Scoring better marks in the Class 10 examination is important for CBSE board students. The RS Aggarwal Class 10 solutions are beneficial for students in fetching more marks in maths subject. The solutions are created according to the latest syllabus of the CBSE. Students should refer to these solutions provided by us to turn challenging topics into an easy one. All the solutions are arranged systematically in a proper step by step format.

Practicing from these solutions will help students to secure higher marks. It will help you in clearing all the doubts of the topics as the solutions cover the complete syllabus. The solutions teach students to solve the questions in a simple and easy way. The RS Aggarwal Solutions Class 10 Chapter 10 – Quadratic Equations Ex 10.6 are beneficial for students to revise the whole syllabus.

## Download PDF of RS Aggarwal Class 10 Solutions Chapter 10â€“Â Quadratic Equations 10F (10.6)

**1. Find the nature of the roots of the following quadratic equations: **

**(i)2x ^{2 }– 8x + 5 = 0 (ii) 3x^{2}– 2\(\sqrt{6}\)x +2 = 0 **

**(iii)5x ^{2}-4x+1=0 (iv) 5x(x – 2) + 6 =0 **

**(v) 12x ^{2} – 4\(\sqrt{15}\)x+5=0 (vi) x^{2} – x + 2 = 0 **

Answer :

(i)The given equation is 2x^{2} – 8x + 5 = 0.

This is of the form ax^{2} + bx + c = 0, where a = 2, b = -8 and c = 5.

Therefore, Discriminant, D = b^{2} – 4ac = (-8)^{2} – 4 x 2 x 5 = 64 – 40 = 24> 0

Hence, the given equation has real and unequal roots.

(ii) The given equation is 3x^{2} â€” 2\(\sqrt{6}\)x + 2 = 0.

This is of the form ax^{2} + bx + c = 0, where a = 3, b = -2\(\sqrt{6}\) and c = 2.

Therefore, Discriminant, D= b^{2} â€“ 4ac = (-2\(\sqrt{6}\))^{2} -4 x 3 x 2 = 24 – 24 = 0

Hence, the given equation has real and equal roots.

(iii) The given equation is 5x^{2} – 4x + 1 = 0.

This is of the form ax^{2} + bx + c = 0, where a = 5, b = -4 and c = 1.

Therefore, Discriminant, D = b^{2} – 4ac = (-4)^{2} – 4 x 5 x 1 = 16 – 20 = -4 < 0

Hence, the given equation has no real roots.

(iv) The given equation is

5x(x – 2) + 6 = 0

\(\Rightarrow\) 5x^{2} – 10x + 6 = 0

This is of the form ax^{2} + bx + c = 0, where a = 5, b = -10 and c = 6.

Therefore, Discriminant, D = b^{2} – 4ac = (-10)^{2} – 4 x 5x 6 = 100 -120 = -20 < 0

Hence, the given equation has no real roots.

(v) The given equation is 12x^{2} – 4\(\sqrt{15}\)x + 5 = 0.

This is of the form ax^{2} + bx + c = 0, where a = 12, b = -4\(\sqrt{15}\) and c = 5.

Therefore, Discriminant, D = b^{2} – 4ac = (-4\(\sqrt{15}\))^{2} – 4 x 12 x 5 = 240 – 240 = 0

Hence, the given equation has real and equal roots.

(vi) The given equation is x^{2 }â€“ x + 2 = 0

This is of the form ax^{2} + bx + c = 0, where a=1, b=-1 and c=2.

Therefore, Discriminant, D = b^{2} – 4ac = (-1)^{2 }â€“ 4 x 1 x2 = 1 â€“ 8 = -7<0

Hence, the given equation has no real roots.

**2. If a and b are distinct real numbers, show that the quadratic equation **

**2(a ^{2}+ b^{2})x^{2} + 2(a+ b)x + 1 =0 has no real roots. **

Answer :

The given equation is 2 (a^{2} + b^{2}) x^{2 }+ 2(a + b)x + 1= 0.

Therefore, D = [2(a+ b)]^{2} – 4 x 2(a^{2} + b^{2}) x 1

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

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

= -4a^{2 }+ 8ab – 4b^{2}

= -4(a^{2} â€” 2ab+ b^{2})

= -4(a â€” b)^{2}<0

Hence. the given equation has no real roots.

