In algebra, a quadratic equation (from the Latin quadratus for "square") is any equation that can be rearranged in standard form as
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0. If a = 0, then the equation is linear, not quadratic, as there is no
a
x
2
{\displaystyle ax^{2}}
term. The numbers a, b, and c are the coefficients of the equation and may be distinguished by calling them, respectively, the quadratic coefficient, the linear coefficient and the constant or free term.The values of x that satisfy the equation are called solutions of the equation, and roots or zeros of the expression on its left-hand side. A quadratic equation has at most two solutions. If there is no real solution, there are two complex solutions. If there is only one solution, one says that it is a double root. A quadratic equation always has two roots, if complex roots are included and a double root is counted for two. A quadratic equation can be factored into an equivalent equation
a
x
2
+
b
x
+
c
=
a
(
x
−
r
)
(
x
−
s
)
=
0
{\displaystyle ax^{2}+bx+c=a(x-r)(x-s)=0}
where r and s are the solutions for x. Completing the square on a quadratic equation in standard form results in the quadratic formula, which expresses the solutions in terms of a, b, and c. Solutions to problems that can be expressed in terms of quadratic equations were known as early as 2000 BC.
Because the quadratic equation involves only one unknown, it is called "univariate". The quadratic equation contains only powers of x that are non-negative integers, and therefore it is a polynomial equation. In particular, it is a second-degree polynomial equation, since the greatest power is two.
I know how to solve similar ODEs like
##
\frac{\partial^2 x}{ \partial t^2} + b \frac{\partial x}{ \partial t} + C x =0
##
Where one can let ## x = e^{rt}##, and the equation becomes
##
r^2 + b r + C =0
##
Which can be solved as a quadratic equation.
But now the problem is that there is...
what i tried to do is to write y=v_0tsin alpha - 1/2gt^2 and x=v_0 cos alpha tand that t=x/v_0 cos alphai plug t in the formula for y and get that y= x tan alpha - gx^2/v_0^2 (tan^2 alpha -1)since jaan klada said there should be a quadratic equation (because its a parabola) i thought that...
I was given a problem to solve that goes like this ##\frac{3}{|x+3|-1}\geq |x+2|## . I got the correct solution for all possible cases and here they are; for ##|x+3|\geq0## and ##|x+2|\geq## i got ##x\epsilon <-2, -2\sqrt{3} ]## and for ##|x+3|\leq0## , ##|x+2|\leq0## I got ##x\epsilon [-5...
https://www.technologyreview.com/2019/12/06/131673/a-new-way-to-make-quadratic-equations-easy/
An interesting article about solving ax2 + bx + c = 0 = (x-R)(x-S), where R and S are the roots.
## x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} ##
In my classes, we were never 'spoon fed' any formula, but...
Can someone please tell me where I am wrong. I am learning how to write ##a^{2} + bx + c## in this form ##f(x)= a(X-X_0)^{2} +Y_0##.
The method used in my textbook is a reduction to the perfect square. And it goes like this:
##f(x)=ax^2+bx+c##
##=a[x^2+\frac{b}{a}x]+c##
##=a\left [...
I am a bit confused, so if anyone can explain to me which way is right I would be very thankful.
I think that the way in pic 1 is right because of the properties written next to the procedure but the professor who posts videos on youtube solved it the way as written in pic 2 where he didn't...
I am refreshing on this...Have to read broadly...i will start with (b) then i may be interested in alternative approach or any correction that may arise from my working. Cheers.
Kindly note that i do not have the solutions to the following questions...
For (b), we know that,
say, if ##x=α##...
Let the roots of the given quadratic equation be ##x=α## and ##x=β## then our quadratic equation will be of the form;
$$x^2-(α+β)x+αβ$$
It follows that ##(α+β)=(r+is)## and ##αβ=4##.
We are informed that ##α^2+β^2=6i ## then $$6i=(r+is)(r+is)-8$$ ... $$8+6i=(r^2-s^2)+2rsi$$
solving the...
For part (i),
##(x-α)(x-β)=x^2-(α+β)x+αβ##
##α+β = p## and ##αβ=-c##
therefore,##α^3+β^3=(α+β)^3-3αβ(α+β)##
=##p^3+3cp##
=##p(p^2+3c)##
For part (ii),
We know that; ##tan^{-1} x+tan^{-1} y##=##tan^{-1}\left[\dfrac...
For part a,
We have ##α+β=b## and ##αβ =c##. It follows that,
##(α^2 + 1)(β^2+1)=α^2β^2+α^2+β^2+1)##
=##α^2β^2+(α+β)^2-2αβ +1##
=##c^2+b^2-2c+1##
=##c^2-2c+1+b^2##...
