Mathematics Stack Exchange
Mathematics Stack Exchange

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 $$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

5400 questions
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Is $x^{\frac{1}{2}}+ 2x+3=0$ a quadratic equation

Is $$x^{\frac{1}{2}}+ 2x+3=0$$ considered a quadratic equation? Should the equation be in the form $$ax^2+bx+c=0$$ to be considered quadratic?
user456
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Why a quadratic equations always equals zero?

On evaluating quadratic equations, It always equals zero: $$ax^2+bx+c=0$$ Why zero? Is it possible to use other number for another purpose?
Red Banana
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Determine all $a,b,c\in \mathbb{Z}$ for which the equation $(a^2+b^2)x^2-2(b^2+c^2)x-(c^2+a^2)=0$ has rational roots.

Determine all $a,b,c\in \mathbb{Z}$ for which the equation $(a^2+b^2)x^2-2(b^2+c^2)x-(c^2+a^2)=0$ has rational roots.. I know $\Delta \ge 0$ and $\sqrt{\Delta}$ must be rational.
piteer
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Prove $ax^2+bx+c=0$ has no rational roots if $a,b,c$ are odd

If $a,b,c$ are odd, how can we prove that $ax^2+bx+c=0$ has no rational roots? I was unable to proceed beyond this: Roots are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ and rational numbers are of the form $\frac pq$.
ABC
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How many fingers do martians have?

Text of problem: It is supposed that we use base 10 as our number system because we have ten fingers. A martian, after seeing the equation $x^2-16x+41=0$ writes the difference of the roots as $10$. End How many fingers do martians have ? Note: For…
Nameless
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How to solve this group of equations?

$$\dfrac{yz(y+z-x)}{x+y+z}=p \tag1$$ $$\dfrac{xz(x+z-y)}{x+y+z}=q \tag2$$ $$\dfrac{xy(x+y-z)}{x+y+z}=r \tag3$$ This group of equations comes from the problem "if the distances of incenter of an triangle to its vertices are $\sqrt{p}$, $\sqrt{q}$ and…
r ne
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Values of $a$ for which $(a+4)x^2-2ax+2a-6 <0$ for all $x \in R$

How can we find all values of $a$ for which the inequality $(a+4)x^2-2ax+2a-6 <0$ is satisfied for all $x \in R$? For the given condition, $D >0$, therefore $ (-2a)^2-4(2a-6)(a+4) >0$. Solving for $a$, I get $(a+6)(a-4) <0$, but the answer is…
Sachin Sharmaa
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Proving a relation related to quadratic equation

Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$. My Attempt: Finding the sum of roots and product of roots for both the equations we…
MrAP
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Proving an alternate quadratic formula

It is well known that the quadratic formula for $ax^2+bx+c=0$ is given by$$x=\dfrac {-b\pm\sqrt{b^2-4ac}}{2a}\tag1$$ Where $a\ne0$. However, I read somewhere else that given $ax^2+bx+c=0$, we have another solution for $x$ as$$x=\dfrac…
Frank
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Why/when did these extraneous solutions appear while solving a quadratic equation?

I am trying to solve the quadratic equation $x^2 + x + 1 = 0$. $x^2 = -1 - x $ $\iff x = -\frac{1}{x} - 1$, assuming $x\neq 0$. Substituting that into the original equation gives $x^2 + (-\frac{1}{x} -1) + 1 = 0$ $x=1$ is a solution to this second…
Ollie
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Why is the '+' solution to the quadratic formula always the one that satisfies my constraint

Let $A,B \in (0,1)$ be known constants, and $C \in (-\infty, \infty)$ be a known constant. Define \begin{equation} \xi(x) = \log \big( x \big) + \log \big( 1+x-A-B \big) - \log \big( A - x \big) - \log \big( B - x \big) - C \end{equation} For a…
Macro
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$y =f(x) =(ax^2 + bx +c)/(dx^2+ex+f)$ We have to find the conditions for this it takes all real values.

$$ y=f(x)=\frac{ax^2+bx+c}{dx^2+ex+f} $$ We have to find the conditions for this it takes all real values. MY solution One approach is to equate it to y and for a quadratic of x and put discriminant greater than equal to 0..That is very lengthy.Is…
maths lover
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Find all real values of the parameter $a$ for which the equation $x(x + 1)(x + a)(x + a+ 1) = a^2$ has four real roots.

Find all real values of the parameter $a$ for which the equation $x(x + 1)(x + a)(x + a+ 1) = a^2$ has four real roots. My Attempt Is my attempt is correct and also is there any other way to solve this problem.
Abhishek Kumar
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Why is my solution incorrect for solving these quadratic equations?

$$\frac2x -\frac5{\sqrt{x}}=1 \qquad \qquad 10)\ \frac3n -\frac7{\sqrt{n}} -6=0$$ I have these two problems. For the first one I create a dummy variable, $y = \sqrt x$ then $y^2 = x$. Substituting this in the first equation, I get: $\displaystyle…
user130306
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7
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4 answers

Find the interval in which $m$ lies so that the expression $\frac{mx^2+3x-4}{-4x^2+3x+m}$ can take all real values, $x$ being real

Find the interval in which $m$ lies so that the expression $\frac{mx^2+3x-4}{-4x^2+3x+m}$ can take all real values, $x$ being real. I don't know how to proceed with this question. I have equated this equation with $y$ to obtain a quadratic…
oshhh
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