A necessary and sufficient condition for the equation AX2 + BX + C = 0 of X to have a root of - 1 is__________ . Better have analysis! ***********It's "one root is - 1", not "only one root is - 1"************

A necessary and sufficient condition for the equation AX2 + BX + C = 0 of X to have a root of - 1 is__________ . Better have analysis! ***********It's "one root is - 1", not "only one root is - 1"************

a-b+c=0
X = - 1, A-B + C = 0
There is a root, so B ^ 2 - 4ac ≥ 0
(a+c)^2 - 4ac ≥0
(a-c)^2 ≥0
This formula always holds
So the condition is A-B + C = 0

It is proved that P: x = 1 is a root of the equation AX2 + BX + C = 0 and is a necessary and sufficient condition for Q: a + B + C = 0

If x = 1, then
ax2+bx+c=0
a+b+c=0
So sufficient conditions
a+b+c=0
ax²+bx+c=0
Subtraction
a(x²-1)+b(x-1)=0
(x-1)(ax-1+b)=0
The root is x = 1
X = (1-B) / a (when a ≠ 0)
So it's necessary
necessary and sufficient condition

Let two of the quadratic equations ax ^ 2 + BX + C = 0 be x1, x2. Let S1 = X1 + 2010x2, S2 = X1 + 2010x2 ^ 2 ,
Sn = X1 ^ n + 2010x2 ^ n, then as2010 + bs2009 + cs2008 =?

First of all, there should be errors in the conditions you give. It should be S1 = X1 + 2010x2, S2 = X1 ^ 2 + 2010x2 ^ 2 Sn = X1 ^ n + 2010x2 ^ n two quadratic equations ax ^ 2 + BX + C = 0 are x1, X2 then a X1 ^ 2 + B X1 + C = 0 and a x2 ^ 2 + B x2 + C = 0as2010 = a X1 ^ 2010 + 2010a x2 ^ 2010bs2009 = B X1 ^ 2

If the solution set of inequality ax2-bx + C > 0 is {x | - 2

It is easy to know from the proposition that ax & sup2; - BX + C = a (x + 2) (x-3) = ax & sup2; - ax-6a  B = a, C = - 6A. A ＜ 0. The inequality CX & sup2; + BX + a ＞ 0 is: - 6AX & sup2; + ax + a ＞ 0. = = = = > 6x & sup2; - X-1 ＞ 0. = = = = > [x - (1 / 2)] [x + (1 / 3)] > 0. = = = > x ＜ - 1 / 3, or X ＞ 1 / 2. The solution set is (-, - 1 /

Solve the inequality ax2-bx + C < 0, the solution set is {x | x < α or X is greater than β} α < β < 0. Find the solution set of inequality CX2 + BX + a
Is it based on the Veda theorem, but the answer is a strange fraction with sign

Because the inequality ax & # 178; - BX + C0
Thus, the solution of inequality CX & # 178; + BX + a > 0 is - 1 / β

In △ ABC, if Sina: SINB: sinc = 6:5:4, then cosa=______ ．

In △ ABC, we know that sina: SINB: sinc = 6:5:4, let the three sides be 6K, 5K and 4K respectively. From the cosine theorem, we can get cosa = B2 + C2 − a22bc = 25 + 16 − 362 × 5 × 4 = 18, so the answer is: 18

1. In △ ABC, cosa = 3 / 5, SINB = 5 / 13, find the value of sinc

The sine of internal angle of triangle is greater than 0
sin²A+cos²A=1
So Sina = 4 / 5
Similarly, SINB = 12 / 13
sinC=sin[180-(A+B)]
=sin(A+B)
=sinAcosB+cosAsinB
=56/65

Triangle ABC, tanb = 1, Tanc = 2, B = 10, a =?
Are you sure?

For any non right triangle, there is always Tana + tanb + Tanc = tanatanbtanc1 + 2 + Tana = 1 * 2 * tanatana = 3sina / cosa = 3. Using sin ^ 2 (a) + cos ^ 2 (b) = 1, we can get Sina = √ (9 / 10) TGB = 1, SINB = √ (1 / 2). Using sine theorem, we can get a = b * Sina / SINB, a = 6 √ 5

In △ ABC, if cosa = 45 and Tan (a-b) = - 12, then the value of Tanc is___ ．

In △ ABC, it is known that cosa = 45, ｜ Sina = 35, Tana = 34. ∵ Tan (a-b) = - 12 = tana-tanb1 + tanatanb = 34-tanb1 + 34tanb, tanb = 2. Then Tanc = Tan (π - a-b) = - Tan (a + b) = Tana + tanbtantanb-1 = 34 + 234 × 2-1 = 112, so the answer is 112

In the triangle ABC, cosa = 4 / 5, tanb = 2, find (1) Tan (a + b); (2) tan2c

tanA=3/4tan(A+B)=(tanA+tanB)/(1-tanAtanB)=（3/4+2）/（1-3/4*2）=-11/2tanC=tan(π-A-B)=-tan（A+B）=11/2tan(2C)=2tanC/(1-tanCtanC)=(2*11/2)/(1-11/2*11/2)=-44/117