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

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

**(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} — 2\(\sqrt{6}\)

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

Therefore, Discriminant, D= b^{2} – 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

\(\Rightarrow\)^{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\(\sqrt{15}\)

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

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 }– 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 }– 4 x 1 x2 = 1 – 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} — 2ab+ b^{2})

= -4(a — 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} ± 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

\(\Rightarrow\)^{2} – 4 x 3 x 27 = 0

\(\Rightarrow\)^{2} – 324 = 0

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

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

\(\Rightarrow\)

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\(\sqrt{5}\)

\(\Rightarrow\)^{2} — 2\(\sqrt{5}\)

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

Therefore, = b^{2 }– 4ac=(-2\(\sqrt{5}\)^{2} – 40K

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

Therefore, 20k^{2} – 40k = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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 — 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

\(\Rightarrow\)

\(\Rightarrow\)

Hence, 4\(\sqrt{3}\)

**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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

But, k \(\neq\)

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} — 4 x (3k + 1) x 1

= 4(k^{2} + 2k + 1) — 4(3k + 1)

= 4k^{2} + 8k + 4 – 12k – 4

= 4k^{2} — 4k

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

Therefore, 4k^{2} – 4k = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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} — (7p+ 2)x + (7p — 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} — 4ac

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

= (49p^{2} + 28p + 4) — 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

\(\Rightarrow\)^{2} — 24p — 16 = 0

\(\Rightarrow\)^{2} – 28p + 4p — 16 = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, 4 and \(-\frac{4}{7}\)

**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} — 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} — 4 x (p + 1) x 3(p + 9)

= 12(p + 1)13(p + 1) — (p + 9)]

= 12(p + 1)(2p — 6)

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

Therefore, 12(p + 1)(2p – 6)=0

\(\Rightarrow\)

\(\Rightarrow\)

But, p \(\neq\)

Thus, the value of p is 3.

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

\(\Rightarrow\)^{2} — 24x + 36 = 0

\(\Rightarrow\)^{2 }— 6x + 9) = 0

\(\Rightarrow\)^{2} = 0

\(\Rightarrow\)

\(\Rightarrow\)

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 —15 = 0.

Therefore, 2(-5)^{2} -p x (-5) — 15 = 0

\(\Rightarrow\)

\(\Rightarrow\)

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

Therefore, D=0

\(\Rightarrow\)

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

\(\Rightarrow\)

\(\Rightarrow\)

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

\(\Rightarrow\)

\(\Rightarrow\)

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

Therefore, D=0

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

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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 —4 is a root of the quadratic equation x^{2} + 2x + 4p = 0.

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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

Therefore, D = 0

\(\Rightarrow\)^{2} — 4 x 1 x 7(3 + 2k) = 0

\(\Rightarrow\)^{2} — 28(3 + 2k) = 0

\(\Rightarrow\)^{2}) — 28(3 + 2k) = 0

\(\Rightarrow\)^{2} — 21 — 14k) =0

\(\Rightarrow\)^{2} — 8k — 20 = 0

\(\Rightarrow\)^{2} — 18k + 10k — 20 = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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

**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} – a^{2}) = 0

Here,

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

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

D = 0

\(\Rightarrow\)^{2 }— 4ac) = 0

\(\Rightarrow\)^{2} — 4 x (1 + m^{2}) x (c^{2} — a^{2}) = 0

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

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

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

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

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

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

Hence proved.

**15. If the roots of the equation (c ^{2} – ab)x^{2} – 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} — ab)x^{2} — 2(a^{2} – bc)x + (b^{2} – ac) = 0

Here,

a = (c^{2} – ab), b = — 2(a^{2} — bc), c = (b^{2} — ac)

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

D=0

\(\Rightarrow\)^{2} — 4ac) = 0

\(\Rightarrow\)^{2} – bc)}^{2} – 4 x (c^{2} – ab) x (b^{2} — ac) = 0

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

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

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

\(\Rightarrow\)^{3} — 3abc + c^{3} + b^{3}) = 0

Now,

a = 0 or a^{3} — 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} — 4 x 2 x 8

= (p^{2} — 64)

If D \(\geq\)

\(\Rightarrow\)^{2} — 64) \(\geq\)

\(\Rightarrow\)

\(\Rightarrow\)

Thus, the roots of the equation are real for p \(\geq\)

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

Answer :

Given:

(a — 12)x^{2} + 2(a — 12)x + 2 = 0

Here,

a = (a — 12), b = 2(\(\alpha\)

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

D = 0

\(\Rightarrow\)^{2} — 4ac) = 0

\(\Rightarrow\)^{2} — 4 x (\(\alpha\)

\(\Rightarrow\)^{2} — 24\(\alpha\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, \(\alpha\)

If the value of a is 12, the given equation becomes non-quadratic. Therefore, the value of \(\alpha\)

**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

\(\Rightarrow\)^{2} — 4ac) = 0

\(\Rightarrow\)^{2} — 4 x 9 x 16 = 0

\(\Rightarrow\)^{2} — 576 = 0

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

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

\(\Rightarrow\)

Therefore, k = 3 or k = —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

\(\Rightarrow\)

\(\Rightarrow\)

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

Therefore, D=(—k)^{2}-4x1x9=k^{2}-36

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

Therefore, k^{2}-36>0

\(\Rightarrow\)

\(\Rightarrow\)

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

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

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

Therefore, 9k^{2} — 144 > 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

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

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

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

Therefore, k^{2}-20 >0

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

\(\Rightarrow\)

\(\Rightarrow\)

**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 — b)x^{2} + 5(a + b)x — 2(a — b) = 0.

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

= 25(a + b)2 + 8(a — b)^{2}

Since a and b are real and a \(\neq\)^{2}> 0 and (a +b)^{2}> 0.

Therefore, 8(a —b)^{2}> 0 ……….. (1) (Product of two positive numbers is always positive)

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

Adding (1) and (2), we get

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

\(\Rightarrow\)

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

**21. If the roots of the equation (a ^{2}+b^{2})x^{2} – 2(ac + bd)x+ (c^{2}+d^{2})=0 are equal, prove that \(\frac{a}{b}\)**

Answer :

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

Therefore, D=0

\(\Rightarrow\)^{2} — 4 (a^{2} + b^{2}) (c^{2}+d^{2}) =0

\(\Rightarrow\)^{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}) = 0

\(\Rightarrow\)^{2} c^{2} + b^{2} d^{2} + 2abcd — a^{2}e — a^{2}d^{2} — b^{2} c^{2} – b^{2}d^{2}) =0

\(\Rightarrow\)^{2}d^{2} + 2abcd — b^{2}c^{2}) = 0

\(\Rightarrow\)^{2}d^{2} — 2abcd + b^{2 }c^{2}) = 0

\(\Rightarrow\)^{2 }= 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence Proved.

**22. If the roots of the equations ax ^{2}+2bx+c=0 and bx^{2}-2\(\sqrt{ac}\)**

Answer :

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

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

\(\Rightarrow\)^{2} — ac) \(\geq\)

\(\Rightarrow\)^{2} — ac \(\geq\)

Also, the roots of the equation bx^{2} — 2 \(\sqrt{acx}\)

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

\(\Rightarrow\)^{2}) \(\geq\)

\(\Rightarrow\)^{2} — ac) \(\geq\)

\(\Rightarrow\)^{2}—ac\(\leq\)

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

b^{2} — ac = 0

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