I subtract 5 from both sides to get 7x^2 = -5 Then I divide both sides by 7 to get -5/7. I then take the square root to get x = sqrt of the imaginary unit i 5/7 then ##\pm { i \sqrt \frac 5 7}##
The quadratic formula on the other hand gets me a different answer, the discriminant = -140 which...
I am confused in (iia) and (iib).
If $x^4 +( \alpha - 1) x^2 + \alpha + 2 = 0$ has real roots that means $y^2 + ( \alpha -1) + \alpha + 2 =0 $ should have at least one non-negative root. This means product of roots of (2) can be greater or less than zero...But I'm not able to comment on sum of...
Summary:: Hi guys, i can't seem to get the correct answer. I'm wondering where did I do wrong. Can someone help me to solve this? I think I need the correct formula to prove the answer :(
Given a root to 𝑥² + 𝑝𝑥 + 𝑞 = 0 is twice the multiple of another. Show that 2𝑝² = 9𝑞. The roots for 𝑥² +...
Hello:
I'm not sure if there's an accepted canonical form for a quadratic equation in two (or more) variables:
$$ax^2+by^2+cxy+dx+ey+f=0$$
Is it the following form? (using the orthogonal matrix Q that diagonalizes the quadratic part):
$$ w^TDw+[d \ \ e]w+f=0$$
$$w^TDw+Lw+f=0$$
where
$$...
Given equation and conditions: ##\boldsymbol{x^2+2(k-3)x+9=0}##, with roots ##\boldsymbol{(x_1,x_2)}##. These roots satisfy the condition ##\boldsymbol{-6<x_1,x_2<1}##.
Question : ##\text{What are the allowable values for}\; \boldsymbol{k}?##
(0) Let me take care of the determinant first...
The equation (a-1)x^2-4ax+4a+7=0 with a is a whole number has positive roots. If x_1>x_2 then x_2-x_1=...
A. –8
B. –5
C. –2
D. 2
E. 8
Since the equation has positive roots then x_1>0 and x_2>0 thus x_1+x_2>0 and x_1x_2>0
x_1+x_2>0
\frac{-(-4a)}{a-1}>0
x_1x_2>0
\frac{4a+7}{a-1}>0
However I...
On simplifying the given equation we get, x^2-x-1=0 and using the quadratic formula we get x=(1+√5)/2 and x=(1-√5)/2
Now, as the formula suggests, there are two possible values for x which satisfies the given equation.
But now, if we follow a process in any general calculator by entering...
Let me start by pasting the question as it appears in the text :My Attempt :
Given equation : ##\boldsymbol{2x^2+mx+m^2-5 = 0}##.
For the roots of this equation to be real, the discriminant : ##m^2-8(m^2-5) \ge 0\Rightarrow 7m^2-40\le 0\Rightarrow -\sqrt{\frac{40}{7}} \le m \le...
Given : The equation ##2x^2-(a^3+8a-1)x+a^2-4a = 0## with roots of opposite signs.
Required : What is the value of ##a## ?
Attempt : The roots of the equation must be of the form ##\alpha, -\alpha##. The sum of the roots ##0 = a^3+8a-1##.
I do not know how to solve this equation.
However...
Given : Equation ##x^2+(2m+1)x+(2n+1) = 0## where ##m \in \mathbb{Z}, n \in \mathbb{Z}##, i.e. both ##m,n## are integers.
To prove : If ##\alpha,\beta## be its two roots, then they are not rational numbers.
Attempt : The discriminant of the equation ##\mathscr{D} = (2m+1)^2 - 4(2n+1) =...
Given : The quadratic equation ##x^2+px+q = 0## with coefficients ##p,q \in \mathbb{Z}##, that is positive or negative integers. Also the roots of the equation ##\alpha, \beta \in \mathbb{Q}##, that is they are rational numbers. To prove that ##\boxed{\alpha,\beta \in \mathbb{Z}}##, i.e. the...
It is given that ##x_1, x_2\; \text{and}\; x_3## are roots of the equation ##ax^2+bx+c=0##, which are pairwise distinct.
If indeed they are roots, we should have ##ax_1^2+bx_1+c= 0 = ax_2^2+bx_2+c= 0 = ax_3^2+bx_3+c= 0##.
On subtracting the first two, we obtain ##a(x_1^2-x_2^2)+b(x_1-x_2) =...
