Let a, b [ ] R, a [ ] 0 be such that the equation, ax2 -- 2bx + 5 = 0 has a repeated root …

Question

Let a, b $$ \in $$ R, a $$ \ne $$ 0 be such that the equation,
ax2 -- 2bx + 5 = 0 has a repeated root $$\alpha $$, which
is also a root of the equation, x2 -- 2bx -- 10 = 0.
If $$\beta $$ is the other root of this equation, then
$$\alpha $$2 + $$\beta $$2 is equal to :

Accepted Answer

Roots of equation ax2
-- 2bx + 5 = 0 are $$\alpha $$, $$\alpha $$.

$$ \therefore $$ $$\alpha $$ + $$\alpha $$ = $${{2b} \over a}$$

$$ \Rightarrow $$ 2$$\alpha $$ = $${{2b} \over a}$$

$$ \Rightarrow $$ $$\alpha $$ = $${{b} \over a}$$ ....(1)

and $$\alpha $$2 = $${5 \over a}$$ ....(2)

From (1) and (2), we get

$${{{b^2}} \over {{a^2}}} = {5 \over a}$$

$$ \Rightarrow $$ b2 = 5a

$$\alpha $$, $$\beta $$ are the roots  of equation x2
-- 2bx -- 10 = 0

$$ \therefore $$ $$\alpha $$ + $$\beta $$ = 2b

and $$\alpha $$ $\beta $= -10

As$\alpha $$ is a root of the equation x2
-- 2bx -- 10 = 0.

$$ \therefore $$\alpha $2 - 2b$\alpha $- 10 = 0$ \Rightarrow $${{{b^2}} \over {{a^2}}} - {{2{b^2}} \over a} - 10 = 0$$ \Rightarrow $${{5a} \over {{a^2}}} - {{10a} \over a} - 10 = 0$$ \Rightarrow $${5 \over a} - 10 - 10 = 0$$ \Rightarrow $$a$=${1 \over 4}$$ \Rightarrow $${\alpha ^2}$= 20$\alpha $$ $\beta $$ = -10

$$ \Rightarrow $$ ${\beta ^2}$= 5$ \therefore $${\alpha ^2} + {\beta ^2}$

Solutions · 1