What are the values of m for which the expression [math]2x^2 + mxy + 3y^2 - 5y - 2[/math] can be factorized in to two linear factors ?
A2A, i have no idea why as I’ve never answered a maths question on Quora before, but thanks i guess, Helps with my prepImagine the equation is quadratic in one of the two variables, say y. It can be written as (ay^2+ by+c) In this situation, a = 3, b = mx - 5, and c = 2x^2 - 2.now a quadratic equation factors into linear factors only when the discriminant b^2 - 4ac is a perfect square trinomial. Applying it to our current equationb^2 - 4ac = (mx - 5)^2 - 4(3)(2x^2 - 2)= (m^2 - 24)x^2 - 10mx + 49= (m^2 - 24)x^2 - 10mx + 7^2Taking it as an equation of the form a^2 +(-) 2ab + b^2, we can write the middle term in the following way= (m^2 - 24)x^2 - 2(7)(5mx/7) + 7^2.In above equation, a^2= (m^2 - 24)x^2, and b^2= 49, and also a= (5mx/7)which means(m^2 - 24)x^2 = (5mx/7)^2 for all x.=> m^2 - 24 = (25/49)m^2,=>49m^2 - (24 * 49) = 25m^2=>24m^2 = 24 * 49=> m^2 = 49=>m = +-7.So the possible answers are (a) and (b).
What does it mean to solve for y in terms of x?
Solve for ‘y' means, solve the equation to get the value of y.And in terms of x , means, value of y not necessarily in pure constant form, but in the form of x.Example: Solve the following for y in terms of x3y +x = 7=> y = (7-x)/3Here, i solved the equation for y, in terms of x.
What is the value of k for which the system of equations Kx+2y=5 and 3x+y=1 has no solution? Later after solving plz tell me what is no solution?
For no solution :Mathematically the equations Written are two linear (straight lines) equations and a solution means A point where these two intersects point is ( values of x,values of y)Since for no solution logically Both the lines should have same slopes i.e. slope Of first line should be equal To slope of the second line Therefore ;Slope of first line = -k/2 = m1Slope of second line = -3 = m2NowPut m1=m2That means K/2 =3 now k =6For k=6 there is no solution.For solution any value Of k except 6 will give the solution.
If log p=x and log q=y,evaluate the following in terms of x and y...continued below?
If log p=x and log q=y, evaluate the following in terms of x and y: (a) log (p^-5q^-8) (b) log (p^-6q^-4)^1/2 (c) log (p^1)/(q^-4) (d) log p^-6/ log q^-2 (e) (log p^3)^-8 Rephrase: expand as much as possible, then replace each log(p) by x and each log(q) by y. Not that it has to be literally log(p) that you change to x, there can be nothing else in the parentheses or else you haven't finished expanding.
How can one find the values of k for which the line 2x -k is tangent to the circle with the equation x^2 + y^2 = 5?
The simple way to do this is to clearly define what it means for tangent so that finding the k values is the easiest. Problem Specific AnswerWe have [math]y = 2x - k[/math] and [math]x^2 + y^2 = 5[/math] and the line is tangent to the circle. What it means for a 2 things to be tangent is that one point and only one point satisfies both equations at the same time. That means that in the above two equations, only the point represented by [math](x,y)[/math] satisfies both equations. We get: [math]y = 2x - k[/math] and [math]x^2 + y^2 = 5 [/math]Substituting, we get [math]x^2 + (2x-k)^2 = 5.[/math] This equation must only have one solution. [math]5x^2 -4kx +(k^2 - 5) = 0.[/math]This quadratic equation must have its determinant equal to 0 in order for the two to be tangent. [math]b^2 - 4ac = 0[/math][math]16k^2 - 20k^2+100 = 0[/math][math]4k^2 = 100[/math][math]\boxed{k = \pm 5}[/math]Generic AnswerWe can generalize this solution for any line and circle.Line: [math]y = mx+b[/math]Circle: [math]x^2 + y^2 = r^2[/math]Substituting we get: [math]x^2 + (mx+b)^2 = r^2[/math][math]\Rightarrow (m^2 + 1)x^2 + 2mbx + (b^2 - r^2) = 0[/math]Taking the discriminant and making it equal to 0, we get[math]4m^2b^2 - 4(m^2 + 1)(b^2 - r^2) = 0[/math][math]4m^2b^2 - 4m^2b^2 + 4m^2r^2 - 4b^2 + 4r^2 = 0[/math][math]m^2r^2 + r^2 = b^2[/math][math]m, r,[/math] and [math]b[/math] must satisfy these equations in order to be tangent. We can check out previous answer by plugging in here: [math]m = 2; r = \sqrt{5}; b = -k.[/math][math]20 + 5 = k^2.[/math]Therefore, [math]\boxed{k = \pm 5}[/math]
Where does the line represented by the equation -2x + y-7=0 intersect the y axis?
U can fine this by 2 methods atleastIf u want to find the intercept on y axis than at y axis x cordinate will be 0 so simply put x=0 so u will get 7Now the other method is to create this equation of line in intercept form that is . (x/a)+(y/b)=1. ———————(A)Wher a and b are intercept on x and y axis respectivelyNow to create this we can write our equation as-2x +y= 7. —————-(B)Now as eqa. A has 1 at right soDivide equa B by 7(-2x/7)+(y/7)=1Now as equation A has coefficients of y and x as 1 so[x/(-7/2)]+(y/7)=1. ——————-(C)So on comparing euation C by AWe geta=(-7/2)b=7So line intersects x axis at (-7/2) and y ant 7…
3x-y=7 ; x+4y=11 solve this equation by cramer's rule?
3x - y = 7 … (1)x + 4y = 11 … (2)Multiply (1) by 4 to get12x - 4y = 28 … (3)x + 4y = 11 … (2)Add (3) and (2), to get13x = 39, or x = 3Put this value x = 3 in (1) and you get9 - y = 7or 9–7 = y or y = 2.Hence x =3, and y =2.
What is the solution for an equation of a line passing through the point of intersection of 2x-3y-5=0 and 7x-5y-2=0 and parallel to the lines 2x-3y+14=0?
Hope you understood… :)
Consider the curve defined by the equation y^5-y-x^2= -1?
a. find the equations of three different lines which are tangent to the curve at x=1 b. find the coordinates of the point(s) at which the tangent line to the curve is vertical. express the coordinates numerically to three decimal places. c. find the coordinates of the point(s) at which the tangent line to the curve is horizontal. express the coordinates numerically to three decimal places. for part a, i got dy/dx= 2x/5y^4-1 then i plugged in 1 into x and got 2/5y^4-1 and set it to 0 for the denominator, i got 5y^2 (y^2-1/5y^2)= 0 and then i got y= + or - the square root of 5 but i can't solve (y^2-1/5y^2)= 0 to find the other y's there's supposed to be three equations of lines so that means there needs to be 3 y's, right? i don't know what to do from here.