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- How do you solve 2x^2 + 7x = 5? - Socratic
Solve using the quadratic formula: #x=(-b+-sqrt(b^2-4ac)) (2a)# Plug in the values from the quadratic formula
- How do you factor and solve 3x^2 – x – 1 = 0? - Socratic
x = "either " 0 3837959396 " " (1 + sqrt13) 12 or = -0 217129273 " " (1 - sqrt13) 12 We can use the quadratic formula to solve this 3x^2 - x- 1 = 0 ax^2 + bx
- How do you find the zeros, real and imaginary, of y=2(x+4)^2+4 using . . .
How do you solve #x^2+10x+9=0# using the quadratic formula? How do you solve #-4x^2+x+1=0# using the quadratic formula? When do you have "no solution" when solving quadratic equations using the quadratic formula?
- How do you find the roots, real and imaginary, of # h=--t^2 . . . - Socratic
How do you solve #x^2+10x+9=0# using the quadratic formula? How do you solve #-4x^2+x+1=0# using the quadratic formula? When do you have "no solution" when solving quadratic equations using the quadratic formula?
- How do you factor #2x^3+x^2-3x-14# - Socratic
So trying the positive possible rational zeros of f(x) we find: f(2) = 2(color(blue)(2))^3+(color(blue)(2))^2-3(color(blue)(2))-14 = 16+4-6-14=0 So x = 2 is a zero and x-2 a factor: 2x^3+x^2-3x-14 = (x-2)(2x^2+5x+7) The quadratic factor is in the form: ax^2+bx+c with a=2, b=5 and c=7 This has discriminant Delta given by the formula: Delta = b^2
- How do you factor 6c^2 + 17c-14=0? - Socratic
6c^2+17c-14 = 6(c+7 2)(c-2 3) First of all, try to solve it using the classical formula \\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} and in your case a=6, b=17 and c=-14 So
- How do you solve #-3x^2 –10 x + 8 = 0 - Socratic
Solve y = -3x^2 - 10x + 8 = 0 Ans: 2 3 and -4 y = - 3x^2 - 10x + 8 = 0 (!) I use the new Transforming Method (Google, Yahoo Search) Transformed y' = x^2 - 10x - 24 = 0 (2) Roots have opposite signs Factor pairs of (-24) --> (-2, 12) This sum is 10 = -b Then, the 2 real roots of (2) are: -2 and 12 Back to equation (1) the 2 real roots are: (-2) -3 = 2 3 and 12 -3 = -4
- How do you find the solution to the quadratic equation
This sum is (4 - 2 = 2 = -b) Then, the 2 real roots are: -2 and 4 to the quadratic equation #x^2-2x-8
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