Given the equation x2 + 3x + k = 0 (1) if the difference between the two equations is 5, find the value of K; (2) if one equation is twice the other, find the product of the two equations

Given the equation x2 + 3x + k = 0 (1) if the difference between the two equations is 5, find the value of K; (2) if one equation is twice the other, find the product of the two equations

When b2-4ac = 9-4k ≥ 0, i.e. K ≤ 94, the equation has a solution. Let x2 + 3x + k = 0 be x1, X2, then there are X1 + x2 = - 3, x1x2 = k, (1) ∵ x1-x2 = 5, (x1-x2) 2 = (x1 + x2) 2-4x1x2 = 9-4k = 25, then the solution is k = - 4; (2) let X1 = 2x2, then x1x2 = 2x22 = k, 3x2 = - 3, then the solution is X1 = - 2, X2 = - 1, then x1x2 = 2

Given the equation x2 + 3x + k = 0 (1) if the difference between the two equations is 5, find the value of K; (2) if one equation is twice the other, find the product of the two equations

When b2-4ac = 9-4k ≥ 0, i.e. K ≤ 94, the equation has a solution. Let x2 + 3x + k = 0 be x1, X2, then there are X1 + x2 = - 3, x1x2 = k, (1) ∵ x1-x2 = 5, (x1-x2) 2 = (x1 + x2) 2-4x1x2 = 9-4k = 25, the solution is k = - 4; (2) let X1 = 2x2, the solution is x1x2 = 2x22 = k, 3x2 = - 3

There are several real roots of the equation x * 3-3x * 2 + 3x-9 = 0 of X

x*3-3x*2+3x-9=0
x²(x-3)+3(x-3)=0
(x²+3)(x-3)=0
So x-3 = 0
x=3
So there is one real root

Fifth grade volume I oral arithmetic questions and answers

One 12 × 3 = 15 × 7 = 10 × 8 △ 16=
Two 56 + 47 = 37 × 3 = 649-112=
Three 75 △ 5 = 17 × 15 = (159 + 97) △ 16=
Four 6 + 7 × 8 = 43 × 9 = (5 △ 7 + 1) × 7=
V 1 △ 2 × 4 = 195 + 624 = 23 × 50=
Six (48 + 8) △ 7 = 12.5 × 8 = 33 △ 12=
Seven 1000-257 = 1001 △ 11 = 34 × 56=
Eight [48 △ (7-4) + 4] × 5 = 47 △ 100 = 347-153 + 47=
IX 512 △ 16 = 21 × 21 = 18 △ 243 △ 27=
X 1 + 2 + 3 + +10= 3×4×5= 73×137=
answer:
（1） 36, 105, 5 (2) 103111537
（3） 15, 255, 16 (IV) 62, 387, 12
（5） 28191150 (VI) 8100,2.75
（7） 743,911904 (8) 100,0.47241
（9） 32, 441, 2 (x) 55, 6010001

In the system of equations x = 2y-t 2x + y = T-3, given Y > 9, find the range of X

The elimination of T leads to y = 3x + 3 > 9
x+1＞3
x＞2

Use 0, 1, 2, 3, 5 to form the multiplication formula of three digits multiplied by two digits. How many can you write? Can you write the formula with the largest product?

102X35= 102X53= 103X25= 103x52= 105x23= 105x32=
201x35= 201x53= 203x15= 203x51= 205x13= 205x31=
.
There are 24 formulas for 0 in the tens of three digits and 24 formulas for 0 in the ones of three digits
There are 24 formulas for 0 in the single digit of two digits, and there are 72 formulas in total
If you want to maximize the product of two factors, you should minimize the difference between the two factors, so choose the product of the smallest three digit 102 and the largest two digit 53, that is, 102x53 = 5406

It is known that the asymptote equation of the hyperbola with the center at the origin is y = - 32x, and the focal length is 213

∵ the asymptote equation of hyperbola is y = - 32x. From the meaning of the problem, we can assume that the hyperbolic equation is x24-y29 = λ (λ≠ 0). When λ & gt; 0, x24 λ - Y29 λ = 1, the focus is on the X axis, ∵ 4 λ + 9 λ = 13, ∵ λ = 1, ∵ hyperbolic equation is x24-y29 = 1. When λ & lt; 0, the equation is y2-9 λ - x2-4 λ = 1, ∵ 4 λ - 9 λ = 13, ∵ λ = - 1 ∵ equation is y29-x24 = 1. In conclusion, the hyperbolic equation is x24-y29 = 1 or y29-x24 = 1

1+2+3-4+9+11-13+99-56=?

52

How can a + B + C, a + B-C, A-B + C judge ＞ 0 or ＜ 0 in quadratic function?
What does this have to do with the value of X? Easy to understand in detail!!!

If a quadratic function is f (x) = ax & sup2; + BX + C, then:
a+b+c=f(1)
a+b-c=f(1)-2f(0)
a-b+c=f(-1)

Use seven fours to make a hundred. Only add and subtract, not multiply and divide?

44+44+4+4+4=100