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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Solving simultaneous equations with multiple variables.

If $(x+a)^2$ is a factor of $x^3 + 6px + k$, show that $k + 2a^3 = 0$ I've tried different ways to solve this, but is it even possible?
Davidovich
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1 answer

If $x^2 + ax - b$ is a factor of $x^3 - 2bx^2 + ax -6$, show that $a = -2b -6/b$.

If $x^2 + ax - b$ is a factor of $x^3 - 2bx^2 + ax -6$, show that $a = -2b -6/b$.
Davidovich
  • 29
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1 answer

solving word problems

a rectangular building is to be placed on a lot that measures 30m by 40m. The building must be placed in the lot so that the width of the lawn is the same on all four sides of the building . Local restrictions state that the building cannot occupy…
Alyana
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1 answer

Help with solution to 2 variable quadratic equation.

Could someone help with an explanation on how to treat the equation below, please? A solution I have read uses the discriminant to find the range i.e. $9-20y^2-16y>or = 0$. To me this means treating the variable Y as a constant , is that…
ralph
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2 answers

If $(a,0)$ lies on the diameter of circle $x^2+y^2=4$ then $x^2-4x-a^2=0$ has

A) Exactly one real root in $(-1,0]$ B) Exactly one real root in $[2,5]$ C) Distinct roots greater than 1 D) Distinct roots less than 5 $$x=\frac{4\pm \sqrt {16+4a^2}}{2}$$ $$x=2\pm \sqrt{4+a^2}$$ Also $a\in [-2,2]$ So $a^2\in [0,4]$ Then $x=4,0$…
Aditya
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Roots of product of two quadratic.

consider the quadratics, $ P(x) = ax^2 + bx +c$ ,and, $Q(x)= -ax^2 + dx+c$ , given that ac $\neq$ 0. What can we say about the roots of PQ? More particularly, does PQ have at least two real roots? my work: I don't really know where to start…
tryst with freedom
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If $x^2+bx+b$ is a factor of $x^3+2x^2+2x+c$, $c \neq 0$, then $b-c=?$

I am not getting where to start! Please help me out
samarth jain
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3 answers

Find the min value of $3a+b$

If $ax^2+bx+c=0$ has no real roots then find min value of $3a+b$ for $c=6$; Please tell me how to proceed , i don't have any clue of what to do.
ABC
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3 answers

Analytical solution for equation with power $n \in [1, 2]$

Does there exist analytical solution to the equation $ax^p + bx + c = 0$ where $p \in [1, 2]$? Please provide references if the answer is affirmative.
hsiaomichiu
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2 answers

What is the maximum value of $x^2+4xy-y^2$ for all $(x,y)$ satisfying $x^2+y^2 = 1$?

Does the trick have something to do with the equation of a circle?
Edward
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1 answer

Condition for both roots be infinity

For what value/s of constant 'p' for which the given quadratic have both roots as infinity. $(2p^3-13p^2+27p-18)x^2 + (2p^2-9p+9)x +2p^2-7p+6=0$ Options are :- $1) 3/2 2) 2 3) 3 4) /phi $ Since both roots are infinite then sum of the roots must be…
Tips
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1 answer

If the roots of the equation $6x^2-7x+K=0$ are rational, then is equal to–

If the roots of the equation $6x^2-7x+K=0$ are rational, then is equal to: $1)$ $-2$ $2)$ $-1,-2$ $3)$ $-2$ $4)$ $1,2$
ANKUR SHUKLA
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6 answers

How would one factorise $m^2 + (2AB)m + B^2 =0$

How would one factorise $m^2 + (2AB)m + B^2 =0$, to go onto solve a second order differential equation
Mr. Janssens
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3 answers

Bi-quadratic equation

Solve for $x$, it has four different solutions: $$x^4 -2x^3-6x^2-2x+1=0$$
Rahul Kumar
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4 answers

How do I solve $5x^2-3=0$?

I tried using the quadratic formula but I don't have a $c$. Do I just put a $1$ in its place or something? The answer is supposed to be $\mp \frac15 \sqrt{15}$.
user591195
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