## anonymous 5 years ago solve by completing the square.

1. anonymous

x$x ^{2}+2x=24$

2. anonymous

move all one side you get$x^{2} +2x -24 =0$

3. anonymous

now you factour(x+6)(x-4)

4. anonymous

x=-6 x=4

5. anonymous

$2x ^{2}-4x-3=0$

6. anonymous

this is quadratic equation form: [-b±√(b^2-4(a)(c))]/(2(a)) a=2 b=-4 c=-3 [-(-4)±√((-4)^2-(4*2*(-3))]/(2*2)) [(4)±√(16+24)]/(4) [(4)±2√(10)]/(4) x =1/2[±√(10)]

7. anonymous

The above method is not "completing the square"

8. anonymous

annette the second you want do complete square?

9. anonymous

Even in first one, it is splitting the middle term method and not completing the square

10. anonymous

Thank you Harkirat for remind me let do again , can you do second one it late I have to go 2/2=1 , 1^2 =1 (x+1)^2=24+1 (x+1)^2=25 x +1 = √(25) x = -1± 5 x=-6 or x= 4

11. anonymous

For the second we proceed as follows: $2x ^{2}-4x-3=0$

12. anonymous

First we take -3 to RHS as 3$2x ^{2}-4x=3$

13. anonymous

Next we divide both sides by 2 (to get rid of the 2 in front of x^2) So we get $x ^{2}-2x=3/2$

14. anonymous

Now to complete the square we add 1^2 to both sides $x ^{2}-2x+1^{2}=3/2 + 1$

15. anonymous

Now left side is a complete square as follows $(x-1)^{2}= 5/2$ This gives $x-1 = \pm \sqrt{5/2}$ Now taking -1 to RHS we get the final answer $x = 1 \pm \sqrt{5/2}$

16. anonymous

So the two roots/zeros of the given quadratic equation are $1+\sqrt{5/2}$ and $1-\sqrt{5/2}$