Let two of the equations 4x ^ 2-7x-3 = 0 be x1, x2 Solving the equation: x1-x2 It's the result of Veda's theorem,

Let two of the equations 4x ^ 2-7x-3 = 0 be x1, x2 Solving the equation: x1-x2 It's the result of Veda's theorem,

x1+x2=7/4
x1*x2=-3/4
(x1+x2)^2=x1^2+x2^2+2x1*x2=49/16
So X1 ^ 2 + x2 ^ 2 = 49 / 4-2x1 * x2 = 49 / 16 + 6 / 4 = 73 / 16
(x1-x2)^2=x1^2+x2^2-2x1*x2
=73/16+6/4=97/16
So x1-x2 = ± √ 97 / 4

Let the two roots of the square-7x-3 = 0 of the equation 4x be X1 and x2. If the roots of the equation are not understood, find (x1-3) (x2-3)

Using the relationship between root and coefficient: ax ^ 2 + BX + C = 0, two x1, X2, then X1 + x2 = - A / B, x1x2 = C / A

If X1 and X2 are the two roots of the equation 2x ^ 2-3x-1 = 0, then X1 ^ 2 + x2 ^ 2

X1, X2 are the two roots of the equation 2x ^ 2-3x-1 = 0,
Then X1 ^ 2 + x2 ^ 2
=(x1+x2)^2-2x1*x2
=(3/2)^2+1
=13/4

Let F 1 and F 2 be the two focal points of the hyperbola, and | F 1, F 2 | = 18. The same branch of the hyperbola crossed by a straight line passing through F 1 lies at two points m and n
If | Mn | = 10 and the perimeter of triangle mf2n is 48, then the standard equation of hyperbola satisfying the condition is

C = 9
|MF1|+|NF1|=|MN|=10
C△MF2N=|MF2| + |NF2| + |MF1| + |NF1|=48
∴|MF2|+|NF2|=48-10=38
According to the first definition of hyperbola, hyperbola is the locus of a point whose absolute value of the distance difference between two fixed points on the plane is a fixed value
|MF2|-|MF1|=2a…… ①
|NF2|-|NF1|=2a…… ②
① + 2
( |MF2|+|NF2| )－( |MF1|+|NF1| ) =4a
That is: 38 - 10 = 4A
a=7
a²=49
b²=c²-a²=81-49=32
The hyperbolic equation is: X & # 178 / 49 - Y & # 178 / 32 = 1 or Y & # 178 / 49 - X & # 178 / 32 = 1

X2 / a2-y2 / B2 = 1, the focus F1, F2, P is the point on the right branch, the triangle pf1f2 inscribed circle center I, tangent to X axis and point a, BF2 vertical, PI to B, e is the eccentricity
Finding the relationship between OA and ob (equality)

Circle I and Pf1, PF2, F1F2 are tangent to D, e, apf1-pf2 = 2apd = PE, F1D = F1A, F2E = f2af1a-f2a = f1d-f2e = pf1-pf2 = 2af1a-f2a = 2af1a = f1o + OA, F2F = of 2-oa of 1 = of 22 OA = 2A OA = a triangle, pf1f2 inscribed center I, Pb is the bisector of angle f1pf2, let the extension of F1B intersect the extension of PF2

F1 and F2 are the left and right focal points of the hyperbola x ^ 2 / 4-y ^ 2 / 5 = 1, point P is the point on the right branch of the hyperbola, I is the inner part of △ pf1f2, and PI intersects X axis at point Q,
If F1q = PF2, then what is the ratio of I to PQ

a=2、c=3
PF1：F1Q=PF2：F2Q
Let F1q = F2P = x, then:
(2a－x)：x=x：(2c－x)
x²－(c－a)x－2ac=0
x²－x－12=0
x=4
Namely: Q (1,0)
F1Q=4、F2Q=2、PF1=2a＋x=8、PF2=x=4
PI：IQ=PF1：F1Q=8：4=2：1
The result is: λ = 2

Let p be a point on the hyperbola x ^ 2-y ^ 2 = 1, and F1 and F2 be the two focuses of the hyperbola. If | Pf1 |: | PF2 | = 3:2, then the area of triangle pf1f2 is (...)
Let p be a point on the hyperbola x ^ 2-y ^ 2 = 1, F1 and F2 are the two focuses of the hyperbola. If | Pf1 |: | PF2 | = 3:2, then the area of the triangle pf1f2 is () solution. Thank you!

Pf1-pf2 = 2A = 1 | Pf1 |: | PF2 | = 3:2 these two conditions can calculate the value of Pf1 and PF2
2C = 2 root sign 2, then cosine theorem solves the value of cos ∠ f1pf2
The area of triangle is equal to 1 / 2pf1pf2sin ∠ f1pf2
Do it yourself. It's easy

It is known that the center of the hyperbola is at the origin, the left and right focal points F1 and F2 are on the x-axis, a (0, √ 2) is the center of the circle, and 1 is the radius, which is tangent to the asymptote of the hyperbola
Point F2 and point a are symmetric with respect to the line y = X
(1) Solving hyperbolic equation
(2) If P is a moving point on the hyperbola, PQ bisects ∠ f1pf2, and passes through F1 as the vertical line of PQ, and the vertical foot is Q, the trajectory equation of Q point is obtained

(1) Let the hyperbolic standard equation be X & # 178 / A & # 178; - Y & # 178 / B & # 178; = 1 ∵ a (0, √ 2), F2 be symmetric with respect to the straight line y = x ∵ F2 (2,0), C = 2, and one of the asymptotes is y = B / ax, that is, BX ay = 0 ∵ a (0, √ 2) is the center of the circle and 1 is the radius, which is tangent to the asymptote of hyperbola ∵ | - a × | / √ (a

If the equation | X-1 | = KX + 2 of X has two different real roots, then the value range of K is the same______ ．

Draw the image of the function y = KX + 2, y = | X-1 | from the image, we can see that only when - 1 < K < 1, the image of the function y = KX + 2, y = | X-1 | has two intersections, that is, the equation KX + 2 = | X-1 | has two real roots. Therefore, the value range of real number k is - 1 < K < 1. So the answer is: - 1 < K < 1

If the equation lx-8l = a of X has two unequal real roots on (6,9), then the value range of real number a-------

From lx-8l = a, we can get (X-8) &# 178; = A & # 178;
That is, X & # 178; - 16x + 64 = A & # 178;
That is X & # 178; - 16x + 64-A & # 178; = 0
So the opening of the parabola X & # 178; - 16x + 64-A & # 178; is upward, and the equation of its axis of symmetry is - A / 2B = 8
Let the equation lx-8l = a have two unequal real roots on (6,9),
Then we need 6 & # 178; - 16x6 + 64-A & # 178; > O and 9 & # 178; - 16x9 + 64-A & # 178; > 0
Simultaneous solution to the problem - 1