Hi , I had to solve a quadratic equation , i got two roots as an answer ( ans= x1 / x2) , and now i need to use one of those answers to complete further tasks like finding y from x+y=c so i need to use x1 and x2 from roots , i was wondering if that's possible and how
I was thinking of this simple equation here, ## x^2 = 4##. Many students present the solution as follows.
$$ x^2 = 4 $$
$$ \therefore x = \sqrt{4} = \pm 2 $$
Now, even though the final answer is correct, there is a mistake in arriving at the solution. Square root symbol means that we have to...
Problem Statement: ##2x^2 + y^2 = 4##
Max and min of
##4x + y^2## ?
Relevant Equations: F'(x) = 0
To find critical point
X = 0 -> y = +- 2
Y =0 -> x = +- ##\sqrt 2##
I input that
Find f(xy) = 4 and f(xy) = +-##4\sqrt 2##
To find with derivative i find both x and y are zero
The answer is wrong
Sample Problem 2.04 Drag race of car and motorcycle
I was following all the way up to using the quadratic equation for this problem...(please see img for a more detailed attempt at a solution.)
So I may have simplified incorrectly here:
but I came up with
2.8t^2-(58.8)t+408.1=0
but when...
Quadratic equation
Ax^2+Bxy+Cy^2+Dx+Ey+F=0
is
(a) elipse when ##B^2-4AC<0##
(b) parabola when ##B^2-4AC=0##
(c) hyperbola when ##B^2-4AC>0##
I found this in Thomas Calculus. However for some values of parameters ##A=17##, ##C=8##, ##B=\sqrt{4 \cdot 17 \cdot 8}##, ##D=E=0##, ##F=20## I got just...
Homework Statement
Go through question number 4
Homework Equations
The Attempt at a Solution
See basically the question is asking us to find the range of the given function x/(x^2+x+1).
So,I began solving it this way...
I am stuck at this step.
I asked my friend for a hint and he told me to...
1. Determine the equation that represents the relationship between the power and the current when the electric potential difference is 24v and the resistance is 1.5 Ω. 2. Draw a graph of the parabola that corresponds to the equation found in (a). 3. Determine the current needed in order for...
Homework Statement
Homework Equations
Discriminant : b^2-4ac
When discriminant = 0 The function has two equal real roots
When discriminant < 0 The discriminant has NO real roots
When discriminant > 0 The function has 2 different real roots
The Attempt at a Solution
Why is it so that...
Hello everyone! Apologies if this is a very repetitive question but I have gone through previous forum posts and am still struggling to understand how to identify which equations are appropriate. In the problem below, I have used the kinematic equation of "v = v(i) + at" but my answer is...
In a graph , straight line intersects the parabola at(-3,9) & (1, 1) Then the equation is
A) x^2-2x+3=0
B) x^2+2x-3=0
C) x^2-3x+2=0
D) x^2-2x-3=0
I know that I can find the answer by substituting the known values to each options, but how to do it the proper way? We need at least three known...
Given that $\alpha$ and $\beta$ are the roots of a quadratic equation, evaluate $\frac{1}{\alpha^2}+\frac{1}{\beta^2}$.
I find this question to be interesting.
The answer based on the answer key is 3 seconds. I used the quadratic equation to solve for t. My question is how do we know what sign to use when solving for the final value? For this problem, I had to use the negative sign, but I knew that I needed to use the negative sign because I already...
If a quadratic equation of two variables represents a conic section (planar intersection of a cone), then does a quadratic equation of three variables represent the complete cone?
@fresh_42 @FactChecker @WWGD
Homework Statement
A 1.2 kg block is dropped from a height of 0.5 m above an uncompressed spring. The spring has a spring constant k = 160 N/m and negligible mass. The block strikes the top end of the spring and sticks to it
Find the compression of the spring when the speed of the block...
Hello.
Assume that I have two polynomials of degree 2, i.e., Quadratic Equations.
1.
Assume that the Quadratic Equation is:
x2 + 7x + 12 = 0
The roots of the Quadratic Equation is -3 and -4.
2.
Assume that there is another Quadratic Equation:
x2 + 8x + 12 = 0
The roots of the Quadratic...
Homework Statement
Homework Equations
for equation which has 2 different solutions, D >0
The Attempt at a Solution
(1)[/B] D > 0
b^2 - 4ac > 0
3 - 4root2.k > 0
k < 3 / ( 4root 2 )
k < ( 3 root 2 ) /8
has solution of sin tetha and cos tetha
sin 0 = 0, cos 0 = 1.
when x = 0, and x = 1 -->...
Verify that the numbers 1 + √5 and 1 - √5 both satisfy the equation x^2 - 2x - 4 = 0.
I believe the question is asking to plug the given numbers into the quadratic equation and evaluate individually.
Let x = 1 + √5 and evaluate.
Let x = 1 - √5 and evaluate.
Both numbers should yield 0 = 0...