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Questions tagged [quadratics]

Questions about quadratic functions and equations, second degree polynomials usually in the forms $y=ax^2+bx+c$, $y=a(x-b)^2+c$ or $y=a(x+b)(x+c)$.

Questions about quadratic functions and equations, second degree polynomials usually in the forms $y=ax^2+bx+c$, $y=a(x-b)^2+c$ or $y=a(x+b)(x+c)$.

The root of $y=ax^2+bx+c$ can be solved by the formula $$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

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Where is my solution wrong for this AIME problem?

Where is my solution to 2018 AIME I Problem 1 wrong? a can be any integer belonging to [1,100], so I count the no.of unordered pairs of positive integers whose sum is less than or equal to 100. i.e., the no.of unordered pairs…
Ram Keswani
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Is there a geometrical meaning of two quadratic equations (in one variable) having a common root?

$p(x)$ = $(x-3)(x-2)$, & $q(x)$ = $(x-2)(x-1)$ Both equations have a common root $x$ = $2$. I was just wondering if there is any geometrical meaning to it. For example, geometrical meaning of roots (or zeroes) of an equation $f(x)$ = 0 is the…
4d_
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Does a rational function with real coefficients always have rational roots?

$a$,$b$,$c$,$p$,$q$,$r$ are real numbers. If ($ax^2$ + $bx$ + $c$)$y$ + ($px^2$ + $qx$ + $r$) = $0$, and $x$ is a rational function of $y$, then prove that $( $ar$ - $pc$)^2$ = ($aq$ - $pb$)($br$ - $qc$). This is a question from my book. It…
4d_
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Why is $ax^2+bx+c=a(x-\alpha)(x-\beta)$

I came across the equation in my maths book. I wonder why it is as such. Can it be equal to $b(x-\alpha)(x-\beta)$ or $c(x-\alpha)(x-\beta)$? Kindly help me out.
Suyash Ishan
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Quadratics Solutions problem

For what values of $m$ does $x^2−4x−m=0$ have no real solutions while $x^2−9x+m^2=0$ has at least one real solution?
IMbADdAtMath
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Quadratic Graphing problem

The vertex of the quadratic $x(x-2a)$ occurs when $x=4$.The vertex of the quadratic $(x+a)(x-3a)$ occurs when $x=L$. What is $L$?
IMbADdAtMath
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Find all real values of $a$ such that $x^4-2ax^2+x+a^2-a=0$

I have to solve this question Find all real values of $a$ such that $$x^4-2ax^2+x+a^2-a=0$$ I have been trying this question for many days but was unable to solve it. I think we can convert this polynomial into a perfect square of some other…
Aryan Singh
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Is this quadratic word problem correct so far?

I'm a little confused as to how to solve this word problem I have. The problem is: A rectangular box (with a top) has a square base. The sum of the lengths of its edges is 8 feet. What dimensions should the box have in order for its surface area…
Someone
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if one root of the equation is the square of the other, then the values of a is

One of the roots of the equation $$8x^2 - 6x - a - 3$$ are the square of the other. Which means if $β\ and \ ⍺$ are the roots then $β = ⍺^2$. Then we have to find a.
weegee
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Profit Maximising output given output and cost

A perfectly competitive firm has a quadratic cost function $$C(q)= 3+7q+q^2$$ and given price for output is 14. What is the maximum profit it can make?
Mbw12
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Solve for $\tau$ where variable inside square root.

How do you solve for $\tau$? $$t = \tau + \frac{\sqrt{A^2+v^2\tau^2}}{c} $$ it might be easy but I just can not see how. Thanks.
nachofest
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Find $y$ if $x^{x+y}=y^n$ and $y^{x+y}=x^{2n}y^n$, where $x,y,n>0$

If $x,y>0$ satisfying the system of equations $x^{x+y}=y^n$ and $y^{x+y}=x^{2n}y^n$, where $n>0$ then prove that $y=\dfrac{1+4n-\sqrt{1+8n}}{2}$ $$ (xy)^{x+y}=(xy)^{2n}\implies x+y=2n\\ x^{2n}=y^n\implies x^2=y\\ x^{2}=2n-x\implies…
Sooraj S
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Quadratic equation with weighted average coefficients

I have tried using the quadratic formula as well as factoring method to solve the following quadratic equation but failed to get the correct answer. The equation is: $$ \theta x^2-x+(1-\theta)=0.$$ Note: the coefficient is theta How to go ahead?…
user508281
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Prove $f(x)=0$ has one root in between roots of $g(x)=0$

$f(x)=(a-b)x^2+(b-c)x+(c-a)$ $g(x)=(b-a)x^2+(a+c-2b)x+(b-c)$ given that $a
emil
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Are these coefficients swapped in this example, or am I wrong?

I believe I'm staring at an error here. Attached is a screenshot. Am I right in understanding that the author swapped the coefficients around in solving q using the quadratic formula? I can see why the mistake was made, if I'm right about the…
amateurjustin
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