I have an algebra problem that I don't get.?

Question

The teacher said to Square then solve, and I have done some work on both othe these problems and she won't tell me where I am making my mistake. Can anyone help me?
Im supposed to square both sides then solve for x.

10. x^2 - 3x - 15 = -5

10. (my attempt)x^2 –3x = 10

(x+9/4)^2 = 49/4

x= 5/4
x= -23/4

11. x^2 - 17x = -10x - 6
11. (my attempt)x^2 –7x +12.25 = 6.25

: (x-49/4)^2 = 25/4
x= 59/4
x= 39/4

I know how to factor it out guys, but my teacher said " Solve the equation by completing the square" That is where I am confused.

Answer 1

As your teacher has asked you to do it by square-root method,let us do these by the same process
10.X^2-3x-15= -5
=>x^2-3x=10
=>x^2-3x+(3/2)^2=10+9/4
=>(x-3/2)^2=49/4
=>x-3/2=+-7/2
=>x=(3/2+7/2) or(3/2-7/2)
Hence x=5 or -2
11. x^2-17x= -10x-6
=>x^2-7x= -6
=>x^2-2.x.7/2 +(7/2)^2=-6+49/4
=>(x-7/2)^2=25/4=(5/2)^2
=>x-7/2=+-5/2
=> x=(7/2+5/2) or (7/2-5/2)
Hence x=6 or 1
However the equations may be solved more quickly by middle-term method.

Answer 2

x² –3x = 10 (continuing from your 1st step)
x² –3x+(3/2)²-(3/2)² =10
x² - 3x +(3/2)² - 9/4 = 10
Refering to your 3rd step, you can't just take the square root of the eqn (x+9/4)^2 = 49/4 to get your answer. This is only applicable if you have say x²= 49 (only a x² term on 1 side and a number on the other side), otherwise you have to make use of the zero product rule.

[x-(3/2)]²- 9/4 -10 = 0
[x-(3/2)]²- 49/4 = 0
[x-(3/2)]²- (7/2)² = 0
[(x-3/2) - (7/2)] [(x-3/2) + (7/2)] =0
(making use of the formula a²-b² = (a+b)(a-b) )

[(x-3/2) - (7/2)] = 0 OR [(x-3/2) - (7/2)] =0
Hence,
x=5 or x=-2

The zero product rule means that you have to make 1 side of the eqn equals to 0. Then factorise the other side of the eqn. This is because there are many numbers that when multiplied can give you an answer (say in this case 49/4), but when the answer is zero, we can know for sure that either multiple can be zero. (eg. [(x-3/2) - (7/2)] = 0 OR [(x-3/2) - (7/2)] =0 as shown above.)

Hope this helps and I didn't end up confusing you instead.

Answer 3

Completing the square

x² - 3x - 15 = - 5

x² - 3x - 15 + 15 = - 5 + 15

x² - 3x = 10

x² - 3x +_____= 10 +______

x² - 3x + 9/4 = 10 + 9/4

(x - 3/2)(x - 3/2) = 40/4 + 9/4

(x - 3/2)² = 40/4 + 9/4

(x - 3/2)² = 49/4

(√x - 3/2)² = ± √49 / √4

x - 3/2 = ± 7/2

x - 3/2 + 3/2 = 3/2 ± 7/2

x = 3/2 ± 7/2

- - - -

Solving for +

x = 3/2 + 7/2

x = 10/2

x = 5

- - - - -

Solving for -

x = 3/2 - 7/2

x = - 4/2

x = - 2

- - - - - - - - s-

Answer 4

Your equation is 10x^2 - 3x -10 = 0
Take that 10 out and you have,
x^2 - (3/10)x -1 = 0
Now you see the x coefficient 3/10? divide it by 2 and get 3/20, now consider the square
S = [x - (3/20)]^2
expanding you have S = x^2 - (3/10)x + 9/400,
basically the part we need x^2 - (3/10)x is equal to [x - (3/20)]^2 - 9/400
so substitute and get,
[x - (3/20)]^2 - 9/400 -1 = 0
[x - (3/20)]^2 = 409 / 400
Now take the square root of both sides,
x - (3/20) =+ V409 / 20 or -V409 / 20
x = 3/20 + V409 / 20 OR 3/20 - V409 / 20

two solutions, real and distinct!

Answer 5

Ok this might be a little hard to understand on hear, but try hard cause i know my stuff

X^2 - 3x - 15 = -5
X^2 - 3x - 10 = 0

Find 2 numbers that go in to A (number in front of X) multiplied by C (number w/out a variable) (so in this case it would be 1 x -10) the 2 numbers must add up to equal the center value (-3)

The 2 numbers are -5 and 2, 2x -5 = -10, 2+ -5 = -3

this brings you to: (X^2 + 2x) (-5x - 10)

Then you factor out the -5 and the X and you get

(X-5)(X+2)=0
set both = 0
X-5 = 0
X+2 = 0

Answers are:
X= 5
X= -2

Answer 6

You're doing fine, except that when you complete the square, you should have (in 10):
(x + 3/2)^2 = 49/4

See why?
If you expand out your version, you would have
(x + 9/4)^2 =
x^2 + 2 * 9/4 * x + (9/4)^2.

In 11, you likewise want
(x - 7/2)^2 = 25/4

Aside from that, you've done fine.

Answer 7

10. x^2 - 3x - 15 = -5

x^2 - 3x = 10

x(x - 3) = 10

x - 3 = 10 / x

"x = 5" because 5 - 3 = 10 / 5 => 2 = 2

11. x^2 - 17x = -10x - 6

x^2 - 7x = -6

x(x - 7) = -6

x - 7 = -6 / x

"x = 1" because 1 - 7 = -6 / 1 => -6 = -6

Answer 8

x^2 - 3x - 10 = 0

(x-5) (x+2) = 0

x=5, x=-2



11. x^2 - 7x +6 = 0

(x-6) (x-1) = 0

x=6, x=1



I used factoring. this may not help but i noticed that the way you use COMPLETING THE SQUARE is incorrect.

Answer 9

you can use quadratic equation instead of solving ax^2+bx+c=0

the formula will be X= (-b+-sqrt(b^2-4ac))/2a

there will be two answers

10. X^2-3X-10=0

X= (-(-3)+-sqrt((-3)^2-4(1)(-10))/2(1)

if positive

X= (-(-3)+sqrt((-3)^2-4(1)(-10))/2(1)

X=5

if negative

X= (-(-3)-sqrt((-3)^2-4(1)(-10))/2(1)

X= - 2

11. X^2-7X+6=0

X= (-(-7)+-sqrt((-7)^2-4(1)(6))/2(1)

if positive

X= (-(-7)+sqrt((-7)^2-4(1)(6))/2(1)

X=6

if negative

X= (-(-7)-sqrt((-7)^2-4(1)(6))/2(1)

X=1

Answer 10

hi buddy this an ordinary question in accordance to linear equations you should use transposition approach in this technique at the same time as a time period is transfered from one area to different its signal turns into opposite answer: x+4-7x=22 =4-7x=22-x (the following x comes from the left area its signal transformations) =4=22-x+7x(the following 7x comes from the left area its signal transformations) =4-22=6x(similar) =-18=6x x=-18/6 =-3 Bye Bye!!!!!!!!!!!!!!!!!!!!!!