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Here we have,

2x² – x + 9 = x² + 4x + 3

⇒ x² – 5x + 6 = 0

LHS = x² – 5x + 6

Substituting x = 2 in LHS, we get

(2)² – 5(2) + 6

⇒ 4 – 10 + 6 = 0 = RHS

⇒ LHS = RHS

Thus, x = 2 is a solution of the given equation.

Similarly,

Substituting x = 3 in LHS, we get

(3)² – 5(3) + 6

⇒ 9 – 15 + 6 = 0 = RHS

⇒ LHS = RHS

Thus, x = 3 is a solution of the given equation.

False, since the discriminant in this case is – 4ac which can still be non-negative if a and c are of opposite signs or if one of a or c is zero.

3x² – 2√6 x + 2 = 0

Compare given equation with the general form of quadratic equation, which is ax² + bx + c = 0

a = 3, b = – 2√6, c = 2

Using Discriminant Formula:

D = b² – 4ac

= (– 2√6)² – 4 x 3 x 2

= 24 – 24

= 0

Roots of equation are real and equal.

(a) \(\frac{-15}{4}\)

Given,α, β, γ are the roots of the 2x³ – 3x² + 6x + 1 = 0

Since S₁ = \(\frac{\text{Coefficient of}\,x^2}{\text{Coefficient of}\,x^3}\) = S₁ = α + β + γ = \(-\big(\frac{-3}{2}\big)\) = \(\frac{3}{2}\)

Since S₂ = \(\frac{\text{Coefficient of}\,x}{\text{Coefficient of}\,x^3}\) = S₂ = αβ + βγ + αγ = \(\frac{6}{2}\) = 3

Since S₃ = \(\frac{\text{Coefficient of constant term}}{\text{Coefficient of}\,x^3}\) = S₃ = αβγ = \(-\frac12\)

Now, α² +β²+ γ² = (α +β + γ)² - 2(αβ + βγ + αγ)

= \(\big(\frac{3}{2}\big)^2\) - 2 x 3 = \(\frac{9}{4}-6\) = \(\frac{-15}{4}\)

(a) \(\frac{2}{3}\)

The equation x² + px + \(\frac{p}{4}+\frac{1}{2}\) = 0  has real roots if the discriminant D ≥ 0.

⇒ p² – 4 \(\bigg(\frac{p}{4}+\frac{1}{2}\bigg)\)≥ 0 ⇒ p² - p - 2 ≥ 0

⇒ p² – 2p + p – 2 ≥ 0 ⇒ p(p – 2) + 1 (p – 2) ≥ 0 

⇒ (p – 2) (p + 1) ≥ 0 

⇒ (p – 2) ≥ 0 and (p + 1) ≥ 0 

⇒ p ≥ 2 or p ≤ –1 

The condition p ≤ –1 is not admissible as 0 ≤ p ≤ 5. 

Now p ≥ 2 ⇒ p can take up the value 2 or 3 or 4 or 5 from the given values. {0, 1, 2, 3, 4, 5} 

∴ Probability (Roots of given equation are real)

= \(\frac{\text{Number of values p can take}}{\text{Given number of values}}\) = \(\frac{4}{6}=\frac{2}{3}.\)

when x > 0 

x² + x-1 = 1 

x² + x - 2 = 0 

(x-1) (x+2) = 0 

x = 1, -2 

when x < 0 

x² - x+1 = 1 

x² – x = 0 

x(x-1) = 0 

x = 0, 1 

hence the given equation has 3 solutions and they are x = 0, 1, -1.

(b) 1

Let the roots of the equation kx² – 5x + 6 = 0 be α and β. 

Then, α + β = \(\frac{5}{k}\)                ...(i) 

αβ = \(\frac{6}{k}\)                                 ...(ii)

Given \(\frac{α}{​​β}\) = \(\frac{2}{3}\) ⇒ α = \(\frac{2}{3}\)β

∴ From (i) and (ii),

  \(\frac{2}{3}\)β + β = \(\frac{5}{k}\) and \(\frac{2}{3}\)β² = \(\frac{6}{k}\)

⇒ \(\frac{5}{3}\)β = \(\frac{5}{k}\) and  β² = \(\frac{9}{k}\) ⇒ β = \(\frac{3}{k}\) and  β² = \(\frac{9}{k}\) 

⇒ \(\frac{9}{k^2}\) = \(\frac{9}{k}\) ⇒ 9k² - 9k = 0 k(k - 1) = 0 ⇒ k = 0 or 1

But k = 0 does not satisfy the condition, so k = 1.

