Quadratic Equation and Inequalities

Let the roots of the quadratic equation be $ \alpha $ and $ \beta $.

From the relation between coefficients and roots, we know that:

$ \alpha + \beta = \frac{-b}{a} = k_1 \quad \Rightarrow \quad b = -a k_1 \, \text{where} \, k_1 \in \mathbb{Z} $

Also,

$ \alpha \beta = \frac{c}{a} = k_2 \quad \Rightarrow \quad c = a k_2 \, \text{where} \, k_2 \in \mathbb{Z} $

Since $ a, b, c $ are in arithmetic progression, we have:

$ c - b = b - a $

Substituting the values of $ b $ and $ c $ from above:

$ a k_2 + a k_1 = 2a k_1 + a \Rightarrow k_2 = -2k_1 - 1 $

Now, the quadratic equation becomes:

$ x^2 - k_1 x + k_2 = 0 $

Substituting the value of $ k_2 $:

$ x^2 - k_1 x - 2k_1 - 1 = 0 $

Since $ \alpha \beta = k_2 $ is odd, both roots $ \alpha $ and $ \beta $ must be odd integers.

Let us test some possible integer values. For example, if $ c = 15 $, the pairs of integer roots could be $ (-1, -15) $ or $ (-3, -5) $.

If the roots are $ -1 $ and $ -15 $, then:
Sum of roots $ = -16 \Rightarrow b = 16, \; a = 17 $
The quadratic becomes $ 17x^2 + 16x + 15 = 0 $.

This does not satisfy the condition because its roots are not the expected $ -1 $ and $ -15 $.

Now, if the roots are $ -3 $ and $ -5 $, then:
Sum of roots $ = -8, \; \text{Product} = 15 $
Hence, the quadratic is $ x^2 + 8x + 15 = 0 $.

This equation fits our condition with integer roots.

For the equation $ x^2 - k_1 x - 2k_1 - 1 = 0 $, substitute $ x = 3 $:

$ 9 - 3k_1 - 2k_1 - 1 = 0 \Rightarrow 8 - 5k_1 = 0 \Rightarrow k_1 = \frac{5}{8} $

Since $ k_1 $ must be an integer, this is not possible. Hence, 3 cannot be a root of the quadratic equation.

To solve for the coefficient of $ x^3 $ in the function $ h(x) = f(x + 1) - g(x + 2) $, we start by expanding both $ f(x + 1) $ and $ g(x + 2) $.

First, expand $ f(x + 1) $:

$ f(x+1) = a_1 + 10(x+1) + a_2 (x+1)^2 + a_3 (x+1)^3 + (x+1)^4 $

From this expansion, the coefficient of $ x^3 $ in $ f(x+1) $ is $ a_3 + 4 $.

Next, expand $ g(x + 2) $:

$ g(x+2) = b_1 + 3(x+2) + b_2 (x+2)^2 + b_3 (x+2)^3 + (x+2)^4 $

Here, the coefficient of $ x^3 $ in $ g(x+2) $ is $ b_3 + 8 $.

Now, calculate the coefficient of $ x^3 $ in $ h(x) = f(x+1) - g(x+2) $:

$ \text{Coefficient of } x^3 \text{ in } h(x) = (a_3 + 4) - (b_3 + 8) = a_3 - b_3 - 4 $

Given that $ f(x) \neq g(x) $ for every $ x \in \mathbb{R} $, it follows that $ f(x) - g(x) \neq 0 $ for any $ x $. This means the equation:

$ (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3 = 0 $

cannot have any real roots. For this polynomial to have no real roots, the degree of the polynomial must be less than or equal to zero, implying that the coefficients of the highest degree terms must equate to zero. Thus, $ a_3 - b_3 = 0 $.

Substituting $ a_3 - b_3 = 0 $ into our earlier expression for the coefficient of $ x^3 $ in $ h(x) $ gives:

$ a_3 - b_3 - 4 = -4 $

Thus, the coefficient of $ x^3 $ in $ h(x) $ is $-4$.

