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Quadratic Equations

Section: Mathematics

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Question : 8 of 51

Marks: +1, -0

Let

(a-3)x^2+12x+(a+6)>0,\ \forall x\in R\ \&\ a\in(\ell,\infty)

\alpha

is the least positive integral value of a , then the roots of

(\alpha-3)x^2+12x+(\ell+2)=0

[AP EAPCET 23 May 2025 S2]

$1,2$
$2,3$
$-1,-2$
$-2,-3$

Solution:

Let

f(x)=(a-3)x^2+12x+(a+6)

to quadratic always be positive

\; a-3>0

\Rightarrow a>3

\; \text{ and }\; D=(12)^2-4(a-3)(a+6)<0

\; =144-4(a^2+3a-18)<0

\Rightarrow a^2+3a-54>0

\Rightarrow a^2+9a-6a-54>0

\Rightarrow (a+9)(a-6)>0

a\in(-\infty,-9)\cup(6,\infty)

Combine with

a>3

\Rightarrow \;\; \;\text{ We take }\; a\in(6,\infty)

Least positive integeral value

\alpha=7,\ l=6

put these values

\; (\alpha-3)x^2+12x+(l+2)

\; =4x^2+12x+8=0

\Rightarrow \;\; x^2+3x+2=0

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

Hence,

x=-1,-2

c

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