(i) \(x^{2}-3x+2=0,x=2,x=1\) 

For x = 2,

2² – 3 × 2 + 2 = 0

⇒ 0 = 0

Thus, x = 2 is a solution.

For, x = 1 

1² – 3 × 1 + 2 = 0 

⇒ 0 = 0 

Thus, x = 1 is a solution.

(ii) \(x^{2}+x+1=0,x=0,x=1\)

For x = 0, 

⇒ 0 + 0 + 1 = 0 

⇒ 1 = 0 which is not true thus x = 0 is not a solution 

For x = 1, 

⇒ 1 + 1 + 1 = 0 

⇒ 3 = 0 which is not true thus x = 1 is not a solution

(iii) \(x^{2}-3\sqrt{3}+6=0,x=\sqrt{3},x=-2\sqrt{3}\)

For x= √3 

⇒ 3 – 3√3 × √3 + 6 = 0 

⇒ 3 – 9 + 6 = 0 

⇒ 0 = 0 

Thus, x = √3 is a solution 

For x = -2√3 

⇒ (-2√3)² – 3√3 × -2√3 + 6 = 0 

⇒ 4 × 3 + 18 + 6 = 0 

⇒ 36 = 0 which is not true, thus x = -2√3 is not a solution

(iv) \(x+\frac{1}{x}=\frac{13}{6},x=\frac{5}{6},x=\frac{4}{3}\)

For x = 5/6

\(\Rightarrow \frac{5}{6}+\frac{6}{5}=\frac{13}{6}\)

\(\Rightarrow \frac{61}{30}=\frac{13}{6}\)

⇒ 61 = 65 which is not true, thus x = 5/6 is not a solution

For x = 4/3

\(\Rightarrow\frac{4}{3}+\frac{3}{4}=\frac{13}{6}\)

⇒ 25/12 = 13/6 

⇒ 25 = 26 which is not true, thus x = 4/3 is not a solution

(v) \(2x^{2}-x+9=x^{2}+4x+3,x=2,x=3\)

For x = 2, 

⇒ 2 × 4 – 2 + 9 = 4 + 4 × 2 + 3 

⇒ 15 = 15, thus x = 2 is a solution. 

For x = 3 

⇒ 2 × 9 – 3 + 9 = 9 + 4 × 3 + 3 

⇒ 24 = 24, thus x = 3 is also a solution

(vi) \(x^{2}-\sqrt{2}x-4=0,=-\sqrt{2},x=-2\sqrt{2}\)

For x = -√2, 

⇒ 2 - √2 × -√2 – 4 = 0 

⇒ 2 + 2 – 4 = 0 

⇒ 0 = 0 

Thus, x = -√2 is a solution 

For x = -2√2 

⇒ 4 × 2 - √2 × -2√2 – 4 = 0 

⇒ 8 + 8 – 4 = 0 

⇒ 12 = 0 which is not true, thus x = -2√2 is not a solution

(vii) \(a^{2}x^{2}-3abx+2b^{2}=0,x=a/b,x=b/a\)

For, x = a/b

\(\Rightarrow a^{2}\times\frac{a^{2}}{b^{2}}-3ab\times\frac{a}{b}+2\times b^{2}=0\)

⇒ a⁴/b² – 3a² + 2b² = 0 which is not true, thus x = a/b is not a solution

For x = b/a

\(\Rightarrow a^{2}\times\frac{a^{2}}{b^{2}}-3ab\times\frac{b}{a}+2 b^{2}=0\)

⇒ b² – 3b² + 2b² = 0 

⇒ 0 = 0 , 

thus x = b/a is a solution

No, since the equation is simplified to x² + 3 = 0 whose discriminant is –12.