Given quadratic polynomials, x² + 20x $-$ 2020 and x² $-$ 20x + 2020 having a, b distinct real and c, d distinct complex roots respectively.

So, a + b = $-$20, ab = $-$2020

and c + d = 20, cd = 2020

Now, ac(a $-$c) + ad(a $-$ d) + bc(b $-$c) + bd(b $-$ d)

= a²(c + d) $-$ a(c² + d²) + b²(c + d) $-$ b(c² + d²)

= (c + d) (a² + b²) $-$ (c² + d²) (a + b)

= (c + d)[(a + b)² $-$ 2ab] $-$ (a + b)[(c + d)² $-$ 2cd]

= 20[(20)² + 4040] + 20[(20)² $-$ 4040]

= 2 $ \times $ 20 $ \times $ (20)² = 40 $ \times $ 400 = 16000

Given quadratic equation

${x^2} - x - 1 = 0$ having roots $\alpha $ and $\beta $, ($\alpha $ > $\beta $)

So, $\alpha = {{1 + \sqrt 5 } \over 2}$ and $\beta = {{1 - \sqrt 5 } \over 2}$

and $\alpha + \beta = 1$, $\alpha $$\beta $ = $ - 1$

$ \because $ ${a_n} = {{{\alpha ^n} - {\beta ^n}} \over {\alpha - \beta }},\,n \ge 1$

So, ${a_{n + 1}} = {{{\alpha ^{n + 1}} - {\beta ^{n + 1}}} \over {\alpha - \beta }}$

${\alpha ^n} + {\alpha ^{n - 1}}\beta + {\alpha ^{n - 2}}{\beta ^2} + ... + \alpha {\beta ^{n - 1}} + {\beta ^n}$

${\alpha ^n} - {\alpha ^{n - 2}} - {\alpha ^{n - 3}}\beta - ... - {\beta ^{n - 2}} + {\beta ^n}$

[as $\alpha $$\beta $ = $ - $1]

= ${\alpha ^n} + {\beta ^n} - ({\alpha ^{n - 2}} + {\alpha ^{n - 3}}\beta + ... + {\beta ^{n - 2}})$

= ${\alpha ^n} + {\beta ^n} - {a_{n - 1}}$

$\left[ {as\,{a_{n - 1}} = {{{\alpha ^{n - 1}} - {\beta ^{n - 1}}} \over {\alpha - \beta }} = {\alpha ^{n - 2}} + {\alpha ^{n - 3}}\beta + ... + {\beta ^{n - 2}}} \right]$

$ \Rightarrow $ ${a_{n + 1}} + {a_{n - 1}} = {\alpha ^n} + {\beta ^n} = {b_n},\,\forall n \ge 1$

So, option (b) is correct.

Now, $\sum\limits_{n = 1}^\infty {{{{b_n}} \over {{{10}^n}}}} = \sum\limits_{n = 1}^\infty {{{{\alpha ^n} + {\beta ^n}} \over {{{10}^n}}}} $

[as, b_n = $\alpha $ⁿ + $\beta $ⁿ]

= $\sum\limits_{n = 1}^\infty {{{\left( {{\alpha \over {10}}} \right)}^n}} + \sum\limits_{n = 1}^\infty {{{\left( {{\beta \over {10}}} \right)}^n}} $

$ \because $ $\left[ {\left| {{\alpha \over {10}}} \right| < 1\,and\,\left| {{\beta \over {10}}} \right| < 1} \right]$

$ = {{{\alpha \over {10}}} \over {1 - {\alpha \over {10}}}} + {{{\beta \over {10}}} \over {1 - {\beta \over {10}}}} = {\alpha \over {10 - \alpha }} + {\beta \over {10 - \beta }}$

$ = {{10\alpha - \alpha \beta + 10\beta - \alpha \beta } \over {(10 - \alpha )(10 - \beta )}}$

$ = {{10(\alpha + \beta ) - 2\alpha \beta } \over {100 - 10(\alpha + \beta ) + \alpha \beta }}$

$ = {{10(1) - 2( - 1)} \over {100 - 10(1) - 1}}$

[as $\alpha $ + $\beta $ = 1 and $\alpha $$\beta $ = $ - $1]

$ = {{12} \over {89}}$

So, option (a) is not correct.

