How do you solve using the completing the square method #2x^2 - 7x - 15 = 0#?

Question

Kennedy Adams · Accepted Answer

#x=-3/2" or "x=5# #"to use the method of "color(blue)"completing the square"# #• " the coefficient of the "x^2" term must be 1"# #rArr2(x^2-7/2x-15/2)=0# #• " add/subtract "(1/2"coefficient of the x-term")^2" to"# #x^2-7/2x# #2(x^2+2(-7/4)xcolor(red)(+49/16)color(red)(-49/16)-15/2)=0# #rArr2(x-7/4)^2+2(-49/16-15/2)=0# #rArr2(x-7/4)^2-169/8=0# #rArr2(x-7/4)^2=169/8# #rArr(x-7/4)^2=169/16# #color(blue)"take the square root of both sides"# #rArrx-7/4=+-sqrt(169/16)larrcolor(blue)"note plus or minus"# #rArrx=7/4+-13/4# #rArrx=7/4-13/4=-3/2" or "x=7/4+13/4=5#

Julia Adams · Answer

undefined To solve the quadratic equation $2x^2 - 7x - 15 = 0$ using the completing the square method:

1. Move the constant term to the other side of the equation: $2x^2 - 7x = 15$.

2. Divide the coefficient of $x^2$ to make it 1: $x^2 - \frac{7}{2}x = \frac{15}{2}$.

3. To complete the square, take half of the coefficient of $x$ ($-\frac{7}{2}$) and square it: $(- \frac{7}{4})^2 = \frac{49}{16}$.

4. Add and subtract the value obtained in step 3 inside the parentheses: $x^2 - \frac{7}{2}x + \frac{49}{16} - \frac{49}{16} = \frac{15}{2}$.

5. Factor the perfect square trinomial: $(x - \frac{7}{4})^2 - \frac{49}{16} = \frac{15}{2}$.

6. Simplify the equation: $(x - \frac{7}{4})^2 = \frac{15}{2} + \frac{49}{16}$.

7. Find a common denominator and combine the fractions: $(x - \frac{7}{4})^2 = \frac{120}{16} + \frac{49}{16} = \frac{169}{16}$.

8. Take the square root of both sides: $x - \frac{7}{4} = \pm \sqrt{\frac{169}{16}}$.

9. Simplify the square root: $x - \frac{7}{4} = \pm \frac{13}{4}$.

10. Solve for $x$: 
    - For $x - \frac{7}{4} = \frac{13}{4}$, $x = \frac{7}{4} + \frac{13}{4} = 5$.
    - For $x - \frac{7}{4} = -\frac{13}{4}$, $x = \frac{7}{4} - \frac{13}{4} = -\frac{3}{2}$.

Therefore, the solutions to the equation $2x^2 - 7x - 15 = 0$ using the completing the square method are $x = 5$ and $x = -\frac{3}{2}$.