No, a quadratic equation with integral coefficients may or may not have integral roots.

Justification

Consider the following equation,

8x² – 2x – 1 = 0

The roots of the given equation are ½ and – ¼ which are not integers.

Hence, a quadratic equation with integral coefficient might or might not have integral roots.

Put x = 3/2 in the polynomial 2x² - 7x + 6

= 2x² - 7x + 6

= 2 (3/2)² - 7(3/2) + 6

= 2 x 9/4 - 21/2 + 6

= 9/2 - 21/2 + 6

= (9 - 21 + 12)/2

= (21-21)/2

= 0

x = 3/2 is a solution of the equation.

The correct option is (A) -b/a

If α and β are the roots of a quadratic equation ax² + bx + c = 0 then α + β = -b/a.

(c) HP

If the roots of the equation a(b – c)x² + b(c – a)x + c(a – b) = 0 are equal, then Discriminant (D) = 0, i.e., 

⇒ b² (c – a)² – 4a(b – c) c(a – b) = 0. 

⇒ b² (c² + a² – 2ac) – 4ac (ab – ca – b² + bc) = 0 

⇒ b²c² + b² a² – 2ab²c – 4a²bc + 4a²c² + 4ab²c – 4abc² = 0 

⇒ a²b² + b²c² + 4a²c² + 2ab²c – 4a²bc – 4abc² = 0 

⇒ (ab + bc – 2ac)² = 0 ⇒ ab + bc – 2ac = 0

⇒ ab + bc = 2ac ⇒ \(\frac{1}{c}+\frac{1}{a}=\frac{2}{b}\) ⇒ a, b, c are in H.P.

(c) ± 2, ± √2 

Let y = 5 + 2√6. Then \(\frac{1}{y}\) = 5 - 2√6. 

Thus the given expression reduces to \(y^{x^2-3}\) + \(\big(\frac{1}{y}\big)^{x^2-3}\) = 10

Again let \(y^{x^2-3}\) = t. Then,

t + \(\frac{1}{t}\) = 10 ⇒ t² - 10t + 1 = 0

⇒ t = \(\frac{10±\sqrt{100-4}}{2}\) = \(\frac{10 ±\sqrt{96}}{2}\)

= \(\frac{10 ±4\sqrt{6}}{2}\) = 5 ± 2√6

\((5+2\sqrt6)^{x^2-3}\) = 5 ± 2√6 = (5 ± 2√6)^±1

⇒ x² – 3 = 1 or x² – 3 = – 1 

⇒ x² = 4 or x² = 2 

⇒ x = ± 2 or x = ± √2

Answer: (A) -b/a

If α and β are the roots of a quadratic equation ax² + bx + c = 0 then α + β = -b/a.

(b) 4

2| \(x\) |² – 5| \(x\) | + 2 = 0 

⇒ (2| \(x\) | – 1) (| \(x\) | – 2) = 0 

⇒ | \(x\) | = \(\frac{1}{2}\), 2 ⇒ x = ± \(\frac{1}{2}\), ± 2 

So, there are 4 solutions.

No, 0.2 is not a root of the equation x² - 0.4 = 0.

If we substitute the value 0.2 in place of x in the equation, x² - 0.4 = 0

⇒ (0.2)² - 0.4 ≠ 0

(b) (ac³)^1/4 + (a³c)^1/4 + b = 0

Let α, β be the roots of the equation ax² + bx + c = 0. Then,

α + β = \(-\frac{b}{a}\)                     .....(i)

αβ = \(\frac{c}{a}\)                            ......(ii)

and   β = α^1/3                .......(iii)

∴ From (ii) and (iii), α. (α)^1/3 = \(\frac{c}{a}\) ⇒ α^4/3 = \(\frac{c}{a}\) ⇒ α = \(\bigg(\frac{c}{a}\bigg)^{3/4}\)

∴ β = \(\bigg(\big(\frac{c}{a}\big)^{3/4}\bigg)^{1/3}\) = \(\bigg(\frac{c}{a}\bigg)^{1/4}\)

∴ Putting these values of a and b in eqn. (i), we have

\(\bigg(\frac{c}{a}\bigg)^{3/4}\) + (c/a)^1/4 = \(-\frac{b}{a}\)

⇒ a. a^–3/4 c^3/4 + a. a^–1/4 c^1/4 = – b 

⇒ a^1/4 c^3/4 + a^3/4 c^1/4 + b = 0 

⇒ (ac³)^1/4 + (a³c)^1/4 + b = 0.

