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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Basic Algebra: How to go backwards?

For a function: $$f(x)=\frac{(1-x^2)}{(1+x^2)-2 x \cos \omega}$$ Let $x=-\frac{1}{3}$ That means that $$f(-\frac{1}{3})=\frac{(\frac{8}{9})}{(\frac{10}{9})+\frac{2}{3} \cos \omega}=\frac{4}{5+3 \cos \omega}$$ If we are instead given $$\frac{4}{5+3…
Astor Florida
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Quadratic equation find all the real values of $x$

Find all real values of $x$ such that $\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ I tried sq both sides by taking 1 in RHS but it didn't worked out well...
user420236
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Positive pairs of integral values satisfying $2xy − 4x^2 +12x − 5y = 11$

The number of positive pairs of integral values of $(x, y)$ that solves $2xy − 4x^2 +12x − 5y = 11$ is? I rearranged it to $(2x-5)(y+1-2x)=6$, which took quite a bit of time. So it can be $2*3$ , $3*2$, $6*1$ or $1*6$ which gives us 2 possible…
user405925
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If $a,b,p,q$ are non-zero real numbers, then the 2 equations $2a^2x^2-2abx+b^2=0$ and $p^2x^2+2pqx+q^2=0$ have...

If $a,b,p,q$ are non-zero real numbers, then the 2 equations $2a^2x^2-2abx+b^2=0$ and $p^2x^2+2pqx+q^2=0$ have: (a)no common root (b)two common roots if $3pq=2ab$ (c)one common root if $2a^2+b^2=p^2+q^2$ (d)two common roots if $3qb=2ap$ …
oshhh
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Quadratic equations with $a$ and $b$

If $a≠b$, and $a²=5a-7$ and $b²=5b-7$, then $a³+b³=?$ How can I find the answer? I found out that $a$ and $b$ are not real numbers. Can anyone please teach me? thank you!!
Estudiantekriss
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Relation between $S_{n-1}$ , $S_n$ and $S_{n+1}$

We know $ \alpha $ and $ \beta $ are roots of $ax^2+bx+c = 0 $. also $S_{n-1} = \alpha^{n-1} + \beta^{n-1}$ , $S_{n} = \alpha^{n} + \beta^{n}$ and $S_{n+1} = \alpha^{n+1} + \beta^{n+1}$. How we can find relation between $S_{n-1}$ , $S_n$ and…
S.H.W
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What is the sum of the squares of the roots of the equation $x^2 − 7[x] + 5 = 0?$ (Here $[x]$ denotes the greatest integer less than or equal to $x$)

I tried many different approaches to this question, $α^2 = 7[α] - 5$ and $β^2 = 7[β] - 5$ So combining both the equations we get, $α^2 + β^2 = 7([α] + [β]) - 10$, Which is the answer. However I am unable to simplify the LHS.
Aradhye Agarwal
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Find all values of $c$ such that solutions of $x^2 + cx + 6 = 0$ are rational.

This is what I've done so far: Using the formula, we get $x = \frac{-c \pm \sqrt{c^2 - 24}}{2}$ For solutions to exist, $c \geq 5$. For solutions to be rational, $c^2 - 24$ has to be a perfect square. I'm kind of stuck now. I'd appreciate if someone…
OinkOink
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Product of roots of $ax^2 + (a+3)x + a-3 = 0$ when these are positive integers

There is only one real value of $'a'$ for which the quadratic equation $$ax^2 + (a+3)x + a-3 = 0$$ has two positive integral solutions.The product of these two solutions is : Since the solutions are positive, therefore the product of roots and sum…
Heisenberg
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Help simplify using quadratic formula

$\dfrac{4\pm\sqrt{28}}{2}=2\pm\sqrt7$ My Question is how did $\dfrac{4\pm\sqrt{28}}{2}$ become simplified as $2\pm\sqrt7$ Can you help me by explaining the steps clearly :) Many Thanks
user360471
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Shortcut for roots of quadratic equation.

I know how to find the roots of a quadratic equation but the process is a bit time consuming.. Is there any way by which we can tell the roots of the quadratic equation just by looking at the equation.
danny
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Geometric interpretation of the coefficients of the quadratic equation.

The quadratic equation has three general forms: $ax^2+bx+c$ $a(x-r_1)(x-r_2)$ $a(x-h)^2+k$ $r_1$ and $r_2$ are the zeroes of the quadratic. $h$ is the horizontal position of the vertex, $k$ is the vertical position of the vertex. Are there…
Frank Vel
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Solve the quadratic equation

$$ \sqrt{a-\sqrt{a+x}}=x $$ This equation contains one variable x we have to find the value of x.I tried to simplify it but it doesn't work....i have also tried the basic concepts of quadratics but it is just waste
Tan_R
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quadratic pre colloge problem

If c is a real number and the negative of one of the solutions of $x^2 -3x +c=0$ is a solution of $x^2 +3x -c=0$ then the solution of $x^2-3x+c=0$ are.... i cant do this problem due to language barrier
Iqbal Jurist
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If $x^4 + 3\cos(ax^2 + bx +c) = 2(x^2-2) $ has two solutions with $a,b,c \in (2,5)$, then find the maximum value of $\frac{ac}{b^2} $

The answer given is 1. i tried like this $3\cos(ax^2 + bx +c) = -x^4 +2x^2-4 = -(x^2 -1)^2 -3 $. The maximum value of $-x^4 +2x^2-4$ is $-3$ so $3\cos(ax^2 + bx +c) =-3$ and the two values of x are $1$ and $-1$. Now $\cos(a + b +c) =-1$ and…
raj
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