**3. Show that the roots of the equation x ^{2} + px-q^{2}= 0 are real for all real values of p and q. **

Answer :

Given:

x^{2} Â± px – q^{2} =0

Here,

a= 1, b = p and c = -q^{2}

Discriminant D is given by :

D = (b^{2} – 4ac)

= p^{2} – 4 x 1 x (-q^{2})

= (p^{2}+4q^{2}) > 0

D > 0 for all real values of p and q .

Thus, the roots of the equation are real.

**4. For what values of k are the roots of the quadratic equation 3x ^{2}+ 2kx + 27 = 0 real and equal? **

Answer :

Given:

3x^{2} + 2kx + 27 = 0

Here,

a = 3, b = 2k and c = 27

It is given that the roots of the equation are real and equal; therefore, we have:

D = 0

\(\Rightarrow\) (2k)^{2} – 4 x 3 x 27 = 0

\(\Rightarrow\)4k^{2} – 324 = 0

\(\Rightarrow\)4k^{2} = 324

\(\Rightarrow\)k^{2} = 81

\(\Rightarrow\)k = Â±9

Therefore, k = 9 or k = -9

**5. For what value of k are the roots of the quadratic equation ix(x-2\(\sqrt{5}\)) + 10 = 0 real and equal? **

Answer :

The given equation is

Kx(x- 2\(\sqrt{5}\)) + 10 = 0

\(\Rightarrow\)kx^{2} â€” 2\(\sqrt{5}\)kx + 10 = 0

This is of the form ax^{2} + bx + c=0, c =0, where a = k, b = -2\(\sqrt{5}\)k and c= 10.

Therefore, = b^{2 }â€“ 4ac=(-2\(\sqrt{5}\)k)^{2} â€“ 40K

The given equation will have reaI and equal roots if D = 0.

Therefore, 20k^{2} – 40k = 0

\(\Rightarrow\) 20k(k – 2) = 0

\(\Rightarrow\) k = 0 or k – 2 = 0

\(\Rightarrow\)k = 0 or k = 2

But, for k = 0, we get 10 = 0, which is not true.

Hence, 2 is the required value of k.

**6. For what values of p are the roots of the equation 4x ^{2} + px + 3 = 0 real and equal? **

Answer :

The given equation is 4x^{2} px + 3 = 0.

This is of the form ax^{2} + bx + c = 0, where a = 4, b = p and c = 3.

Therefore, D = b2 â€” 4ac = p^{2} – 4 x 4 x 3 =p^{2} – 48

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

Therefore, p^{2} – 48 = 0

\(\Rightarrow\)p2 = 48

\(\Rightarrow\)p= Â±4\(\sqrt{3}\)

Hence, 4\(\sqrt{3}\) and -4\(\sqrt{3}\) are the required values of p.

**7. Find the nonzero value of k for which the roots of the quadratic equation **

** 9x ^{2}-3kx + k = 0 are real and equal. **

Answer :

The given equation is 9x^{2} – 3kx + k = 0.

This is of the form ax^{2} + bx + c = 0, where a = 9, b = -3k and c = k.

Therefore, D = b^{2} – 4ac (-3k)^{2} – 4 x 9 x k = 9k^{2} – 36k

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

Therefore, 9k^{2} – 36k = 0

\(\Rightarrow\)9k(k – 4) = 0

\(\Rightarrow\)k = 0 or k – 4 = 0

\(\Rightarrow\)k = 0 or k = 4

But, k \(\neq\) 0 (Given)

Hence, the required value of k is 4.

**8. Find the values of k for which the quadratic equation (3k+ 1)x ^{2} + 2(k + 1)x + 1 = 0 has real and equal roots. **

Answer :

The given equation is (3k + 1)x^{2} + 2(k + 1)x + 1= 0.

This is of the form ax^{2 }+ bx+ c= 0, where a = 3k +1, b = 2(k + 1) and = 1.

Therefore, D = b^{2} – 4ac

= [2(k +1)]^{2} â€” 4 x (3k + 1) x 1

= 4(k^{2} + 2k + 1) â€” 4(3k + 1)

= 4k^{2} + 8k + 4 – 12k – 4

= 4k^{2} â€” 4k

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

Therefore, 4k^{2} – 4k = 0

\(\Rightarrow\)4k(k -1) = 0

\(\Rightarrow\)k = 0 or k â€”1= 0

\(\Rightarrow\)k= 0 or k =1

Hence, 0 and 1 are the required values of k.

**9. Find the values of p for which the quadratic equation **

** (2P+1)x ^{2} -(7p + 2) x + (7p – 3) = 0 has real and equal roots. **

Answer :

The given equation is (2p+ 1)x^{2} â€” (7p+ 2)x + (7p â€” 3) = 0.