(c) \(\frac{1}{2}\bigg(\frac{b}{a}-\frac{q}{p}\bigg).\)

As α, β are the roots of the equation ax² + bx + c = 0, so

α + β = \(-\frac{b}{a}\), αβ = \(\frac{c}{a}\)

Also, (α + x), (β + x) are the roots of the equation 

px² + qx + r = 0, then α + \(x\) + β + \(x\) = – \(\frac{q}{p}\)

and (α + x) (β + x) = \(\frac{r}{p}\) ⇒ α + β + 2x = – \(\frac{q}{p}\)

⇒ \(\frac{-b}{a}\) + 2x = – \(\frac{q}{p}\) ⇒ K = \(\frac{1}{2}\bigg(\frac{b}{a}-\frac{q}{p}\bigg).\)

Correct option is (A) ± 2√6

For equal roots, we have

\(b^2-4ac=0\)

\(\Rightarrow k^2-4\times2\times3=0\)

\(\Rightarrow k^2=24\)

\(\Rightarrow k=\pm\sqrt{24}\)

\(=\pm\sqrt{4\times6}=\pm2\sqrt6\)

Correct option is C) ± 2√6

Correct option is (C) p² – 2q

Given that a and b are roots of equation

\(x^2-px + q = 0\)

\(\therefore\) Sum of roots \(=\frac{-(-p)}1=p\)

\(\Rightarrow\) a+b = p   _____________(1)

And product of roots \(=\frac q1=q\)

\(\Rightarrow\) ab = q    _____________(2)

Now, \(a^2+b^2=(a+b)^2-2ab\)

= \(p^2-2q\)

Correct option is C) p² – 2q

Correct option is (D) q²/4p

Given that quadratic equation \(Px^2+qx+r=0\) has equal roots.

Then its discriminant must be zero.

i.e., D = 0

\(\Rightarrow(-q)^2-4\times p\times r=0\)

\(\Rightarrow q^2-4pr=0\)

\(\Rightarrow4pr=q^2\)

\(\Rightarrow\) \(r=\frac{q^2}{4p}\)

Correct option is D) q²/4p

The correct answer is (B) a, b, c are real numbers and a ≠ 0.

A quadratic equation in the variable x is of the form ax² + bx + c = 0 where a, b, c are real numbers and a ≠ 0.

If roots of given equation are real and distinct then D = b² – 4ac > 0 

Here a = 1, b = 4 and c = k 

So, 

4² – 4 (1) (k) >0 

16 – 4k > 0 

16 > 4k 

K< 4

Answer: (A) k < 4

The given equation is x² + 4x + k = 0

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

D = b² – 4ac

= (4)² – 4 x 1 x k

= 16 – 4k

The given equation will have real and distinct roots, if

D > 0

16 – 4k > 0

16 > 4k

4k < 16

k < 16/4

k < 4

Yes, the roots will be equal and opposite in sign.

Given, x² + bx + c = 0, b = 0, c < 0

⇒ x² – c = 0

\(x = \pm \sqrt c\)

Hence, the roots are equal and opposite in sign.

(c) 3 : 1 and 1 : 2

2x² – 7xy + 3y₂ = 0

⇒ 2\(\bigg(\frac{x}{y}\bigg)^2\) - 7\(\bigg(\frac{x}{y}\bigg)\) + 3 = 0

∴ \(\frac{x}{y}\) = \(\frac{-b±\sqrt{b^2-4ac}}{2a}\) = \(\frac{7±\sqrt{49-24}}{4}\) = \(\frac{7±5}{4}\) = 3, \(\frac{1}{2}\)

∴ x : y = 3 : 1 and 1 : 2.

