$2a^2 – bc – 9a + 10 = 0$ then the range of $‘a’$ is?

Question

Let $a$, $b$ and $c$ be real numbers such that $2a^2 – bc – 9a + 10 = 0 $ and $4b^2 + c^2 + bc – 7a – 8 = 0,$ then the range of $‘a’$ is?

How to do we proceed ? Should I make quadratic in $a$ ?

Ans: $a \in [1,4.2]$

Answer 1

Hint: $$(2b – c)^2 + 5bc – 7a – 8 = 0\quad...(i) \\ 2a^2 – bc – 9a + 10 = 0\quad...(ii) $$ Now, $5(ii) + (i) = ? $.

Answer 2

When using the quadratic formula you have the part under the square root, so a is real only if the value of the square root is real, so all you need to do (in the first equation) is to plug in the numbers under the square root and check when it is larger or equal to zero. For the second equation, you need to to change it so you have $a =\dots$ and then check when the left side of the equation is real And now you can take the intersection of the two ranges

Answer 3

From the first equality

$bc=2a^2-9a+10$

$\Rightarrow (4b)c=8a^2-36a+40 \quad (1)$

Substituting in the $2$nd expression,

$4b^2+c^2-9a+2a^2+10-7a-8=0 $

$\Rightarrow 4b^2+c^2=-2a^2+16a+8 \quad (2)$

Adding $(1)$ and $(2)$, we get

$(2b+c)^2=6a^2-20a+48=2(3a^2-10a+24)$

Note that $3a^2-10a+24\gt 0 ,\forall a$

Similarly on subtracting $(1)$ from $(2)$, we get

$(2b-c)^2=-10a^2+52a-32=-2(5a^2-26a+16)$

Find the range of $a$ where $5a^2-26a+16 \le 0$

This range will give the answer