This is of the form ax^{2} + bx + c = 0, where a = 2p +1, b = -(7p + 2) and c = 7p- 3.

Therefore, D = b^{2} â€” 4ac

= [â€”(7p + 2)]^{2} â€” 4 x (2p + 1) x (7p â€” 3)

= (49p^{2} + 28p + 4) â€” 4(14P^{2} + p – 3)

=49p^{2}+28p+4-56p^{2}-4p+12

= -7p^{2} + 24p+ 16

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

Therefore, -7p^{2} + 24p+ 16 = 0

\(\Rightarrow\)7p^{2} â€” 24p â€” 16 = 0

\(\Rightarrow\)7p^{2} – 28p + 4p â€” 16 = 0

\(\Rightarrow\)7p(p â€” 4) + 4(p â€” 4) = 0

\(\Rightarrow\) (p â€” 4)(7p+ 4) = 0

\(\Rightarrow\)p-4 =0 or 7p+4=0

\(\Rightarrow\)p=4 p=\(-\frac{4}{7}\)

Hence, 4 and \(-\frac{4}{7}\) are the required values of p.

**10. Find the values of p for which the quadratic equation **

** (9+1)x ^{2} – 6(p + 1) x + 3(p + 9) = 0, p \(\neq\) -1 has equal roots. Hence, find the root of **

**the equation. **

Answer :

The given equation is (p + 1)x^{2} â€” 6(p + 1)x + 3(p + 9) = 0.

This is of the form ax^{2} + bx + c = 0, where a = p +1, b = -6(p + 1) and c = 3(p + 9).

Therefore, D = b^{2} – 4ac

= [-6(p + 1)]^{2} â€” 4 x (p + 1) x 3(p + 9)

= 12(p + 1)13(p + 1) â€” (p + 9)]

= 12(p + 1)(2p â€” 6)

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

Therefore, 12(p + 1)(2p â€“ 6)=0

\(\Rightarrow\)p+1 =O or 2p-6=0

\(\Rightarrow\)p=-1 o p=3

But, p \(\neq\) -1 (Given)

Thus, the value of p is 3.

Putting p = 3, the given equation becomes 4x^{2} â€” 24x + 36 = 0.

\(\Rightarrow\)4x^{2} â€” 24x + 36 = 0

\(\Rightarrow\)4(x^{2 }â€” 6x + 9) = 0

\(\Rightarrow\)(x â€” 3)^{2} = 0

\(\Rightarrow\)x â€” 3 = 0

\(\Rightarrow\)x = 3

Hence, 3 is the repeated root of this equation.

**11. If -5 root of the quadratic equation 2x ^{2} + px – 15 = and the quadratic equation p (x^{2} + x) + k = 0 has equal roots, find the value of k. **

Answer :

It is given that -5 is a root of the quadratic equation 2x^{2}+ px â€”15 = 0.

Therefore, 2(-5)^{2} -p x (-5) â€” 15 = 0

\(\Rightarrow\)-5p+ 35 = 0

\(\Rightarrow\)p = 7

The roots of the equation px^{2} – px + k = 0 = 0 are equal.

Therefore, D=0

\(\Rightarrow\)p2 – 4pk = 0

\(\Rightarrow\) (7)^{2}-4 x 7 x k=0

\(\Rightarrow\)49 – 28k = 0

\(\Rightarrow\)k=\(\frac{49}{28}\)= \(\frac{7}{4}\)

Thus, the value of k is \(\frac{7}{4}\)

**12. If 3 Is a root of the quadratic equation x ^{2} – x + k = 0, find the value of P so that the root of the equation x^{2} + k (2x + k + 2)+p=0 are equal. **

Answer :

It is given that 3 is a root of the quadratic equation x^{2} – x + k = 0.

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

\(\Rightarrow\)k+6=0

\(\Rightarrow\)k = -6

The roots of the equation x^{2} + 2kx + (k^{2 }+ 2k +p) = 0 are equal.

Therefore, D=0

\(\Rightarrow\) (2k)^{2} – 4 x 1 x (k^{2} + 2k +p) = 0

\(\Rightarrow\) 4k^{2} – 4k^{2} – 8k- 4p = 0

\(\Rightarrow\) -8k- 4p= 0

\(\Rightarrow\)p= \(\frac{8k}{-4}\) =-2k

\(\Rightarrow\)p=-2x(-6)=12

Hence, the value of p is 12.