Correct answer is Yes

Correct option is (C) 2

Let x = \(\sqrt{2+\sqrt{2+\sqrt{2\,+......}}}\)

\(\Rightarrow x=\sqrt{2+x}\)

\(\Rightarrow x^2=2+x\)      (By squaring both sides)

\(\Rightarrow x^2-x-2=0\)

\(\Rightarrow x^2-2x+x-2=0\)

\(\Rightarrow\) x (x - 2) + 1 (x - 2) = 0

\(\Rightarrow\) (x + 1) (x - 2) = 0

\(\Rightarrow\) x = -1 or x = 2

\(\because\) \(\sqrt{2+\sqrt{2+\sqrt{2\,+......}}}\) \(>\sqrt2>0\)

\(\therefore\) \(x\neq-1\)

\(\therefore\) x = 2

Hence, \(\sqrt{2+\sqrt{2+\sqrt{2\,+......}}}=2\)

Correct option is C) 2

Correct option is (B) q² < 4 pr

The quadratic equation \(px^2+qx+r=0\) has imaginary roots if \(q^2-4\,pr<0\)

\(\Rightarrow\) \(q^2<4\,pr\)

Correct option is B) q² < 4 pr

Correct option is (A) ± 2

Given quadratic equation is \(x^2+px-3=0.\)

Let \(\alpha\;and\;\beta\) are roots.

\(\therefore\) Sum of roots = -p

\(\Rightarrow\) \(\alpha+\beta\) = -p    _____________(1)

And product of roots = -3

\(\therefore\) \(\alpha\beta\) = -3         _____________(2)

Now, \(\alpha^2+\beta^2=10\)    (Given)

\(\Rightarrow\) \((\alpha+\beta)^2-2\alpha\beta=10\)

\(\Rightarrow(-p)^2-2\times-3=10\)

\(\Rightarrow p^2\) = 10 - 6 = 4

\(\Rightarrow\) p = \(\pm\,2\)

Correct option is A) ± 2

(c) ab + b² = 2ac.

Let α, β be the roots of the equation ax² + bx + c = 0. Then, 

α + β = –\(\frac{b}{a}\), αβ = \(\frac{c}{a}\) 

Given, α + β = α² + β² 

i.e., α + β = (α² + β)² – 2αβ

⇒ \(\frac{-b}{a}\) = \(\frac{b^2}{a^2}\) - \(\frac{2c}{a}\)

⇒ \(-\frac{ab}{a^2}\) = \(\frac{b^2}{a^2}\) - \(\frac{2ac}{a^2}\) ⇒ ab + b² = 2ac.

Let α, β be the roots of the equation 7x² + 12x + 18 = 0.

\(\bigg[\)For a quadratic equation ax² + bx + c = 0, sum of roots = \(-\frac{a}{b}\), product of roots = + \(\frac{c}{a}\)\(\bigg]\)

∴ α + β = \(\frac{12}{7}\) and αβ = \(\frac{18}{7}\)

⇒ (α + β)² = \(\bigg(\frac{-12}{7}\bigg)^2\) ⇒ α² + β² + 2αβ = \(\frac{144}{49}\)

⇒ α² + β² = \(\frac{144}{49}\) - \(\frac{36}{7}\) = \(\frac{-108}{49}\)

∴ Required ratio = α² + β² : αβ = \(\frac{​​\frac{-108}{49}}{\frac{18}{7}}\) = \(-\frac{6}{7}\) = – 6 : 7.

i. x² – 4x + 4= 0 

Comparing the above equation with

ax² + bx + c = 0, we get

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

∴ ∆ = b² – 4ac 

= (- 4)² – 4 × 1 × 4 

= 16 – 16 

∴ ∆ = 0

∴ Roots of the given quadratic equation are real and equal.

ii. 2y² – 7y + 2 = 0

Comparing the above equation with

ay² + by + c = 0, we get

a = 2, b = -7, c = 2

∴ ∆ = b² – 4ac 

= (- 7)² – 4 × 2 × 2 

= 49 – 16 

∴ ∆ = 33 

∴ ∆ > 0

∴ Roots of the given quadratic equation are real and unequal.

iii. m² + 2m + 9 = 0

Comparing the above equation with

am² + bm + c = 0, we get 

a = 1, b = 2, c = 9 

∴ ∆ = b² – 4ac 

= (2)² – 4 × 1 × 9

= 4 – 36 

∴ ∆ = -32 

∴ ∆ < 0

∴ Roots of the given quadratic equation are not real.

