Quadratic formula for 2x2-3x-2=0? or no solution?
This is not a quadratic equation...it is a simple equation: 2x2-3x-2=0 = 4-3x-2=0 = 4-2=3x = 2=3x : x=2/3 a quadratic equation requires a squared and a simple unknown eg: aX^2 + bX + c = 0....which reads: a times X squared plus b times X plus c equals zero then you can use the formula: X =[ -b^2 +or- sq.rt (b^2 -4ac)] /2a which reads: X equals minus b squared plus or minus the square root of (b squared minus 4 times a times c) ALL divided by two times a Quadratic equations always have two roots/ solutions which is why you have plus or minus in the formula. The equation sould be solved for BOTH plus and minus the square root function. it is always difficult to explain formulae in plain text but I hope this has helped. APOLOGIES.... I didn't realise you were using 2x2 to represent 2x^2......as said notation in plain text are difficult. 2x^2 - 3x - 2 =0 will factorize (2x +1)(x - 2) = 0 so either (2x +1) = 0 or (x - 2) = 0 so 2x = -1 ie x = -1/2 or x = +2 However the formula given above will work for all quadratic equations, and is especially useful when the factorisation is not obvious.
If the difference between the roots of the equation x^2-px+8=0 is 4, then what is the value of ‘p’?
Well, although others have already posted the correct answer to this question, yet I'm gonna answer it anyway because I have a different and easier (according to me) method, than the ones mentioned. Here we go!Given polynomial p(x) = x²-px+8Let the zeroes be α and β respectively.We know that α + β = -(coefficient of x)÷coefficient of x²Therefore, α + β = - (-p)÷1=> α + β = p …(I)Also, it's given that, α - β = 4 …(II)Also, αβ = constant term÷coefficient of x²Therefore, αβ = 8÷1 = 8 …(III)Now, squaring both sides in (I):=> (α + β)² = p²=> α² + β² + 2αβ = p²=> (α - β)² + 2αβ + 2αβ = p² [∵α² + β² = (α-β)² + 2αβ]=> (α - β)² + 4αβ = p²=> (4)² + (4 × 8) = p² [By (II) and (III)]=> 16 + 32 = p²=> p = ±√48=> p = 4√3 or -4√3Cheers! Further queries are appreciated.
Find the value of K, if the quadratic equation 3x^2-K root3x+4=0 has equal roots?
We can solve this problem using two different approaches.Using the sum and product of roots of a quadratic equationUsing the discriminant b^2 -4ac of the quadratic formulaMethod 1:Let us say that the two real and equal roots of the given quadratic function are “P”We observe that in comparison with standard form,a= 3b= -ksqrt3c =4.The Sum of the roots of a quadratic function is given by the formula -b/a and the product of roots is given by the formula c/a.For more information on sum and product of roots, you can please refer to the article below.Sum and product of the roots of a quadratic equation.Sum of the roots = P+P = - (-ksqrt3 /3)=>2p= k (sqrt3 )/3 ( dividing by 2 both sides)=>P = k (sqrt3 )/6————→equation-1The product of roots is p*p = c/a =4/3P^2 = 4/3—- Equation-2let us square the equation-1 both sides P^2= [k (sqrt3 )/6]^2 =k^2*3/36 =k^2/12therefore P^2= k^2/12—→ Equation -3P^2 = k^2/12 and from Equation-3 and we already have P^2 = 4/3So equting equation-2 & Equation-3 we havek^2 /12 =4/3K^2= 16K = +/-4Therefore the value of would be 4 or -4 for the quadratic to have equal roots.Method 2:Discriminant b^2 -4ac =0 when a quadratic function has equal roots.For more explanation for this statement, you can please refer to the article belowFinding roots using the Quadratic Formula--Conceptmathhelp.comwe hav ethe values of a,b,c on comparision with standard form andtherefore (-ksqrt3 )^2 - 4(3)(4) =0=>k^2 *3–48 =0=> 3k^2 =48=>k^2 =48/3= 16=> k= sqrt16 = +/-4Therefore k = 4 or -4 for the given quadratic to have equal roots.RegardsSaiFounder conceptmathhelp.com
If one root of quadratic equation is given how to find other root?
