Question

If both roots of the equation 4x² - 12ax + (9a² - 4a + 8) = 0 exceed 3, then the set of exhaustive values of a is​

Answer 1

Step-by-step explanation:

We have,

$\sf{4 {x}^{2} - 12ax + \left( 9 {a}^{2} - 4a + 8 \right) = 0 }$

Let its roots be $\alpha$ and $\beta$

$\tt{ \dagger \: \: Sum \: \: of \: \: roots \: , \: \: \alpha + \beta = - \dfrac{ - 12a}{4} = 3a}$

$\tt{ \dagger \: \: Product \: \: of \: \: roots \: , \: \: \alpha \beta = \dfrac{ 9 {a}^{2} - 4a + 8}{4} }$

Now,

$| \alpha - \beta| = \sqrt{( \alpha + \beta)^{2} - 4 \alpha \beta }$

$\implies | \alpha - \beta| = \sqrt{( 3a)^{2} - ( 9 {a}^{2} - 4a + 8)}$

$\implies | \alpha - \beta| = \sqrt{9a^{2} - 9 {a}^{2} + 4a - 8}$

$\implies | \alpha - \beta| = \sqrt{ 4a - 8}$

By the given condition,

$\implies | \alpha - \beta| < 3$

$\implies \sqrt{ 4a - 8} < 3$

$\implies 4a - 8 <9$

$\implies 4a <17$

$\implies a < \dfrac {17}{4}$

Answer 2

We have,

$4 x^{2}-12 a x+\left(9 a^{2}-4 a+8\right)=0$

As here both roots of the equations is 0 and 3

So, let its roots be $\alpha and beta$

As we know that,

Sum of roots will be given by,

$\alpha+\beta=-\frac{-12 \mathrm{a}}{4}=3 \mathrm{a}$

We know that,

Product of the roots are given by

, $\alpha \beta=\frac{9 a^{2}-4 a+8}{4}$

Now applying the rule in the given question,

We will get like,

$\begin{array}{l}|\alpha-\beta|=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta} \\\Longrightarrow|\alpha-\beta|=\sqrt{(3 a)^{2}-\left(9 a^{2}-4 a+8\right)} \\\Longrightarrow|\alpha-\beta|=\sqrt{9 a^{2}-9 a^{2}+4 a-8} \\\Longrightarrow|\alpha-\beta|=\sqrt{4 a-8}\end{array}$

Now Applying the obtained expression in the given question,

Using the condition given in the question,

We will get that,

$\begin{array}{l}\Longrightarrow|\alpha-\beta|<3 \\\Longrightarrow \sqrt{4 a-8}<3 \\\Longrightarrow 4 a-8<9 \\\Longrightarrow 4 a<17 \\\Longrightarrow a<\frac{17}{4}\end{array}$

Hence, above will be required answer.