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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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Find the quadratic equation whose roots are the x and y intercepts of the line passing through (1,1) and making a triangle of area A with the axes

I do not how to solve this, can such equation even exist? For the root to lie on the y intercept, the line would have to pass through origin, which means one root will be 0, breaking down the whole the thing. Am I missing something here?
Aditya
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The roots of the equation $(a^4+b^4)x^2+4abcd x + (c^4+d^4)$will be?

The discriminant shall be $$16(abcd)^2-4a^4c^4-4a^4d^4-4b^4c^4-4b^4d^4$$ $$-4(b^2d^2+a^2c^2)^2-4(a^2d^2+b^2d^2)^2$$ which is clearly a negative value. So the roots should be imaginary. But the answer given is real and equal. How is that possible?
Aditya
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The number of solutions of $\sqrt {3x^2+x+5}=x-3$ is

Squaring on both sides we get the equation $$2x^2+7x-4=0$$ Then $$x=\frac 12$$ and $$x=-4$$ They do indeed satisfy the original equation but aren’t a part of the answer. I get that it must have to be the $\pm \sqrt {a}$ but I would still like a…
Aditya
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If $\frac{1}{2-i}$ is a root of $ax^2+bx+c=0$ and $\frac{1}{3-2\sqrt 2} $ is a root of $px^2+dx+q=0$

find the inequality relation between $a, b, c$ and $d$ The roots can be written as $$\frac{2+i}{5}$$ and $$3+2\sqrt 2$$ Then $$\frac{-b}{a}=\frac 45, \frac ca = \frac 15$$ And $$\frac{-d}{p}=6, \frac qp=1$$ It’s clear that $a>c>b$ but I can’t…
Aditya
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Find the equation which roots are $\alpha, -\beta$ if $\alpha, \beta$ be the roots of $ax^2+bx+c=0$.

Let $\alpha, \beta$ be the roots of $ax^2+bx+c=0$ with $a>0$ where $a, b, c\in \mathbb{R}$. We need to find an equation which roots are $\alpha, -\beta$. My Trial: Now $\alpha+\beta=-\frac{b}{a}$ and $\alpha\beta=\frac{c}{a}$. Since…
KON3
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Equations $x^2+bx+ca=0$ and $x^2+cx+ab=0$ have a common root. Prove that equation $x^2+ax+bc=0$ contains two other roots

Two quadratic equations are, $$x^2+bx+ca=0$$ $$x^2+cx+ab=0$$ where $b$ and $c$ are not equal to zero, have one common root. Prove that the equation containing their other roots is $$x^2+ax+bc=0$$
Bitthal Maheshwari
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How to find a square root

Let $m$ be a positive square-free integer, and a number $z\in Q(\sqrt{m})$ is given. How to find a square root of $z$, if it exists in $Q(\sqrt{m})$? Here $Q$ is the field of rational numbers. I am asking for a reference or an explanation of an…
Alexandre Eremenko
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Graph of $x^2$ + $y$ $=$ $0$ is an upward or a downward opening parabola?

That is my exact question. If we graph $x^2$ + $y$ $=$ $0$, do we get a downward opening parabola? Let me explain what actually got me confused. I know that $y$ = $x^2$ is an upward opening parabola because its leading coefficient is positive, or…
4d_
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Quadratic equation roots

How do I get the roots of the quadratic equation? $dx^2+(d ^2−d+1)x+d−1 = 0$ The solution is supposed to be $1-d$ and $-1/d$ but I have no idea how to get to it.
AdamPedrovec
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A Question on Quadratic Equations.

Show that the expression $\frac{(ax-b)(dx-c)}{(bx-a)(cx-d)}\\$ will be capable of all values when x is real, if $a^2-b^2$ and $c^2-d^2$ have the same sign. Here's my approach: I tried equating it with y which formed another Quadratic Equation, then…
Crocogator
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Quadratic Equation - Nature of roots

What is the product of real roots of the equation $t^2x^2+|x|+q=0$ Since the complex equation is positive so sum of the roots are positive, here I am having four option as answers : $>0$ $<0$ not exist irrelevant
Sachin Sharmaa
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The nature of quadratic equation

Given that $p, q$ are real numbers. Prove that if the equation $2x^2+2(p+q)x+p^2+q^2=0$ have real roots, $p$ must be equal to $q$. What I tried: If the equation has equal roots, then it can be rewritten as: $$2x^2 + 4px + 2p^2 = 0$$ Coefficients of…
Aleksandra Asanin
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Show that $\{a,b\}=\{c,d\}$ when $a+b+\sqrt{(a+b+1)^2-4ab}=c+d+\sqrt{(c+d+1)^2-4cd}$

My approach is the following: If $a+b=c+d$, then we obtain $ab=cd$ and the result follows. If $a+b=c+d+p$, $p\ne 0$, then I obtain a complicated expression, from which I cannot deduce anything. Any help is appreciated.
baronbrixius
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$3(2x^2+x-2)^2=8x^2+4x-9$

I should solve the following biquadratic equation with changing of variables (e.g., $x^2=z$). I do not see how to do it, and I think the answer in my book does not cover all solutions: $1$ and $-\dfrac{3}{2}$. $3(2x^2+x-2)^2=8x^2+4x-9$
keylin.rion
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Location of roots (quadratic Equation). Conditions on $a$

Find set of values of $a$ If $($$x^2$+$x$)$^2$+ $a$$($$x^2$+$x$) + $4$ $=$ $0$ has all four real roots where two of them are equal. Here’s how I tried solving it : I assumed $($$x^2$+$x$$)$ = $t$ $\Rightarrow$ $t$ $\in$ $[$$\frac{-1}{4}$,…
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