If the unary quadratic equation ax is about x? +bx+c=0, and there are two equal ones, then △=b? -4ac=0
Because the equation (b-c)x? The two roots of +(c-a)x+(a-b)=0 are equal.
So (c-a)? -4(b-c)(a-b)=0
Answer? -2ac+c？ -4(ab-b？ -ac+bc)=0
Answer? -2ac+c？ -4ab+4b？ +4ac-4bc=0
Answer? +4b？ +c？ +2ac-4ab-4bc=0
The following are the efforts to decompose the factors.
(a？ -4ab+4b？ )+(2ac-4bc)+c？ =0
(a-2b)？ +2×c×(a-2b)+c？ =0
(a-2b+c)？ =0
That is, a-2b+c=0 and a+c=2b.
2) If Equation 2x? -(k+ 1)x+k+3=0, and the difference between them is 1, so the value of k is _ _ _ _.
Vieta's Theorem: On the Quadratic Equation ax of X? +bx+c of two x 1 equals 0, and x2 satisfies.
x 1+x2=-b/a，x 1？ x2=c/a
Solution: Let 2x? -(k+ 1)x+k+3=0 two x 1, x2.
Then x 1+x2=(k+ 1)/2, x 1? x2=(k+3)/2
So (x 1-x2)?
=x 1？ -2x 1？ x2+x2？
=(x 1？ +2x 1？ x2+x2？ )-4 x 1？ x2
=(x 1+x2)？ -4x 1x2
=(k+ 1)？ /4- 2(k+3)
Because equation 2x? -(k+ 1)x+k+3=0, and the difference between them is 1.
So (x 1-x2)? = 1
So (k+ 1)? /4- 2(k+3)= 1
(k+ 1)？ -8(k+3)-4=0
k？ +2k+ 1-8k-24-4=0
k？ -6k-27=0
(k-9)(k+3)=0
k 1=9，k2=-3
So the value of k is 9 or -3.
3) Let x 1 and x2 be the equation x? +px+q=0, and the two real roots of x 1+ 1 and x2+ 1 are equations about x? +qx+two real roots of p = 0, then p = _____ _ _, and q = _ _ _ _.
Solution: x 1, x2 is the equation x? Two real roots of +px+q=0
From Vieta theorem, x 1+x2 =-p...①, x 1? x2=q……②
Is X 1+ 1, x2+ 1 an equation about x? Two real roots of +qx+p=0
From Vieta theorem, x 1+ 1+x2+1=-q ... ③, (x1+1)? (x2+ 1)=p……④
① Substitute ③ to get -p+2 =-q. Q-p =-2...⑤.
Substituting ④, x1x 2+x1+x2+1= p, we get q-p+ 1=p, that is, Q-2p =- 1...⑥.
Simultaneous ⑤, ⑤, p=- 1, q=-3.
4) The equation about x is known. x？ -(k+ 1)x+ 1/4k？ The two roots of+1=0 are the lengths of two sides of a rectangle. When the diagonal length of the rectangle is root number 5, find the value of k.
Solution: let x be the equation x? -(k+ 1)x+ 1/4k？ +1=0 Two x 1, x2
According to Vieta theorem, x 1+x2=k+ 1, x 1? x2= 1/4k？ + 1
So x 1? +x2？
=(x 1+2x 1？ x2+x2？ )-2x 1？ x2
=(x 1+x2)？ - 2x 1？ x2
=(k+ 1)？ -? k？ -2
Is it because of x 1 again? +x2？ =(√5)? =5
So (k+ 1)? -? k？ -2=5
k？ +2k-6=0
k？ +4k- 12=0
(k+6)(k-2)=0
k 1=-6，k2=2
When k=-6, the original equation becomes x? -5x+ 10=0, △ = 25-40 =- 15 < 0, no real root (omitted).
When k=2, the original equation becomes x? -3x+2=0，x 1= 1，x2=2。
So the value of k is 2.
I hope it helps you.