The correct answer is (A) k < 4.

The given equation is x² + 4x + k = 0

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

D = b² – 4ac

= (4)² – 4 x 1 x k

= 16 – 4k

The given equation will have real and distinct roots, if

D > 0

16 – 4k > 0

16 > 4k

4k < 16

k < 16/4

k < 4

Answer: (D) x² – 5x + 3 = 0

Clearly (x² – 5x + 3) is a quadratic polynomial.

x² – 5x + 3 = 0 is a quadratic equation.

Answer: (A) k < 4

The given equation is x² + 4x + k = 0

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

D = b² – 4ac

= (4)² – 4 x 1 x k

= 16 – 4k

(d) b = 0. but a ≠ 0, c ≠ 0.

Given equation is ax² + bx + c = 0.

∴Roots are x = \(\frac{-b±\sqrt{b^2-4ac}}{2a}\)

Given \(\bigg[\frac{-b+\sqrt{b^2-4ac}}{2a}\bigg]\) = - \(\bigg[\frac{-b-\sqrt{b^2-4ac}}{2a}\bigg]\)

⇒ – b + \(\sqrt{b^2-4ac}\) = b + \(\sqrt{b^2-4ac}\)

⇒ 2b = 0 ⇒ b = 0. but a ≠ 0, c ≠ 0.

Correct option is (B) 34

Given quadratic equation is \(x^2+8x+15=0\)

Let roots are \(\alpha\;and\;\beta.\)

\(\therefore\) Sum of roots \(=\frac{-8}1=-8\)

\(\Rightarrow\) \(\alpha+\beta\) = -8         _____________(1)

And product of roots \(=\frac{15}1=15\)

\(\Rightarrow\) \(\alpha\beta=15\)            _____________(2)

Now, \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\)

\(=(-8)^2-4\times15\)

= 64 - 60

= 4 \(=2^2\)

\(\therefore\) \(\alpha-\beta=2\)         _____________(3)

By adding equations (1) & (3), we get

\((\alpha+\beta)+(\alpha-\beta)=-8+2\)

\(\Rightarrow\) \(2\alpha=-6\)

\(\Rightarrow\) \(\alpha=\frac{-6}2=-3\)

\(\therefore\) \(\beta=-8-\alpha\)        (From (1))

= -8 - (-3)

= -8+3 = -5

\(\therefore\) \(\alpha^2+\beta^2=(-3)^2+(-5)^2\)

= 9+25 = 34

Alternative :-

\(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\)

\(=(-8)^2-2\times15\)

= 64 - 30 = 34

Correct option is B) 34

Correct option is (C) ± 8

For equal roots, we have D = 0

\(\Rightarrow p^2-4\times1\times16=0\)

\(\Rightarrow p^2=64\)

\(\Rightarrow p=\pm8\)

Correct option is C) ± 8

given α² + β² = 9 

(α + β)²- 2 αβ = 9

Given x²- px + 16 = 0 and α, β are roots of the equation then 

Sum of roots α + β = p 

Product of roots α β = 16 

Substituting these in (α + β)²- 2 αβ = 9 we get, 

p² – 2(16) = 9 

p² = 41 

P = ±√41

i. x² +7x – 1 = 0

Comparing the above equation with

ax² + bx + c = 0, we get

a = 1, b = 7, c = -1

∴ b² – 4ac = (7)² – 4 × 1 × (-1)

