) 2x² + kx + 2 = 0**

**Solution:**

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

D=b

^{2}-4ac=(k)

^{2}-4*2*2=k

^{2}-16Roots are real

D≥0⇒k

^{2}-16≥0k

^{2}≥16⇒(k)^{2}≥(±4)^{2}k≥4 or k≤-4

**(iii) 4x² – 3kx +1=0**

**Solution:**

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

D=b

^{2}-4ac=(-3k)

^{2}-4*4*1=9k

^{2}-16Roots are real

D≥0⇒9k

^{2}-16≥09k

^{2}≥16⇒k^{2}≥16/9(k)

^{2}≥(±)^{2}k≥4/3 or k≤-4/3

**(iv) 2x² + kx – 4 = 0**

**Solution:**

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

D=b

^{2}-4ac=k

^{2}-4*2*(-4)=k

^{2}+32The roots are real

D≥0⇒k

^{2}+32≥0k

^{2}+32≥0 for all value ofk ∈ R

### Question 17. If the roots of the equation (b – c) x² + (c – a) x + (a – b) = 0 are equal, then prove that 2b = a + c. [CBSE 2002C]

**Solution:**

(b – c) x² + (c – a) x + (a – b) = 0

Here a=b-c, B=c-a, c=a-b

D=b

^{2}-4ac=(c-a)

^{2}-4(b-c)(a-b)The roots are equal

D=0

(c-a)

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

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

^{2}+a^{2}-2ca-4ab+4b^{2}+4ca-4bc=0a

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

^{2}=0⇒a-2b+c=0=> a + c = 2b

=> 2b = a + c

Hence proved.

### Question 18. If the roots of the equation (a² + b²) x² – 2 (ac + bd) x + (c² + d²) = 0 are equal. prove that ab = cd

**Solution:**

Here a=a² + b², b= – 2 (ac + bd), c=c² + d²

D=b

^{2}-4ac=[-2(ac+bd)]

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

^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2}]=4[a

^{2}c^{2}+b^{2}d^{2}+2abcd]-4[a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2}]=4a

^{2}c^{2}+4b^{2}d^{2}+8abcd-4a^{2}d^{2}-4b^{2}c^{2}-4b^{2}d^{2}=8abcd-4a

^{2}d^{2}-4b^{2}c^{2}The roots are equal

D=0

8abcd-4a

^{2}d^{2}-4b^{2}c^{2}=0a

^{2}d^{2}+b^{2}c^{2}-2abcd=0 ———–(Dividing by -4)(ad-bc)

^{2}=0⇒ad-bc=0ad=bc⇒a/b=c/d

### Question 19. If the roots of the equations ax² + 2bx + c = 0 and bx² – 2√ac x + b = 0 are simultaneously real, then prove that b² = ac

**Solution: **

ax

^{2}+2bx+c=0 ———–(i)and bx

^{2}-2x+b=0 ———–(ii)Let D

_{1}and D_{2}are the discriminants of there simultaneous equation (i) and (ii)D1=(2b)

^{2}-4*a*c=4b^{2}-4acand D

_{2}=(-2)^{2}-4*b*b=4ac=4b

^{2}These have real roots

D

_{1}≥0⇒4b^{2}-4ac≥0⇒4b2≥4ac⇒b2≥ac ————-(i)

and D

_{2}≥0Therefore, 4ac-4b

^{2}≥0 ⇒4ac≥4b^{2}ac ≥ b

^{2}——————(ii)ac≥b

^{2}≥acb

^{2}=ac

### Question 20. If p, q are real and p ≠ q, then show that the roots of the equation (p – q) x² + 5(p + q) x – 2(p – q) = 0 are real and unequal.

**Solution:**

Here a=p-q, b=5(p+q), c=-2(p-q)

D=b

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

^{2}-4*(p-q)*-2(p-q)=25(p+q)

^{2}+8(p-q)^{2}p and q are real and p≠q

25(p+q)

^{2}+8(p-q)≥0The given quadratic equation has real and unequal roots.

### Question 21. If the roots of the equation (c² – ab) x² – 2 (a² – bc) x + b² – ac = 0 are equal, prove that either a = 0 or a3 + b3 + c3 = 3abc.

**Solution:**

Here a=c

^{2}-ab, b=-2(a^{2}-bc), c=b2-acD=b

^{2}-4ac=[-2(a

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

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

^{4}+4b^{2}c^{2}-8a^{2}bc-4b^{2}c^{2}+4ac^{3}+4ab^{3}-4ac^{2}bc=4a

^{4}+4ab+4ac^{3}-12a^{2}bc=4a[a

^{3}+b^{3}+c^{3}-3abc]D=0

4a(a

^{3}+b^{3}+c^{3}-3abc)=0a(a

^{3}+b^{3}+c^{3}-3abc)=0Either a=0

or a

^{3}+b^{3}+c^{3}=3abc=0a

^{3}+b^{3}+c^{3}=3abc

### Question 22. Show that the equation 2 (a² + b²) x² + 2 (a + b) x + 1 = 0 has no real roots, when a ≠ b.

**Solution:**

Here a=2(a+b2), b=2(a+b), c=1

D=b

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

^{2}-4*2*(a^{2}+b^{2})*14(a+b

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

^{2}+4b^{2}+8ab-8a^{2}-8b^{2}=-4a

^{2}-b^{2}+8ab-4[a

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

^{2}D<0

Roots are not real

### Question 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.

**Solution:**

Here a=3, b=-2(a+b+c), c=ab+ac+ca

D=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}+2ab+2bc+2ca-3ab-3bc-3ca]=2[(a-b)]

^{2}+(b-c)^{2}+(c-a)2]Clearly, D≥0

Roots are real

If roots are equal, then

D=0

(a-b)

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

a=b=, b=c, c=a

a=b=c

Hence proved

### Question 24. If a, b, c are real numbers such that ac ≠ 0, then show that at least one of the equations ax² + bx + c = 0 and – ax² + bx + c = 0 has real roots.

**Solution:**

a,b,c are real number

and ac≠0

ax

^{2}+bx+c=0 ———-(i)-ax+bx+c=0 ——–(ii)

Let D

_{1}and D_{2 }be the discriminants of the two equation (i) and (ii)D

_{1}= b^{2}-4ac and D_{2}=b^{2}-4acIf equation (i) has real roots, then

D

_{1}≥0b

^{2}-4ac≥0b

^{2}-≥acNow D

^{2}=b^{2}+4ac4ac≤b

^{2}b

^{2}+4ac≥0D≥0

Both the equation has real roots

Hence proved

### Question 25. If the equation (1 + m²) x² + 2mcx + (c² – a²) = 0 has equal roots, prove that c² = a² (1 + m²).

**Solution: **

Here a=1+m

^{2}, b=2mc, c=c^{2}-a^{2}D=b

^{2}-4ac=(2mc)

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

^{2}c^{2}-4c^{2}+4a-4m^{2}c^{2}+4m^{2}a^{2}=4a

^{2}+4m^{2}a^{2}-4c^{2}Root are equal

D=0⇒4a

^{2}+4m2a^{2}-4c^{2}=0a

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

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

^{2}(a+m^{2})=c^{2}c

^{2}=a^{2}(^{1}+m^{2})Hence proved

## Please

Loginto comment...