Question

If a, ß are the roots of x+2(a - 5)x +3 = 0. Find all values of a such that both 1 and 5 lie between
alpha and beta

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: \alpha , \beta \: are \: roots \: of \: {x}^{2} + 2(a - 5)x + 3 = 0$

Now, we have to find value of 'a' such that both the roots lie in between 1 and 5.

We know,

Conditions for both the roots of the quadratic equation f(x) = ax² + bx + c = 0 to be lies between ( h, k ) are

1. a f( h ) > 0

2. a f( k ) > 0

3. h < x - coordinate of vertex ( -b / 2a ) < k

4. Discriminant > = 0

Now, According to given data, we have

• h = 1

• k = 5

• a = 1

• b = 2( a - 5 )

• c = 3

Now,

We take first condition

$\rm :\longmapsto\:af(h) > 0$

$\rm :\longmapsto\:1 \times f(1) > 0$

$\rm :\longmapsto\: f(1) > 0$

$\rm :\longmapsto\: {1}^{2} + 2(a - 5)(1) + 3 > 0$

$\rm :\longmapsto\: 1 + 2(a - 5)+ 3 > 0$

$\rm :\longmapsto\: 4 + 2a - 10 > 0$

$\rm :\longmapsto\: 2a - 6> 0$

$\rm :\longmapsto\: 2a > 6$

$\bf\implies \:a > 3 - - - (1)$

Now, we take second condition

$\rm :\longmapsto\:af(k) > 0$

$\rm :\longmapsto\:1 \times f(5) > 0$

$\rm :\longmapsto\: f(5) > 0$

$\rm :\longmapsto\: {5}^{2} + 2(a - 5)(5) + 3 > 0$

$\rm :\longmapsto\: 25 + 10a - 50 + 3 > 0$

$\rm :\longmapsto\: 10a - 22> 0$

$\rm :\longmapsto\: 10a > 22$

$\rm :\longmapsto\: 5a > 11$

$\bf\implies \:a > \dfrac{11}{5} - - - (2)$

We take third condition

$\rm :\longmapsto\:h < - \: \dfrac{b}{2a} < k$

$\rm :\longmapsto\:1 < - \: \dfrac{2(a - 5)}{2} < 5$

$\rm :\longmapsto\:1 < - a + 5< 5$

$\rm :\longmapsto\:1 - 5< - a < 5 - 5$

$\rm :\longmapsto\: - 4< - a < 0$

$\bf :\longmapsto\:0< a < 4- - - (3)$

We take fourth condition

$\rm :\longmapsto\:Discriminant \geqslant 0$

$\rm :\longmapsto\: {4(a - 5)}^{2} - 4 \times 3 \times 1 \geqslant 0$

$\rm :\longmapsto\: {(a - 5)}^{2} - 3 \geqslant 0$

$\rm :\longmapsto\: {(a - 5)}^{2} \geqslant 3$

$\bf :\implies\:a \leqslant - \sqrt{3} + 5 \: \: or \: \: a \geqslant \sqrt{3} + 5 - - - (4)$

So, From equation (1), (2), (3) and (4), we concluded that

$\rm :\longmapsto\: 3 < a \leqslant 5 - \sqrt{3}$