**13. If -4 is a root of the equation x ^{2} + 2x + 4p = 0, find the value of k for which the quadratic equation x2 + px +(1+ 3k) + 7(3 + 2k)= 0 has equals roots. **

Answer :

It is given that â€”4 is a root of the quadratic equation x^{2} + 2x + 4p = 0.

Therefore, (-4)^{2} +2 x (-4)+4p=0

\(\Rightarrow\)16â€” 8+4p= 0

\(\Rightarrow\)4p+8= 0

\(\Rightarrow\)p=-2

The equation x^{2}+px(1 + 3k) + 7(3 + 2k) = 0 has equal roots.

Therefore, D = 0

\(\Rightarrow\) [p(1 + 3k)]^{2} â€” 4 x 1 x 7(3 + 2k) = 0

\(\Rightarrow\) [-2(1 + 3k)]^{2} â€” 28(3 + 2k) = 0

\(\Rightarrow\)4(1 + 6k + 9k^{2}) â€” 28(3 + 2k) = 0

\(\Rightarrow\)4(1 + 6k + 9k^{2} â€” 21 â€” 14k) =0

\(\Rightarrow\)9k^{2} â€” 8k â€” 20 = 0

\(\Rightarrow\)9k^{2} â€” 18k + 10k â€” 20 = 0

\(\Rightarrow\)9k(k â€” 2) + 10(k â€” 2) = 0

\(\Rightarrow\) (k â€” 2)(9k + 10) = 0

\(\Rightarrow\)k â€” 2=0 or 9k+ 10 =0

\(\Rightarrow\)k = 2 or k = â€”\(-\frac{10}{9}\)

Hence. the required value of k is 2 or â€”\(-\frac{10}{9}\)

**14. If the quadratic equation (1+m ^{2})x^{2 }+ 2mcx +c^{2} – a^{2}= 0 has equal roots.prove that c^{2} = a^{2}(1 + m^{2}).**

Answer :

Given:

(1 + m^{2})x^{2} + 2mcx + (c^{2} â€“ a^{2}) = 0

Here,

a= (1 + m^{2}), b = 2mc and c = (c^{2} â€” a^{2})

It is given that the roots of the equation are equal; therefore, we have:

D = 0

\(\Rightarrow\) (b^{2 }â€” 4ac) = 0

\(\Rightarrow\) (2mc)^{2} â€” 4 x (1 + m^{2}) x (c^{2} â€” a^{2}) = 0

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

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

\(\Rightarrow\) -4c^{2} + 4a^{2} + 4m^{2}a^{2} = 0

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

\(\Rightarrow\)a^{2} (1 + m^{2}) = c^{2 }

\(\Rightarrow\)c^{2} = a^{2}(1 + m^{2})

Hence proved.

**15. If the roots of the equation (c ^{2} – ab)x^{2} â€“ 2(a^{2}-bc)x+(b^{2}-ac) x (b2 – ac)= 0 are real and equal, show that either a = 0 or (a^{3} + b^{3} + c^{3}) = 3abc . **

Answer :

Given:

(c^{2} â€” ab)x^{2} â€” 2(a^{2} – bc)x + (b^{2} – ac) = 0

Here,

a = (c^{2} – ab), b = â€” 2(a^{2} â€” bc), c = (b^{2} â€” ac)

It is given that the roots of the equation are real and equal; therefore, we have:

D=0

\(\Rightarrow\) (b^{2} â€” 4ac) = 0

\(\Rightarrow\){-2(a^{2} – bc)}^{2} – 4 x (c^{2} – ab) x (b^{2} â€” ac) = 0

\(\Rightarrow\)4(a^{4} â€” 2a^{2}bc + b^{2}c^{2}) – 4(b^{2}c^{2} – ac^{3} â€” ab^{3} + a^{2}bc) = 0

\(\Rightarrow\)a^{4} â€” 2a^{2}bc – b^{2}c^{2} â€” b^{2}c^{2} + ac^{3} + ab^{3 }â€” a^{2}bc = 0

\(\Rightarrow\)a^{4} â€” 3a^{2}bc + ac^{3} + ab^{3} = 0

\(\Rightarrow\)a(a^{3} â€” 3abc + c^{3} + b^{3}) = 0

Now,

a = 0 or a^{3} â€” 3abc + c^{3} + b^{3} = 0

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

**16. Find the values of p for which the quadratic equation 2x ^{2} + px +8=0 has real roots. **

Answer :

Given:

2x^{2} + px + 8 = 0

Here,

a = 2, b = p and c = 8

Discriminant D is given by :

D = (b^{2} – 4ac)

= p^{2} â€” 4 x 2 x 8

= (p^{2} â€” 64)

If D \(\geq\) 0, the roots of the equation will be real.

