Solve by completing the square.
5x^2+10x-7=0
My answer: x= -1+ or - 2sqrt.of 10 all over 5
divide by 5
x^2 + 2x + ... = 7 + ...
x^2 + 2x + 1= 7 + 1
(x+1)^2 = 8
x+1 = ± √8 = ± 2√2
x = -1 ± 2√2
I do not get it. Here let me show my work.
5x^2+10x-7=0
5(x^2+2x+1)-7-1
5(x+1)^2-8=0
5(x+1)^2=8
(x+1)^2=+/- sqrt of 8/5
(x+1)=+/- 2sqrt of 2 over sqrt of 5
(x+1)=+/- 2sqrt of 10 over 5
x= -1+/- 2sqrt of 10 all over 5
WAIT NEVER MIND.....
MY FINAL ANSWER IS X=-1+/-2 SQRT OF 15 ALL OVER 5
But why is your answer x=-5+/- 2sqrt of 15 all over 5?
To solve the quadratic equation 5x^2 + 10x - 7 = 0 by completing the square, follow these steps:
1. Move the constant term to the other side of the equation:
5x^2 + 10x = 7
2. Divide the equation by the coefficient of x^2 to make the leading coefficient equal to 1:
x^2 + 2x = 7/5
3. Take half of the coefficient of x (which is 2) and square it:
(2/2)^2 = 1
4. Add the result from step 3 to both sides of the equation:
x^2 + 2x + 1 = 7/5 + 1
x^2 + 2x + 1 = (7+5)/5
x^2 + 2x + 1 = 12/5
5. Factor the left side of the equation:
(x + 1)(x + 1) = 12/5
6. Simplify the right side of the equation:
(x + 1)^2 = 12/5
7. Take the square root of both sides of the equation, considering the positive and negative square roots:
x + 1 = ±√(12/5)
8. Subtract 1 from both sides of the equation, considering both positive and negative roots:
x = -1 ± √(12/5)
Remember to simplify the square root if possible. In this case, you can simplify √(12/5) by simplifying the inside of the square root:
x = -1 ± √(12)/√(5)
x = -1 ± √(12)/√(5) * √(5)/√(5)
x = -1 ± √(60)/√(25)
x = -1 ± √(60)/5
Further simplification is possible by factorizing 60:
x = -1 ± √(4*15)/5
x = -1 ± 2√(15)/5
Hence, the solutions to the quadratic equation are:
x = (-1 + 2√(15))/5
x = (-1 - 2√(15))/5
forgot to divide the 7 by 5
x^2 + 2x + 1 = 7/5 + 1 = 12/5
(x+1)^2 = ±√12/√5 = ± √60/5 = ± 2√15/5
x = -1 ± 2√15/5
x = (-5 ± 2√15)/5