solving quadratic equations help?

Question

1. x^2-8x+16=0
2. 3u^2=-4u+15
3. 2x^2=200
4. 4(t+6)^2=160


please and thank u

Answer 1

1. x^2-8x+16
We are looking for 2 number whose sum is -8 and whose product is 16.

We can factor this into (x-4)(x-4)=16
Solving for x, we get x=4.

2. 3u^2=-4u+15
3u^2+4u-15 a b c d
Using similar concepts, (3u-5)(u+3)
We want the first number in each bracket (a,c) to equal 3 (3x1) and the product of the sums of the middle numbers and outer numbers to equal 4 (-5+9) and lastly the product of the last numbers in the brackets (b,d) to equal -15 (-5x3)

3. 2x^2=200
x^2=100
x= +/- 10

4. 4(t+6)^2=160
(t+6)^2=40
we can square root each side
t+6={square root}40 this can be written as square root of (4x10) and 4 goes out of the square root to become 2x(square root of 10)
t=2 x (square root of 10) - 6
t is roughly 0.324

Answer 2

(x-4)^2 = 0 x=4
3u^2 + 4u - 15 = 0
(3u-5)(u+3)=0 u=5/3 or -3
x^2 = 100 x= +/- 10
(t+6)^2 = 40 t= -6 +/- 2sqrt(10)

Answer 3

First arrange the equations to the form ax² + bx + c = 0

Q4. 4(t+6)² = 160
(t²+12t+36) = 40
t² + 12t - 4 = 0

x = [-(12) ± √(12² - 4(1)(-4))]/2(1)
= [-12 ± √(160)]/2
= -6 ± 2√(10)
x₊ = -6 + 6.325 = 0.325
x₋ = -6 - 6.325 = -12.325