\(\Rightarrow\) (p^{2} â€” 64) \(\geq\) 0

\(\Rightarrow\) (p + 8) (p â€” 8) \(\geq\) 0

\(\Rightarrow\)p \(\geq\)8 and p \(\leq\)-8

Thus, the roots of the equation are real for p \(\geq\) 8 and p \(\leq\) -8.

**17. Find the value of a for which the equation (a – 12)x ^{2} + 2 (a â€“ 12)x+2=0 has equal roots. **

Answer :

Given:

(a â€” 12)x^{2} + 2(a â€” 12)x + 2 = 0

Here,

a = (a â€” 12), b = 2(\(\alpha\) â€” 12) and c = 2

It is given that the roots of the equation are equal; therefore, we have:

D = 0

\(\Rightarrow\) (b^{2} â€” 4ac) = 0

\(\Rightarrow\) {2(\(\alpha\) â€” 12)}^{2} â€” 4 x (\(\alpha\) â€” 12) x 2 = 0

\(\Rightarrow\) 4{\(\alpha\)^{2} â€” 24\(\alpha\) + 144) â€” 8(\(\alpha\) â€” 12) = 0

\(\Rightarrow\)4\(\alpha ^{2}\) â€” 96\(\alpha\) + 576 â€” 8\(\alpha\) + 96 = 0

\(\Rightarrow\)4\(\alpha ^{2}\) â€” 104\(\alpha\) + 672 = 0

\(\Rightarrow\) \(\alpha ^{2}\) â€” 26\(\alpha\) + 168 = 0

\(\Rightarrow\) \(\alpha ^{2}\) â€” 14\(\alpha\) â€” 12\(\alpha\) + 168 = 0

\(\Rightarrow\) \(\alpha\) (\(\alpha\) â€” 14) â€” 12(\(\alpha\) â€” 14) =0

\(\Rightarrow\) (\(\alpha\) â€” 14)( \(\alpha\) â€” 12) =0

Therefore, \(\alpha\) = 14 or \(\alpha\) = 12

If the value of a is 12, the given equation becomes non-quadratic. Therefore, the value of \(\alpha\) will be 14 for the equation to have equal roots.

**18. Find the value of k for which the roots of 9x ^{2} + 8kx + 16 = 0 are real equal. **

Answer :

Given:

9x^{2} + 8kx + 16 = 0

Here,

a = 9, b = 8k and c = 16

It is given that the roots of the equation are real and equal; therefore, we have:

D = 0

\(\Rightarrow\) (b^{2} â€” 4ac) = 0

\(\Rightarrow\) (8k)^{2} â€” 4 x 9 x 16 = 0

\(\Rightarrow\)64k^{2} â€” 576 = 0

\(\Rightarrow\)64k^{2} = 576

\(\Rightarrow\)k^{2} = 9

\(\Rightarrow\)k= Â±3

Therefore, k = 3 or k = â€”3

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

**(i) kx ^{2}+6x + 1 = 0 (ii) x^{2} – kx + 9 = 0 **

**(iii)9x ^{2}+3kx + 4 = 0 (iv) 5x^{2}– kx + 1 =0 **

Answer :

(i) The given equation is kx^{2 }-I- 6x +1= 0

Therefore, D=6^{2}-4xkx1=36-4k

The given equation has real and distinct roots if D > 0.

Therefore, 36-4k>0

\(\Rightarrow\)4k < 36

\(\Rightarrow\)k<9

(ii) The given equation is x^{2} â€” kx + 9 = 0.

Therefore, D=(â€”k)^{2}-4x1x9=k^{2}-36

The given equation has real and distinct roots if D > 0.

Therefore, k^{2}-36>0

\(\Rightarrow\) (k â€” 6)(k + 6) > 0

\(\Rightarrow\)k<-6 or k>6

(iii) The given equation is 9x^{2} +3kx-1+4 = 0.

Therefore, D = (3k)^{2} â€” 4 x 9 x 4 = 9k^{2 }â€” 144

The given equation has real and distinct roots if D> 0.