$ \because $ ${\alpha ^2} = \alpha + 1$ and ${\beta ^2} = \beta + 1$

$ \Rightarrow $${\alpha ^{n + 2}} = {\alpha ^{n + 1}} + {\alpha ^n}$ and ${\beta ^{n + 2}} = {\beta ^{n + 1}} + {\beta ^n}$

$ \Rightarrow $ $({\alpha ^{n + 2}} + {\beta ^{n + 2}}) = ({\alpha ^{n + 1}} + {\beta ^{n + 1}}) + ({\alpha ^n} + {\beta ^n})$

$ \Rightarrow {a_{n + 2}} = {a_{n + 1}} + {a_n}$

Similarly, ${a_{n + 1}} = {a_n} + {a_{n - 1}}$

${a_n} = {a_{n - 1}} + {a_{n - 2}}$

..............

............

${a_3} = {a_2} + {a_1}$

On adding, we get

${a_{n + 2}} = ({a_n} + {a_{n - 1}} + {a_{n - 2}} + ... + {a_2} + {a_1}) + {a_2}$

$ \because $ $\left[ {{a_2} = {{{\alpha ^2} - {\beta ^2}} \over {\alpha - \beta }} = \alpha + \beta = 1} \right]$

So, ${a_{n + 2}} - 1 = {a_1} + {a_2} + {a_3} + ...... + {a_n}$

So, option (c) is also correct.

And, now $\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{{10}^n}}}} = \sum\limits_{n = 1}^\infty {{{{\alpha ^n} - {\beta ^n}} \over {(\alpha - \beta ){{10}^n}}}} $

$ = {1 \over {\alpha - \beta }}\left[ {\sum\limits_{n = 1}^\infty {{{\left( {{a \over {10}}} \right)}^n}} - \sum\limits_{n = 1}^\infty {{{\left( {{\beta \over {10}}} \right)}^n}} } \right]$

$ = {1 \over {\alpha - \beta }}\left[ {{{{\alpha \over {10}}} \over {1 - {\alpha \over {10}}}} - {{{\beta \over {10}}} \over {1 - {\beta \over {10}}}}} \right]$,

$\left[ {as\left| {{\alpha \over {10}}} \right| < 1\,and\,\left| {{\beta \over {10}}} \right| < 1} \right]$

$ = {{10(\alpha - \beta )} \over {(\alpha - \beta )[100 - 10(\alpha + \beta ) + \alpha \beta ]}}$

$ = {{10} \over {100 - 10 - 1}} = {{10} \over {89}}$

Hence, options (b), (c) and (d) are correct.

$\alpha $² = $\alpha $ + 1

$\beta $² = $\beta $ + 1

a_n = p$\alpha $ⁿ + q$\beta $ⁿ

= p($\alpha $^n$-$1 + $\alpha $^n$-$2) + q($\beta $^n$-$1 + $\beta $^n$-$2)

= a_n$-$1 + a_n$-$2

$ \therefore $ a₁₂ = a₁₁ + a₁₀

$\alpha = {{1 + \sqrt 5 } \over 2}$,

$\beta = {{1 - \sqrt 5 } \over 2}$

${a_4} = {a_3} + {a_2}$

$ = 2{a_2} + {a_1}$

$ = 3{a_1} + 2{a_0}$


$28 = p(3\alpha + 2) + q(3\beta + 2)$

$28 = (p + q)\left( {{3 \over 2} + 2} \right) + (p - q)\left( {{{3\sqrt 5 } \over 2}} \right)$

