How to solve the square of 2x - 9x + 8 = 0

How to solve the square of 2x - 9x + 8 = 0

X square + 4x-3 = 0 find 2x cube + 9x square - 2x-8

2X cube + 9x square - 2x-8
=(2x^3+8x^2-6x)+(x^2+4x-3)-5
=2x(x^2+4x-3)+0-5
=0*2x-5
=-5

(x + 2) (x-3) - 2 (X-6) (x + 5) - 3 (xsquare - 5x + 17), x = 5.5

(x + 2) (x-3) - 2 (X-6) (x + 5) - 3 (xsquare - 5x + 17)
=(x²-x-6)-2(x²-x-30)-3(x²-5x+17)
=x²-x-6-2x²+2x+60-3x²+15x-51
=16x+3
=16*5.5+3
=91

If AB is opposite to each other and CD is reciprocal to each other, then the third power of (a + b) - the fourth power of 3 (CD) =?

A + B = 0, so the third power of (a + b) = 0, CD = 1, so the fourth power of 3 (CD) = 3, so 0-3 = - 3

A and B cars run from both places at the same time. Car a runs 45 kilometers per hour, and car B runs 42 kilometers per hour. The two cars meet 12 kilometers away from the destination. How many hours do the two cars meet after they leave at the same time

Midpoint: 12 kilometers
therefore
12×2÷（45-42）
=24÷3
=8 hours

It is known that the solution X of the equation MX + 2 = 2m-2x satisfies 2x-3 = 5, and the value of M is obtained

2x-3 = 5, x = 4
Substitute x = 4 into MX + 2 = 2m-2x, that is 4m + 2 = 2m-8, M = - 5

On the same road, a and B go in the same direction from a and B which are 5km apart. A's speed is 5km / h, B's speed is 3km, and a takes a dog,
When a overtakes B, the dog first overtakes B, then returns to meet a, then returns to overtake B, and then repeats until a overtakes B. the known speed is 15km / h. what is the total distance in this process?

When a overtakes B, the dog first overtakes B, then returns to meet a, and then returns to overtake B, and repeats in turn until a overtakes B. the known speed is 15km / h. what is the total distance in this process
5km/(5km/h-3km/h)=2.5h
15km/h*2h=30km
The total distance is 30km

When someone reads a decimal point, he loses the decimal point. As a result, he reads it as two hundred and forty-five. The original decimal part reads only one zero. What is the original number

200045
The original number is 200.045

The two trains of Party A and Party B depart from ab at the same time and meet for the first time at 75KM away from A. after meeting, the two trains continue to move forward and arrive at the destination
Return immediately, the second meeting is 45 kilometers away from B. how many kilometers is the distance between AB and B. please tell me the most detailed idea

Two cars a and B met twice and walked three whole journey
When car a and car B met once, they walked 75 kilometers, that is to say, the whole journey, car a walked 75 meters
Now it's three full walks. A total of 75 × 3 = 225 km
It's 45 kilometers longer than a whole journey
The whole journey is 225 - 45 = 180 km

In this paper, we prove the existence of the limit of a monotone bounded sequence by using the existence criterion of the limit of a monotone bounded sequence
The sequence is: √ 2, √ (2 + √ 2), √ (2 + √ (2 + √ 2))

A (n + 1) = √ (2 + an) mathematical induction assumes that increasing sequence is a (n + 1) ana1 = √ 2n = 2 A2 = √ (2 + √ 2) A2 > A1N = Ka (K + 1) > AKN = K + 1a (K + 2) = √ (2 + a (K + 1)) > A (K + 1) = √ (2 + AK), so it is increasing sequence a (n + 1) = √ (2 + an) > an2 + an > an & # 178; - 1