How do you solve by completing the square: #2x^2  7x 15 = 0#?
By signing up, you agree to our Terms of Service and Privacy Policy
To solve the quadratic equation 2x^2  7x  15 = 0 by completing the square, follow these steps:
 Move the constant term to the other side of the equation: 2x^2  7x = 15.
 Divide all terms by the coefficient of x^2 to make the leading coefficient equal to 1: x^2  (7/2)x = 15/2.
 To complete the square, take half of the coefficient of x, square it, and add it to both sides of the equation: x^2  (7/2)x + (7/4)^2 = 15/2 + (7/4)^2.
 Simplify the left side of the equation: (x  7/4)^2 = 15/2 + 49/16.
 Find a common denominator for the terms on the right side and combine them: (x  7/4)^2 = (120/16 + 49/16).
 Add the terms on the right side: (x  7/4)^2 = 169/16.
 Take the square root of both sides of the equation: x  7/4 = ±√(169/16).
 Simplify the square root: x  7/4 = ±(13/4).
 Add 7/4 to both sides of the equation: x = 7/4 ± 13/4.
 Simplify the expressions: x = (7 ± 13)/4.
Therefore, the solutions to the equation 2x^2  7x  15 = 0 by completing the square are x = (7 + 13)/4 and x = (7  13)/4, which simplify to x = 5 and x = 3/2.
By signing up, you agree to our Terms of Service and Privacy Policy
To solve the quadratic equation (2x^2  7x 15 = 0) by completing the square:

Move the constant term to the other side: [2x^2  7x = 15]

Divide all terms by the coefficient of (x^2) (in this case, 2): [x^2  \frac{7}{2}x = \frac{15}{2}]

To complete the square, take half of the coefficient of (x), square it, and add it to both sides of the equation: [x^2  \frac{7}{2}x + \left(\frac{7}{4}\right)^2 = \frac{15}{2} + \left(\frac{7}{4}\right)^2]

Simplify the equation: [x^2  \frac{7}{2}x + \frac{49}{16} = \frac{15}{2} + \frac{49}{16}]

Combine the constants on the right side: [x^2  \frac{7}{2}x + \frac{49}{16} = \frac{120}{16} + \frac{49}{16}] [x^2  \frac{7}{2}x + \frac{49}{16} = \frac{169}{16}]

Rewrite the left side as a perfect square: [\left(x  \frac{7}{4}\right)^2 = \frac{169}{16}]

Take the square root of both sides: [x  \frac{7}{4} = \pm \sqrt{\frac{169}{16}}] [x  \frac{7}{4} = \pm \frac{13}{4}]

Solve for (x): [x = \frac{7}{4} \pm \frac{13}{4}]

Simplify the solution: [x = \frac{7 \pm 13}{4}]
So, the solutions are: [x_1 = \frac{7 + 13}{4} = \frac{20}{4} = 5] [x_2 = \frac{7  13}{4} = \frac{6}{4} = \frac{3}{2}]
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a onesided 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 onesided 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 onesided 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 onesided 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.
 How is the graph of #y=1/3x^24# related to the graph of #f(x)=x^2#?
 What is the vertex of #y=x^2+12x+26#?
 A principal of $500 is invested in an account at 7% per year compounded annually. What is the total amount of money in the account after 5 years?
 How do you find the vertex of #f(x) = x^2 + 3#?
 How do you solve #3x^2+5x+2=0# by factoring?
 98% accuracy study help
 Covers math, physics, chemistry, biology, and more
 Stepbystep, indepth guides
 Readily available 24/7