Can you solve these equation by completing the square
2x^2+7x-4=0
3x^2x-1=0
I'll do one, and you try the other.
2x^2+7x = 4
(x^2 + 7/2 x) = 2
(x^2 + 7/2 x + (7/4)^2) = 2 + (7/4)^2
(x + 7/4)^2 = 81/16
x + 7/4 = ±9/4
x = -7/4 ± 9/4
x = -4 or 1/2
Just recall that (x+a)^2 = x^2 + 2ax + a^2
That's why I added (7/4)^2 to both sides.
The second one has a typo, so fix it and follow the same logic.
Sure! Let's solve each equation step-by-step using the method of completing the square.
1. 2x^2 + 7x - 4 = 0
Step 1: Move the constant term to the other side of the equation:
2x^2 + 7x = 4
Step 2: Divide the entire equation by the coefficient of the x^2 term to make the coefficient 1:
x^2 + (7/2)x = 2
Step 3: Take half of the coefficient of the x term, square it, and add it to both sides of the equation:
x^2 + (7/2)x + (7/4)^2 = 2 + (7/4)^2
x^2 + (7/2)x + 49/16 = 32/16 + 49/16
x^2 + (7/2)x + 49/16 = 81/16
Step 4: Write the left side as a perfect square trinomial:
(x + 7/4)^2 = 81/16
Step 5: Take the square root of both sides of the equation:
x + 7/4 = ±√(81/16)
Step 6: Simplify the square root on the right side:
x + 7/4 = ±(9/4)
Step 7: Solve for x by subtracting 7/4 from both sides:
x = -7/4 ± 9/4
Therefore, the solutions to the equation 2x^2 + 7x - 4 = 0 are:
x = -4/2 or x = 1/2
2. 3x^2x - 1 = 0
I assume there is an error in the equation you provided. Please double-check your equation and resubmit it.
Yes, I can help you solve these equations by completing the square. To complete the square, we need to manipulate the equation so that one side is a perfect square trinomial. Here's how you can solve each equation step by step:
Equation: 2x^2 + 7x - 4 = 0
Step 1: Move the constant term to the other side of the equation.
2x^2 + 7x = 4
Step 2: Divide the entire equation by the coefficient of x^2 to make the coefficient 1.
x^2 + (7/2)x = 2
Step 3: Take half of the coefficient of x, square it, and add it to both sides of the equation.
x^2 + (7/2)x + [(7/2)/2]^2 = 2 + [(7/2)/2]^2
x^2 + (7/2)x + (7/4)^2 = 2 + (7/4)^2
x^2 + (7/2)x + 49/16 = 2 + 49/16
Step 4: Simplify the equation.
x^2 + (7/2)x + 49/16 = 32/16 + 49/16
x^2 + (7/2)x + 49/16 = 81/16
Step 5: Write the left side of the equation as a squared binomial.
(x + 7/4)^2 = 81/16
Step 6: Take the square root of both sides of the equation and remember to consider both positive and negative roots.
x + 7/4 = ±√(81/16)
Step 7: Simplify the equation.
x + 7/4 = ±(9/4)
Step 8: Solve for x.
x = -7/4 ± 9/4
Therefore, the solutions for the equation 2x^2 + 7x - 4 = 0 are x = -4/2 or x = 1.
Now let's move on to the next equation:
Equation: 3x^2 + x - 1 = 0
This equation is already in the form of ax^2 + bx + c = 0, so we can directly proceed to completing the square.
Step 1: Move the constant term to the other side of the equation.
3x^2 + x = 1
Step 2: Divide the entire equation by the coefficient of x^2 to make the coefficient 1.
x^2 + 1/3x = 1/3
Step 3: Take half of the coefficient of x, square it, and add it to both sides of the equation.
x^2 + 1/3x + [(1/3)/2]^2 = 1/3 + [(1/3)/2]^2
x^2 + 1/3x + (1/6)^2 = 1/3 + (1/6)^2
x^2 + 1/3x + 1/36 = 1/3 + 1/36
Step 4: Simplify the equation.
x^2 + 1/3x + 1/36 = 12/36 + 1/36
x^2 + 1/3x + 1/36 = 13/36
Step 5: Write the left side of the equation as a squared binomial.
(x + 1/6)^2 = 13/36
Step 6: Take the square root of both sides of the equation and remember to consider both positive and negative roots.
x + 1/6 = ±√(13/36)
Step 7: Simplify the equation.
x + 1/6 = ±(√13/6)
Step 8: Solve for x.
x = -1/6 ± √13/6
Therefore, the solutions for the equation 3x^2 + x - 1 = 0 are x = (-1 ± √13)/6.