= 49 + 4

∴ b² – 4ac = 53

ii. 2y² – 5y + 10 = 0 

Comparing the above equation with

ay² + by + c = 0, we get 

a = 2, b = -5, c = 10 

∴ b² – 4ac = (-5)² -4 × 2 × 10 

= 25 – 80 

∴ b² – 4ac = -55

iii. √2 x² + 4x + 2√2 = 0

Comparing the above equation with

ax + bx + c = 0, we get

a = √2, b = 4, c = 2√2 

∴ b² – 4ac = (4)² – 4 × √2 × 2√2 

= 16 – 16 

∴ b² – 4ac = 0

Answer: (A) 1

The given equation is 2x² – 7x + 6 = 0

Here, a = 2, b = -7 and c = 6

Discriminant (D) = b² – 4ac

= (-7)² – 4 x 2 x 6

= 49 – 48

= 1

2x² – 7x + 6 = 0

Compare given equation with the general form of quadratic equation, which is

ax² + bx + c = 0

Here, a = 2, b = -7 and c = 6

Discriminant formula: D = b² – 4ac

(-7)² – 4 x 2 x 6

= 1

(a)  5\(\sqrt{10}\).

Given, the roots of the given equation 12x² – mx + 5 = 0 are in the ratio 3 : 2. Let the roots of the given equation be 3α and 2α. Then, 

Sum of roots = 3α + 2α = \(\frac{m}{12}\) ⇒ 5α = \(\frac{m}{12}\)            ........(i)

and (3α)(2α) = \(\frac{5}{12}\) ⇒ 6α² = \(\frac{5}{12}\) ⇒ α² = \(\frac{5}{72}\)

⇒ α = \(\sqrt{\frac{5}{12}}\)

∴ From (i) and (ii)

5. \(\sqrt{\frac{5}{12}}\) = \(\frac{m}{12}\) ⇒ m = 60\(\sqrt{\frac{5}{12}}\) = 60.\(\frac{\sqrt5}{6\sqrt2}\) = 10\(\sqrt{\frac{5}{2}}\)

= 10.\(\frac{\sqrt5}{\sqrt2}\).\(\frac{\sqrt2}{\sqrt2}\) = \(\frac{10}{2}\). \(\sqrt{10}\) = 5\(\sqrt{10}\).

Equating the expression with 0,

3x² + 5x + 2 = 0

On factorising it further,

3x² + 3x + 2x + 2 = 0

3x(x + 1) + 2(x + 1) = 0

(3x + 2) (x + 1) = 0

x = -2/3 or x = -1

The zeroes of 3x² + 5x + 2 are – 2/3 and -1.

The sum of the two zeros of the quadratic equation is given by \(-b/a\)

Here it’s given \(-b/a\) = 2\(\sqrt{3}\)

The product of the quadratic equation is \(c/a\)

Here \(c/a\) = 2

the quadratic equation is of the form ax² + b x + c = 0

or x² + (sum of the roots) x + product of the roots = 0

\(=\text{x}^2-2\sqrt{3}\) x + 2

f(x) = k(x² – \(2\sqrt{3}\) x + 2), where k is any real number

Equating the expression with 0,

2x² - 5x + 2 = 0

On factorising it further,

2x² - 4x- x + 2 = 0

2x(x - 2) -1(x - 2) = 0

(2x - 1) (x - 2) = 0

x = 1/2 or x = 2

The zeroes of 2x² - 5x + 2 are 1/2 and 2.

Correct option is (C) 0

Sum of roots of \(x^ 2-16=0\) is \(\frac{-b}a=\frac{-0}1=0\)

(By comparing \(x^ 2-16=0\) with \(ax^2+bx+c=0,\) we get a = 1, b = 0, c = -16)

Correct option is C) 0

(i) 2x² - 7x + 6 = 0

Here 

a = 2

b = - 7

c = 6

Discriminant D is diven by: 

D = b² - 4ac

= (-7)² - 4 x 2 x 6

= 49 - 48

= 1

(ii) 3x² - 2x + 8 = 0

Here

a = 3,

b = -2,

c = 8

Discriminant D is given by:

D = b² - 4ac

= (-2)² - 4 x 3 x 8

= 4 - 96

= - 92

(iii) 2x² - 5\(\sqrt{2x}\) + 4 = 0

Here

a = 2

b = - 5√2.

c = 4

Discriminant D is given by:

D = b² - 4ac

= (- 5√2)² - 4 x 2 x 4

= (25 x 2) - 32

= 50 - 32

= 18