$ \therefore $ p $-$ q = 0

and $(p + q) \times {7 \over 2} = 28$

$ \Rightarrow $ p + q = 8

$ \Rightarrow $ p = q = 4

$ \therefore $ p + 2q = 12

Given, first equation $x^2-2 x \sec \theta+1=0$

Using quadratic equation formula we get,

$ \begin{aligned} x & =\frac{-(-2 \sec \theta) \pm \sqrt{(-2 \sec \theta)^2-4}}{2} \\\\ \Rightarrow & x=\frac{2 \sec \theta \pm \sqrt{4 \sec ^2 \theta-4}}{2} \\\\ \Rightarrow & x=\frac{2 \sec \theta \pm 2 \tan \theta}{2} \\\\ \Rightarrow & x=\sec \theta \pm \tan \theta \text { as } \theta \in\left(\frac{-\pi}{6}, \frac{-\pi}{2}\right) \\\\ \Rightarrow & \alpha_1=\sec \theta-\tan \theta \text { and } \beta_1=\sec \theta+\tan \theta \end{aligned} $

Now, for the equation $x^2+2 x \tan \theta-1=0$

$ \begin{aligned} & x=\frac{-2 \tan \theta \pm \sqrt{4 \tan ^2 \theta+4}}{2} \\\\ \Rightarrow & x=\frac{-2 \tan \theta \pm 2 \sec \theta}{2} \\\\ \Rightarrow & x=-\tan \theta \pm \sec \theta \\\\ \Rightarrow & x=(\sec \theta-\tan \theta),-(\sec \theta+\tan \theta) \end{aligned} $

Given, that $\alpha_2$ and $\beta_2$ are the roots of equation and $\alpha_2>\beta_2$

$ \begin{aligned} &\Rightarrow \alpha_2 =\sec \theta-\tan \theta, \beta_2=-(\sec \theta+\tan \theta) \\\\ &\Rightarrow \alpha_1+\beta_2 =(\sec \theta-\tan \theta)-(\sec \theta+\tan \theta) \\\\ & =\sec \theta-\tan \theta-\sec \theta-\tan \theta \\\\ &\Rightarrow \alpha_1+\beta_2 =-2 \tan \theta \end{aligned} $

Given, x₁ and x₂ are roots of

$\alpha {x^2} - x + \alpha = 0$

$\therefore$ ${x_1} + {x_2} = {1 \over \alpha }$ and ${x_1}{x_2} = 1$

Also, $\left| {{x_1} - {x_2}} \right| < 1$

$ \Rightarrow {\left| {{x_1} - {x_2}} \right|^2} < 1 \Rightarrow {({x_1} - {x_2})^2} < 1$

or, ${({x_1} + {x_2})^2} - 4{x_1}{x_2} < 1$

$ \Rightarrow {1 \over {{\alpha ^2}}} - 4 < 1$ or ${1 \over {{\alpha ^2}}} < 5$

$ \Rightarrow 5{\alpha ^2} - 1 > 0$

or, $(\sqrt 5 \alpha - 1)(\sqrt 5 \alpha + 1) > 0$

$\therefore$ $\alpha \in \left( { - \infty , - {1 \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},\infty } \right)$ .....(i)

Also, $D > 0$

$ \Rightarrow 1 - 4{\alpha ^2} > 0$ or $\alpha \in \left( { - {1 \over 2},{1 \over 2}} \right)$ ...... (ii)

$\alpha \in \left( { - {1 \over 2},{{ - 1} \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},{1 \over 2}} \right)$

Given, a quadratic equation $p(x)=0$ with real coefficients has purely imaginary roots.

Let $i \lambda$ and $-i \lambda$ are the roots of $p(x)=0$ where $i=\sqrt{-1}$ and $\lambda$ is a real number except zero.

