How do you solve #2x^2=6x-2# using the quadratic formula?

Answer 1

To solve the equation (2x^2 = 6x - 2) using the quadratic formula, follow these steps:

  1. Rewrite the equation in standard form: (2x^2 - 6x + 2 = 0).
  2. Identify the coefficients: (a = 2), (b = -6), (c = 2).
  3. Substitute the coefficients into the quadratic formula: (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}).
  4. Calculate the discriminant: (b^2 - 4ac = (-6)^2 - 4(2)(2) = 36 - 16 = 20).
  5. Substitute the discriminant into the formula: (x = \frac{{6 \pm \sqrt{20}}}{4}).
  6. Simplify the square root of 20: (x = \frac{{6 \pm 2\sqrt{5}}}{4}).
  7. Simplify the fraction: (x = \frac{{3 \pm \sqrt{5}}}{2}).

Therefore, the solutions to the equation (2x^2 = 6x - 2) using the quadratic formula are (x = \frac{{3 + \sqrt{5}}}{2}) and (x = \frac{{3 - \sqrt{5}}}{2}).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

#x=(3+sqrt(5))/2# or #(3-sqrt(5))/2#

Using quadratic formula, solution of the equation #ax^2+bx+c=0# is given by #x=(-b+-sqrt(b^2-4ac))/(2a)#. Note that the quadratic equation is in its general form.
Hence converting the equation to its general form, the equation #2x^2=6x−2# this becomes #2x^2-6x+2=0# or dividing by #2#, #x^2-3x+1=0# and as such #a=1, b=-3# and #c=1#.
Hence, solution is given by #x=(-(-3)+-sqrt((-3)^2-4*1*1))/(2*1)# or
#x=(3+-sqrt(9-4))/2# or #x=(3+-sqrt(5))/2# i.e.
#x=(3+sqrt(5))/2# or #(3-sqrt(5))/2#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To solve the equation 2x^2 = 6x - 2 using the quadratic formula, follow these steps:

  1. Rewrite the equation in the standard form: 2x^2 - 6x + 2 = 0.
  2. Identify the values of coefficients a, b, and c: a = 2, b = -6, and c = 2.
  3. Substitute the values into the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
  4. Plug in the values: x = (-(−6) ± √((-6)^2 - 4(2)(2))) / (2(2)).
  5. Simplify: x = (6 ± √(36 - 16)) / 4.
  6. Further simplify: x = (6 ± √20) / 4.
  7. Split into two solutions: x = (6 + √20) / 4 and x = (6 - √20) / 4.
  8. Simplify the radicals if possible: x = (6 + 2√5) / 4 and x = (6 - 2√5) / 4.
  9. Reduce the fractions: x = (3 + √5) / 2 and x = (3 - √5) / 2.

So, the solutions to the equation 2x^2 = 6x - 2 using the quadratic formula are ( x = \frac{3 + \sqrt{5}}{2} ) and ( x = \frac{3 - \sqrt{5}}{2} ).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7