F(x)=x^2 +6x-8 Show that if the equation f(x+a)=a has no real root then a^2<4(b-a)?

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Answer 1 · 2018-05-08T19:09:02

$(a+17) lt 0, or a lt -17$ .

Prerequisite :

The quadr. eqn. $ax^2+bx+c=0$ has no real roots.

$iff b^2-4ac lt 0$ .

We see that, for $f(x)=x^2+6x-8$ ,

$f(x+a)=a rArr (x+a)^2+6(x+a)-8=a$

$rArr x^2+(2a+6)x+(a^2+5a-8)=0$ .

Since this quadr. eqn. has no real roots,

$(2a+6)^2-4(1)(a^2+5a-8) lt 0$ .

$:. 4(a+3)^2-4(a^2+5a-8) lt 0$ .

$:. 4{(a^2+6a+9)-(a^2+5a-8)} lt 0, i.e.,$

$4(a+17) lt 0$ .

$:. (a+17) lt 0, or a lt -17$ .