$\begin{aligned} & \therefore p(x)=a(x-i \lambda)(x+i \lambda) \\ & \Rightarrow p(x)=a\left(x^2+\lambda^2\right) \end{aligned}$

Now, $\quad p(p(x))=0$

$\Rightarrow \quad a\left((p(x))^2+\lambda^2\right)=0$

$\Rightarrow a\left[a^2\left(x^2+\lambda^2\right)^2+\lambda^2\right]=0$

$\Rightarrow a^2\left(x^2+\lambda^2\right)^2+\lambda^2=0$

$\Rightarrow \quad\left(x^2+\lambda^2\right)^2=-\frac{\lambda^2}{a^2}$

$\Rightarrow \quad x^2+\lambda^2= \pm i \frac{\lambda}{a}$

$\Rightarrow \quad x^2= \pm i \frac{\lambda}{a}-\lambda^2$

$\Rightarrow \quad x= \pm \sqrt{ \pm i \frac{\lambda}{a}-\lambda^2}$

Hence, $p(p(x))=0$ has four roots but all the roots are neither purely real nor purely imaginary.

Hint:

(i) In a quadratic equation imaginary roots are always in conjugate pair i.e, if one root is $p+i q$, then other root must be $p-i q$

(ii) A quadratic equation with real coefficients has purely imaginary roots, then consider $\pm i \lambda$ are the roots of the given quadratic, where $\lambda \in \mathrm{R}-\{0\}$.

Given, $3^x=4^{x-1}$

Taking $\log$ on both side with base 3

$\begin{aligned} & \Rightarrow \log _3 3^x=\log _3\left(4^{x-1}\right) \\ & \Rightarrow x \log _3 3=(x-1) \log _3 4 \\ & \Rightarrow \quad x .1=(x-1) \log _3 4 \\ & \Rightarrow \quad x=\frac{\log _3 4}{\log _3 4-1} \\ & \Rightarrow \quad x=\frac{1}{1-\frac{1}{\log _3 4}} \\ \end{aligned}$

$\text { According to base changing rule } \frac{1}{\log _3 4}=\log _4 3$

$\begin{array}{ll} \therefore & x=\frac{1}{1-\log _4 3} \quad \text{... (i)}\\ \Rightarrow & x=\frac{1}{1-\log _{2^2} 3} \\ \Rightarrow & x=\frac{1}{1-\frac{1}{2} \log _2 3} \quad\left[\log _{a^m} x=\frac{1}{m} \log _a x\right] \\ \Rightarrow & x=\frac{2}{2-\log _2 3} \quad \text{... (ii)} \end{array}$

$\begin{aligned} & \Rightarrow \quad x=\frac{2}{2-\frac{1}{\log _3 2}} \\ & \Rightarrow \quad x=\frac{2 \log _3 2}{2 \log _3 2-1} \quad \text{... (iii)} \end{aligned}$

From equation (i), (ii) and (iii), it is clear that option (A), (B) and (C) are correct.

Hints:

(i) Base changing Rule $\log _b^a=\frac{1}{\log _a^b}$

(ii) $\log _a x^n=n \log _a x$

(iii) $\log _{a^m} x=\frac{1}{m} \log _a x$

Let a + 1 = t⁶. Thus, when a $\to$ 0, t $\to$ 1.