If the given root is complex, then it’s conjugate will be the other root. So, if [math]a+ib[/math] is a root, then [math]a-ib[/math] is also a root.If the given root is a surd of the form [math]a+\sqrt{b}[/math], then [math]a-\sqrt{b}[/math] is also a root.If the root given is rational, then we can use the product of the roots rule.The product of the roots of the quadratic equation [math]ax^{2}+bx+c=0[/math] is [math]\frac{c}{a}[/math].So, in this case, we can find the other root by dividing this product by the given root.If the co-efficient of [math]x^{2}[/math] is ‘1’ ([math]i.e.a=1[/math]), then this becomes much easier. You just need to divide the constant term of the quadratic by the known root.Example: If one root of the quadratic equation [math]x^{2}-14x+45=0[/math] is given as 5, then the other root is [math]\frac{45}{5}=9[/math].
What are the real or complex roots of [math]4x^5-20x^4+3x^3-31x^2-9x+1=0[/math]?
Looks like a HW Q, no? It's of odd degree, so it has at least one real root; use the rational root theorem to list candidates and then check them, eg, using synthetic division. If it's a HW problem, at least one of the roots is rational, most likely; once you've found it, write down the linear binomial factor it implies, and figure out the quartic polynomial that's the other factor. There is a complicated formula for the roots of a general quartic, but unless the lesson was on its use, there are probably two more rational roots (or perhaps one of multiplicity greater than one) you are also meant to find and factor out their linear binomial factors 'til the remaining factor is a quadratic, and then use the quadratic formula to find the last two. Oh, but before you do any of that, might I suggest that you graph the thing (i.e., y=the thing) using a graphing calculator: that will tell you, if you've been paying attention in class, where to look for the real roots, if any such roots have multiplicity greater than one, and how many--hint: an even number--complex roots to expect. If all this is Greek to you, either go back and study some more, or I'm afraid you're in way over your head. (But you've suggested which Chapter of Sullivan & Sullivan III I should first post online: thanks!)
A polynomial leaves remainder [math]2[/math] when divided by [math]x-1[/math] and remainder [math]1[/math] when divided by [math]x-2[/math]. If the polynomial is divided by [math](x-1)(x-2)[/math], then what would be the remainder?
By remainder theorem, “When f(x) is divided by (x−1), the remainder is 2” can be translated to:-[math]f(1)=2[/math] ….(Equation-1)Similarly, we have:-[math]f(2)=1[/math].......(Equation-2)When [math]f(x)[/math] is divided by [math](x−1)(x−2)[/math], then basically we have:-[math]f(x)=(x−1)(x−2)[/math]Quotient + Remainder……(Equation-3)Since the degree of the remainder must be one lower than that of the divisor, [math]=2[/math] from [math](x−1)(x−2)[/math]], the remainder can have degree = 1 (or lower) only. The remainder should then take the form [math]ax+b[/math], which is a general expression of degree 1 (or lower if a = 0) in x. Therefore, (3) becomes:-[math]f(x)=(x−1)(x−2)Q(x)+(ax+b)[/math].....(Equation-4)(1) and (2) can be used to find the values of [math]a[/math] and [math]b[/math] from (4).f(1) = a+b and f(2) = 2a+b…. (From equation 1,2 and 4)a+b = 2 and 2a+b=1Solving the given equation,a= -1 and b=3Thus, Remainder = ax+b = -x+3
1. Which describes the polynomial 3x^3 - 2x + 1? (best answer to whoever does it in 20 mins.) =)?
1.) cubic trinomial 2.) A.17x^3 -15x^2 + 2x + 3 3.) x^2 -2x -9 4.) 6x^2 + 2x + 6 <--- can't read the question but can guess from the hints from the question 5.) 12x^2- 11x + 2 7.) 2x(2x^2 + 9x + 1) 8.) (2x + 3)(2x- 3) 9.) (x- 4)^2 10.)(x- 5)(x + 12) 12.) p(x) = 6x^3 13.) (x^2- 4y)(x^4 + 4x^2y + 16y^2) 14.) f(x) = x^2
This is simple math. When 3/(x+2) = x/2, how do I find what x =?
I’m going to solve in a different manner than most other answers here.3/(x+2)=x/2Multiply both sides by 2(x+2), assuming x=-2 is not a solution.6=x(x+2)Distribute:6=x^2+2xNow here is where my answer differs.Most people here will use the quadratic formula at this point, which is not only tedious and hard to memorize, but also prone to mistakes. Here is an easier method:Completing the Square.Add 1 to the equation.x^2+2x+1=7(x+1)^2=7x+1= plus or minus sqrt(7)x = -1 plus or minus sqrt(7)Now this seems simpler than the quadratic formula. Follow the KISS principle: Keep It Simple, Stupid.