Solve for x: ax2 + 4a2 – 3b x – 12ab 0
WebFeb 13, 2015 · question_answer Answers (1) person. Raghunath Reddy. Given 6a 2 x 2 - 7abx - 3b 2 = 0. Compare to ax 2 + bx + c = 0. a = 6a 2 , b = -7ab , c = - 3b 2. The quadratic formula to find the roots of quadratic equation are −b±√b2−4ac 2a - b ± b 2 - 4 a c 2 a , where b 2 - 4ac is called the discriminant of the quadratic equation. WebTo solve an equation of type √ (ax + b) = cx + d or √ (ax 2 + bx + c) = dx + e, square both the sides. To solve √ (ax + b) ± √ (cx + d) = e, transfer one of the radical to the other side and square both the sides. Keep the expression with radical sign on one side and transfer the remaining expression on the other side.
Solve for x: ax2 + 4a2 – 3b x – 12ab 0
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WebFeb 16, 2024 · Solve the following quadratic equations by factorization: ax2+(4a2-3b)x-12ab=0. LIVE Course for free. Rated by 1 million+ students Get app now Login. … WebOct 10, 2024 · a x ( x + 4 a) − 3 b ( x + 4 a) = 0. ( a x − 3 b) ( x + 4 a) = 0. a x − 3 b = 0 or x + 4 a = 0. a x = 3 b or x = − 4 a. x = 3 b a or x = − 4 a. The roots of the given quadratic equation are 3 b a and − 4 a. Updated on 10-Oct-2024 10:20:37. Previous Page Print Page Next Page.
WebFeb 4, 2012 · 12abx2-(9a2-8b2)x-6ab=0 Solve for x. Asked by 04 Feb, 2012, 07:30: AM Expert Answer 12abx 2 - 9a 2 x + 8b 2 x -6ab = 0. 3ax(4bx - 3a ... (2x + 2) days and (x - y) days, respectively. Reena and Chhaya together can complete the whole work in 18 days and Jaya and Payal together can complete the whole work in 24 days. Payal and ... WebFeb 19, 2013 · x=ax 2 +(4a 2-3b)x-12ab=0 =ax 2 +4ax 2-3bx-12ab=0 =ax(x+4a)-3b(x+4a)=o =(ax-3b)(x+4a)=o =ax-3b=0,x+4a=0 =ax=3b,x=-4a =x=3b/a,x=-4a. 57
WebOct 15, 2024 · 2a + 3b = 8 . ab = 5. Formula used: (x + y) 2 = x 2 + y 2 + 2xy. Calculations: Squaring the given equation, ⇒ (2a + 3b) 2 = 64. ⇒ 4a 2 + 9b 2 + 12ab = 64. Substituting the value of ab, ⇒ 4a 2 + 9b 2 + (12 × 5) = 64. ⇒ 4a 2 + 9b 2 … WebOct 10, 2024 · Given: Given quadratic equation is $ax^2\ +\ (4a^2\ –\ 3b)x\ –\ 12ab\ =\ 0$. To do: We have to solve the given quadratic equation by factorization.
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WebIn math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0. How do you know if a quadratic equation has two solutions? A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive. phoenix 10 condo rentals orange beachWebSep 17, 2024 · 2.5: Solving Matrix Equations AX=B. T/F: To solve the matrix equation A X = B, put the matrix [ A X] into reduced row echelon form and interpret the result properly. T/F: The first column of a matrix product A B is A times the first column of B. Give two reasons why one might solve for the columns of X in the equation A X = B separately. ttc supply chainWebMar 18, 2024 · Solve the following simultaneous linear equations by using Cramer's rule 2x - y + 3z = 8, -x+2y+z= 4, 3x + y - 4z = 0 1. four times a number minus twenty-nine is eleven. … ttc subway timingsWebMar 14, 2013 · Below is the Program to Solve Quadratic Equation. For Example: Solve x2 + 3x – 4 = 0. This quadratic happens to factor: x2 + 3x – 4 = (x + 4) (x – 1) = 0. we already know that the solutions are x = –4 and x = 1. # import complex math module import cmath a = 1 b = 5 c = 6 # To take coefficient input from the users # a = float (input ... ttc subway wifiWebSolve for x. a x − a 2 = b x − b 2. Collect all terms with x on one side of the equation. a x − b x = a 2 − b 2. Factor both sides of the equation. ( a − b) x = ( a + b) ( a − b) Divide both sides of the equation by the coefficient of x (which is a − b) x = a + b. (where a ≠ b since this would mean dividing by 0) ttc subway station mapsWebThe solution (s) to a quadratic equation can be calculated using the Quadratic Formula: The "±" means we need to do a plus AND a minus, so there are normally TWO solutions ! The blue part ( b2 - 4ac) is called the "discriminant", because it can "discriminate" between the possible types of answer: when it is negative we get complex solutions. phoeniverWebMy first thoughts were finding the properties of a derivative by using the conditions given above and possibly getting a system to solve. Nevertheless, as of now, I don't find any way how to develop this and then find the constants. phoenicurus erythronotus