$\therefore$ $({t^2} - 1){x^2} + ({t^3} - 1)x + (t - 1) = 0$

$ \Rightarrow (t - 1)\{ (t + 1){x^2} + ({t^2} + t + 1)x + 1\} = 0$,

as t $\to$ 1

$2{x^2} + 3x + 1 = 0$

$ \Rightarrow 2{x^2} + 2x + x + 1 = 0$

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

Thus, x = $-$1, $-$1/2

or, $\mathop {\lim }\limits_{a \to {0^ + }} \alpha (a) = - {1 \over 2}$

and $\mathop {\lim }\limits_{a \to {0^ + }} \beta (a) = - 1$

We have,

${(2x)^{\ln 2}} = {(3y)^{\ln 3}}$ ...... (1)

${3^{\ln x}} = {2^{\ln y}}$ ....... (2)

$ \Rightarrow (\log x)(\log 3) = (\log y)\log 2$

$ \Rightarrow \log y = {{(\log x)(\log 3)} \over {\log 2}}$ ........ (3)

Taking log both sides of Eq. (1), we get

$(\log 2)\{ \log 2 + \log x\} = \log 3\{ \log 3 + \log y\} $

${(\log 2)^2} + (\log 2)(\log x) = {(\log 3)^2} + {{{{(\log 3)}^2}(\log x)} \over {\log 2}}$ from Eq. (3)

$ \Rightarrow {(\log 2)^2} - {(\log 3)^2} = {{{{(\log 3)}^2} - {{(\log 2)}^2}} \over {\log 2}}(\log x)$

$ \Rightarrow - \log 2 = \log x$

$ \Rightarrow x = {1 \over 2} \Rightarrow {x_0} = {1 \over 2}$.

We have, ${a_n} = {\alpha ^n} - {\beta ^n}$

${\alpha ^2} - 6\alpha - 2 = 0$

Multiplying with $\alpha$⁸ on both sides, we get

${\alpha ^{10}} - 6{\alpha ^9} - 2{\alpha ^8} = 0$ ..... (1)

Similarly, ${\beta ^{10}} - 6{\beta ^9} - 2{\beta ^8} = 0$ ..... (2)

From Eqs. (1) and (2), we get

${\alpha ^{10}} - {\beta ^{10}} - 6({\alpha ^9} - {\beta ^9}) = 2({\alpha ^8} - {\beta ^8})$

$ \Rightarrow {a_{10}} - 6{a_9} = 2{a_8} \Rightarrow {{{a_{10}} - 2{a_8}} \over {2{a_9}}} = 3$.

The given equations are

${x^2} + bx - 1 = 0$

${x^2} + x + b = 0$ ....... (1)

Common root is $(b - 1)x - 1 - b = 0$.

$ \Rightarrow x = {{b + 1} \over {b - 1}}$

This value of x satisfies Eq. (1), we get

${{{{(b + 1)}^2}} \over {{{(b - 1)}^2}}} + {{b + 1} \over {b - 1}} + b = 0$

$ \Rightarrow b = i\sqrt 3 ,\, - i\sqrt 3 ,\,0$

Sum of roots = ${{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}$ and product = 1

Given, $\alpha$ + $\beta$ = $-$ p and $\alpha$³ + $\beta$³ = q

$ \Rightarrow (\alpha + \beta )({\alpha ^2} - \alpha \beta + {\beta ^2}) = q$

$\therefore$ ${\alpha ^2} + {\beta ^2} - \alpha \beta = {{ - q} \over p}$ ..... (i)

and ${(\alpha + \beta )^2} = {p^2}$

$ \Rightarrow {\alpha ^2} + {\beta ^2} + 2\alpha \beta = {p^2}$ ..... (ii)

From Eq. (i) and (ii), we get

${\alpha ^2} + {\beta ^2} = {{{p^3} - 2q} \over {3p}}$

and $\alpha \beta = {{{p^3} + q} \over {3p}}$

$\therefore$ Required equation is

${x^2} - {{({p^2} - 2q)x} \over {({p^3} + q)}} + 1 = 0$

$ \Rightarrow ({p^3} + q){x^2} - ({p^3} - 2q)x + ({p^3} + q) = 0$

As a, b, c, p, q $\in$ R and the two given equations have exactly one common root.

$\Rightarrow$ Either both equations have real roots.

Or both equations have imaginary roots.

