$(p+r)^2>q^2+(p-r)^2$

Does this imply that $(p+r)^2>q^2$ @LeakyNun when $(p+r)>(p-r)$

square must be non-negative

$a^2 \ge 0$

$(p-r)^2 \ge 0$

$(p+r)^2 > q^2 + (p-r)^2 \ge q^2 + 0 = q^2$

$(p+r)^2>q^2$

the condition $(p+r)>(p-r)$ is useless

@LeakyNun p and r are both positive

@Abcd useless condition

@LeakyNun How 0?

Oh. I see. Nice @LeakyNun !

@LeakyNun But that zero is minimum value

@Abcd so?

@LeakyNun We can't ignore their maximum value. Rather we should consider only their maximum value IMO

@Abcd why?

@LeakyNun why not?

@Abcd just point out which step is wrong

@LeakyNun This one

@Abcd which step?

@LeakyNun Substituting (p-r)^2 = 0

$(p-r)^2 \ge 0$

add $q^2$ to both sides

done

what do you get

@LeakyNun Understood

Find the value of a for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all values for all real values of x

$\Delta$ is same for both numerator and denominator @LeakyNun

Let $y=\dfrac{ax^2+3x-4}{a+3x-4x^2}$ (introducing yet another variable :D)

@LeakyNun why?

Then, $(-4y-a)x^2 + (3y-3)x +(ay+4) = 0$

@LeakyNun How?

@Abcd simple algebraic manipulations

@LeakyNun No, please tell.

@Abcd just multiply both sides by $a+3x-4x^2$

Then, since $y$ takes all values, the above quadratic solution has solutions for all $y$, meaning that for all $y$, the discriminant is non-negative.

$\Delta = (3y-3)^2 + 4(4y+a)(ay+4) \ge 0$ for all $y$

$9y^2-18y+9+16ay^2+(64+4a^2)y+16a\ge0$

$(9+16a)y^2 + (46+4a^2)y+(16a+9) \ge 0$

what am I doing with my life

since that has to hold for all $y$, the graph of that (where $y$ is the independent variable) must not cross through the horizontal axis, so the determinant is smaller than or equal to 0.

$\Delta_y = (46+4a^2)^2 - 4(9+16a)(16a+9) \le 0$

@LeakyNun 3- y instead of 3y-3

@Abcd how?

@LeakyNun $ax^2+3x-4 = y(a+3x-4x^2)$

$\implies (a+4)x^2 +x(3-3y)-+(-4-ay)=0$ @LeakyNun

@LeakyNun Never mind, I was wrong. Wait a minute. Reading the next stuff you have written.

$(23+2a^2)^2 \le (16a+9)^2$

Please wait

you can always read later

$-16a-9 \le 23+2a^2 \le 16a+9$

$2a^2+16a+32 \ge 0$ and $2a^2-16a+14 \le 0$

$2(a+8)^2 \ge 0$ and $2(a-1)(a-7) \le 0$

true and $1 \le a \le 7$

$1 \le a \le 7$

@LeakyNun Didn't get the discriminant is non negative part.

@LeakyNun Correct answer.

@Abcd for all $y$, there is root for $x$, so $\Delta_x \ge 0$ for all $y$

@LeakyNun You mean $y \epsilon (-\infty, +\infty)$? and therefore y must intersect X axis too?

@Abcd no, stop thinking that $y$ must be the dependent variable

where $\epsilon$ = belongs to

it's just another parameter

@LeakyNun Is y a function of x?

...

yes, it is

no it isn't

@LeakyNun What do you mean by parameter?

what the question exactly meant by saying that $a$ is a parameter

@LeakyNun i guess this word's the reason I couldn't understand the previous question.

take it to mean "constant".

Oh and I think this is the reason I couldn't understand family of lines. PARAMETER!

@LeakyNun okay, then?

then what

y can take all real values

@LeakyNun y can't take any complex value?

what the

@LeakyNun Hmm

@LeakyNun I know you are angry but why can't y take any complex value? I am damn sorry for irritating you sir.

If $x$ is real and $a$ is real then $y$ how complex?

@LeakyNun Right. Got it.

@LeakyNun ?

@Abcd ignore it

@LeakyNun Didn't understand the sudden shift to this part.

for all $y$

so if you view this as a quadratic in $y$

@LeakyNun yes

then its discriminant would have to be smaller than or equal to 0

@LeakyNun Why

or else $f(y)$ would cross through the horizontal axis and be negative

quadratiception

Why can't f(y) be negative?

Got it.

good

@LeakyNun Also 16a+9 >0

@Abcd right

@LeakyNun How $2(a+8)^2>=0$. I am getting (a+4)^2>=0

$2a^2+16a+32=2(a^2+8a+16)=2(a+4)^2$

sorry

the point stands.

Find the solution for $x \epsilon (-2,3)$ of the equation:$(x^2+x+1)^2+1= (x^2+x+1)(x^2-x-5))$. I got $(-2,-1]$ but answer is no solution @LeakyNun

how do you get an entire interval as solution?

@LeakyNun i just used the fact that $x^2- x-5>0$ then did intersection of sets.

But you are right. Equation should have only 4 solutions.

Should I use same property for other (x^2+x+1) also?

$(x^2+x+1)(2x+6)+1=0$

$x^2+x+1>0$

$2x+6 > 0$

therefore there is no solution

@LeakyNun How?

@Abcd basic distribution laws

yeah, got it. done.

$x^2-11x+a= 0$

$x^2 - 14x+2a=0$

How do I determine the condition for common roots?

solve them together?

@LeakyNun a=-3x?

right

then?

solve them

@LeakyNun Solution says: $x^2/-8a = x/-a = 1/-3$ Why?

no idea

it's the same with what you wrote

@LeakyNun How to get that 8a condition./