Therefore, 9k^{2} â€” 144 > 0

\(\Rightarrow\)9 (k2 â€” 16) > 0

\(\Rightarrow\) (k â€” 4)(k + 4) > 0

\(\Rightarrow\)k < â€”4 or k > 4

(iv) The given equation is 5x^{2} â€” kx+ 1= 0.

Therefore, D=(â€”k)^{2}-4x 5 x 1=k^{2} â€”20

The given equation has real and distinct roots if D> 0.

Therefore, k^{2}-20 >0

\(\Rightarrow\)k^{2} – (2\(\sqrt{5}\))^{2}> 0

\(\Rightarrow\) (k-2\(\sqrt{5}\))(k+2\(\sqrt{5}\)) >0

\(\Rightarrow\)k < â€”2\(\sqrt{5}\) or k > 2\(\sqrt{5}\)

**20. If a and b are real and a \(\neq\) b then show that the roots of the equation (a-b)x ^{2}+5(a +b)x – 2(a – b) = 0 are real and unequal.**

Answer :

The given equation is

(a â€” b)x^{2} + 5(a + b)x â€” 2(a â€” b) = 0.

Therefore, D = [5(a + b)]^{2} â€” 4 x (a â€” b) x [ â€”2(a â€” b)]

= 25(a + b)2 + 8(a â€” b)^{2}

Since a and b are real and a \(\neq\) b, so (a â€” b)^{2}> 0 and (a +b)^{2}> 0.

Therefore, 8(a â€”b)^{2}> 0 â€¦â€¦â€¦.. (1) (Product of two positive numbers is always positive)

Also, 25(a + b)^{2}> 0 â€¦â€¦â€¦â€¦(2) (Product of two positive numbers is always positive)

Adding (1) and (2), we get

25(a + b)^{2} + 8(a â€” b)^{2 }> 0 (Sum of two positive numbers is always positive)

\(\Rightarrow\)D > 0

Hence, the roots of the given equation are real and unequal.

**21. 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 \(\frac{a}{b}\)= \(\frac{c}{d}\).**

Answer :

It is given that the roots of the equation (a^{2} + b^{2})x^{2} â€” 2(ac + bd)x + (c^{2} + d^{2}) = 0 are equal.

Therefore, D=0

\(\Rightarrow\) [-2(ac+ bd)]^{2} â€” 4 (a^{2} + b^{2}) (c^{2}+d^{2}) =0

\(\Rightarrow\) 4(a^{2}c^{2} +b^{2}d^{2} + 2abcd) â€” 4(a^{2 }c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2}) = 0

\(\Rightarrow\) 4 (a^{2} c^{2} + b^{2} d^{2} + 2abcd â€” a^{2}e â€” a^{2}d^{2} â€” b^{2} c^{2} – b^{2}d^{2}) =0

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

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

\(\Rightarrow\) (ad – bc)^{2 }= 0

\(\Rightarrow\)adâ€” bc = 0

\(\Rightarrow\)ad= bc

\(\Rightarrow\) \(\frac{a}{b}\)= \(\frac{c}{d}\)

Hence Proved.

**22. If the roots of the equations ax ^{2}+2bx+c=0 and bx^{2}-2\(\sqrt{ac}\)x + b = 0 are simultaneously real then prove that b^{2}=ac.**

Answer :

It is given that the roots of the equation ax^{2} + 2bx + c = 0 are real.

Therefore, D1=(26)^{2}-4xaxc\(\geq\)0

\(\Rightarrow\)4(b^{2} â€” ac) \(\geq\) 0

\(\Rightarrow\)b^{2} â€” ac \(\geq\) 0 â€¦â€¦â€¦â€¦. (1)

Also, the roots of the equation bx^{2} â€” 2 \(\sqrt{acx}\) +b = 0 are real.

Therefore, D_{2}=(-2\(\sqrt{ac}\)^{2}-4xbxb \(\geq\)0

\(\Rightarrow\)4(ac â€” b^{2}) \(\geq\)0

\(\Rightarrow\)â€”4 (b^{2} â€” ac) \(\geq\) 0

\(\Rightarrow\)b^{2}â€”ac\(\leq\)0 â€¦â€¦â€¦â€¦â€¦â€¦..(2)

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if

b^{2} â€” ac = 0

\(\Rightarrow\)bb^{2}= ac

### Key Features of RS Aggarwal Class 10 Solutions Chapter 10 â€“Â Quadratic Equations Ex 10F ( 10.6)

- It provides easy solutions to difficult and tricky questions.
- It is explained in a step by step manner for better understanding.
- It helps in clearing your doubts and concepts.
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