$\Rightarrow$ Either $\Delta_1\ge0$ and $\Delta_2 > 0$ or

$\Delta_1 < 0$ and $\Delta_2 < 0$

$\Rightarrow~p^2-q\ge0$ and $b^2-ac\ge0$

or $p^2-q\le0$ and $b^2-ac\le0$

$\Rightarrow~(p^2-q)(b^2-ac)\ge0$

$\therefore$ Statement 1 is true.

Also, we have $\alpha\beta=q$ and ${\alpha \over \beta } = {c \over a}$

$\therefore$ ${{\alpha \beta } \over {{\alpha \over \beta }}} = {q \over c} \times a \Rightarrow {\beta ^2} = {{qa} \over c}$

As $\beta \ne 1$ or $-1$

$ \Rightarrow {\beta ^2} \ne 1$

$ \Rightarrow {{qa} \over c} \ne 1$ or $c \ne qa$

Again as exactly one root $\alpha$ is common and $\beta\ne1$

$\therefore$ $\alpha + \beta \ne \alpha + {1 \over \beta } \Rightarrow {{ - 2b} \over a} \ne - 2P$

$ \Rightarrow b \ne ap$

$\therefore$ Statement 2 is correct.

But Statement 2 is not correct explanation of Statement 1.

Since $\alpha,\beta$ are the roots of

${x^2} - px + r = 0$

$\therefore$ $\alpha + \beta = p$ and $\alpha \beta = r$

We know that

${\alpha \over 2}$ and 2$\beta$ are the roots of the equation ${x^2} - qx + r = 0$

$\therefore$ ${\alpha \over 2} + 2\beta = q$ and ${\alpha \over 2} \times 2\beta = r$

Solving $\alpha + \beta = p$ and ${\alpha \over 2} + 2\beta = q$

We get $\alpha = {2 \over 3}(2p - q)$ and $\beta = {1 \over 3}(2q - p)$

$\alpha \beta = r$

$r = {2 \over 9}(2p - q)(2q - p)$

$x^{2}+2(a+b+c) x+3 \lambda(a b+b c+c a)=0$

For the given quadratic equation has real roots, we must have

$\{2(a+b+c)\}^{2}-4 \times 3 \lambda(a b+b c+c a) \geq 0$

i.e, Discriminant $\geq 0$

$\Rightarrow 4(a+b+c)^{2} \geq 12 \lambda(a b+b c+a c)$

$\Rightarrow \lambda \leq \frac{4}{12} \frac{(a+b+c)^{2}}{(a b+b c+a c)}$

$\Rightarrow \lambda \leq \frac{(a+b+c)^{2}}{3(a b+b c+a c)}=\frac{\left(a^{2}+b^{2}+c^{2}\right)}{3(a b+b c+a c)}+\frac{2}{3}$ ...... (i)

$\left(\because(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c\right)$

In $\triangle \mathrm{ABC}$

$|a-b| < c$

$\Rightarrow \quad(a-b)^{2} < c^{2}$

$\Rightarrow \quad a^{2}+b^{2}-2 a b < c^{2}$ .... (ii)

Similarly $b^{2}+c^{2}-2 b c < a^{2}$ ..... (iii)

$a^{2}+c^{2}-2 a c < b^{2}$ ..... (iv)

Adding (ii), (iii) and (iv)

$2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+a c) < $

$\left(a^{2}+b^{2}+c^{2}\right)$

$\Rightarrow a^{2}+b^{2}+c^{2} < 2(a b+b c+a c)$

$\Rightarrow \frac{a^{2}+b^{2}+c^{2}}{a b+b c+a c} < 2$

From (i), we have

$\lambda \leq \frac{a^{2}+b^{2}+c^{2}}{3(a b+b c+a c)}+\frac{2}{3}$

$\begin{array}{ll} \Rightarrow \lambda < \frac{2}{3}+\frac{2}{3} \quad \because \frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a} < 2 \\ \Rightarrow \lambda < \frac{4}{3} \end{array}$