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Anonymous users2024-02-111. Solution: From the meaning of the question: two equal real roots, that is, =0=k -4 1 1=0 k =4 k= 2Answer: When k= 2, there are two equal real roots.
1) Solution: Two unequal real numbers with the root 0
2k）²-4×1×（k-2）²＞0
4k²-4k²+16k-16＞0
16k-16＞0
k 1 answer: When k 1, there are two unequal real roots.
2) Solution: Two equal real roots =0
2k）²-4×1×（k-2）²=0
4k²-4k²+16k-16=0
16k-16=0
k 1 answer: When k 1, there are two unequal real roots.
3) Solution: 0
2k）²-4×1×（k-2）²＜0
4k²-4k²+16k-16＜0
16k-16＜0
k 1 When k 1, there are two unequal real roots.
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Anonymous users2024-02-101。(-k+1)‘2-4k=0 k'2-6k+1=0 k=3+(-2*root number 2.)
2-4*(k-2)'2=4*2*(2k-2)=16(k-1)1),k-1>0 k>1
2),k-1=0 k=1
3), k-1<0 k<1 (there should be no real root, because it can be virtual).
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Anonymous users2024-02-09Simplify: x - (k-1) x + k = 0, (k-1) - 4k = 0, k = 3 plus or minus 2 root number 2
Mostly Delta.
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Anonymous users2024-02-08The answer to the first question is: 3 plus 2 and the sum of the root number 2 (I don't know how to play the root number).
The second question is:
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Anonymous users2024-02-07The formula for δ is: δ=b -4ac.
The discriminant expression of a quadratic equation is usually represented by the Greek letter δ (pronounced "Deta").
There are three cases of the roots of a quadratic equation ax +bx+c=0(a≠0): there are two equal real roots, two unequal real roots, and no real roots. Because there is a special relationship between the roots and the coefficients of a quadratic equation, we don't need to solve the equation to make a judgment about the root.
The general form of a quadratic equation is ax +bx+c=0, then δ=b -4ac.
If δ 0, then the unary quadratic equation has two unequal real roots;
If δ=0, then this unary quadratic equation has two equal real roots;
If δ 0, then this unary quadratic equation has no real roots.
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Anonymous users2024-02-06If the unary quadratic equation ax +bx+c=0(a≠0), the discriminant is: b -4ac
1. When >0, the equation has two unequal real roots;
2. When =0, the equation has two equal real roots;
3. When <0, the equation has no real roots, but has 2 conjugate multiple roots. Pei Nian Gai.
Real numbers include positive, negative, and 0. Positive numbers include: positive integers and positive fractions; Negative matching between:
Negative integers and negative fractions. Real numbers also include rational and irrational numbers; Rational numbers include: integers and fractions.
Integers include: positive integers, 0, negative integers. Scores include:
Positive scores, negative scores;
The second classification method of fractions: including finite decimals, infinite cyclic decimals; Irrational numbers include: positive irrational numbers, negative irrational numbers. Infinite non-cyclic decimal numbers are called irrational numbers, and they are expressed as 2 and 3.
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Anonymous users2024-02-05Hello! -b+√(b^2-4ac)]/2a
b-√(b^2-4ac)]/2a
Binary equation: ax 2 + bx + c = 0
A is not equal to 0).
The formula for finding the root is: x1=[-b+under the root number(b 2-4ac)] 2abx2=[-b-under the root number(b 2-4ac)] 2ab only represents a personal opinion, don't spray if you don't like it, thank you.
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Anonymous users2024-02-04The discriminant formula of the unary quadratic equation ax +bx+c=0 =b -4ac is derived from the root finding formula of the equation, because ax +bx+c=0===>a(x+b 2a) -b 4a+c=0===>x=[-b (b -4ac)] 2a
From the root finding formula, it can be seen that the result of b -4ac determines whether the equation has a real root, or what kind of real root it has, so it is called b -4ac as the discriminant formula and symbol of a quadratic equation
1) When =0, the equation has one real root (or two equal real roots) (2) When <0, the equation has no solution.
3) When >0, the equation has two unequal real roots, and Vedic theorem is derived from the root finding formula and discriminant formula.
Suppose a quadratic equation has two real roots x1 and x2, then the relation between these two real roots is:
x1+x2=[-b+ 2a+[-b- 2a=-b ax1x2=[-b+ 2a [-b- 2a=c aOf course, the first condition for the above conditions to be true (including discriminants) is a≠0
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Anonymous users2024-02-03A perfectly squared formula is formed, and finally a square is equal to a formula, which must be greater than zero, and the discriminant formula is obtained by deformation.
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Anonymous users2024-02-02f(x)=ax 2+bx+c, if there is a root x1, x2, then x1+x2=-b a, x1 *x2=c a
If f(x)=0 exists, then there must be f'(x1)*f'(x2) 0, i.e., (2ax1+b)(2ax2+b) 0, i.e., 4a 2*x1*x2-2ab(x1+x2)+b 2 0, i.e., 4a 2*(c a)-2ab(-b a)+b 2 0, which is simplified to obtain b 2 4ac
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Anonymous users2024-02-01b -4ac>0, the equation has two unequal real roots.
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Anonymous users2024-01-31Hello! The discriminant expression of a quadratic equation is usually represented by the Greek letter δ (pronounced "Deta").
The general form of a quadratic equation is ax +bx+c=0, then δ=b -4ac.
If δ 0, then the unary quadratic equation has two unequal real roots;
If δ=0, then this unary quadratic equation has two equal real roots;
If δ 0, then this unary quadratic equation has no real roots.
Note: Since the formula for finding the root of a quadratic equation is.
x1,2=(-b root number b -4ac) 2a so when b -4ac 0, then this unary quadratic equation has two unequal real roots;
When b -4ac=0, then this unary quadratic equation has two equal real roots;
When b -4ac 0, then this unary quadratic equation has no real roots.
Hope it helps!
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Anonymous users2024-01-30The general form of a quadratic equation is the square of ax + bx + c 0
The discriminant formula of the root is b-4ac
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Anonymous users2024-01-29x 2-2bx-(a-2b 2)=0 has a real solution, then =(-2b) 2+4(a-2b 2) 0-4b 2+4a 0
a≥b^22a^2-ab^2-5a+b^2+4=0
2a^2+b^2(1-a)-5a+4=0
Because a 1, 1-a 0
Because B2A
So b 2(1-a) a(1-a).
2a^2+a(1-a)-5a+4≤0
a^2-4a+4≤0
a-2)^2≤0
So a-2=0
a=2 is substituted into 2a2-ab 2-5a+b 2+4=0,8-2b 2-10+b 2+4=0
b^2+2=0
b^2=2a^2=2^2=4
a^2+b^2=6
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Anonymous users2024-01-28OK. The formula method solves all one-dimensional quadratic equations.
First of all, we must determine how many roots a quadratic equation has by the discriminant expression of the root of δ=b 2-4acWhen δ=b 2-4ac<0 x no real root (juniary) 2When δ=b 2-4ac=0, x has two identical real roots, i.e., x1=x23
When δ=b 2-4ac>0, x has two different real roots, and when the judgment is completed, if the equation has a root, the root of the equation can be found according to the formula: x= 2a.
Guo Dunyun: The original system of equations is.
y =x +1 (1)36+x =r (2)(r y) 2=5 y (3) is obtained by (1), x =y 1 (4), substituting (2) obtains, y +35=r (5), substituting (3) obtains, y y+35) 2=5 y, y(y y+35)=10,y3 >>>More
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Solution: In the form of an +bn + c, it can be matched into a(n+b 2a) +4ac-b ) 4a, and the square term in front can determine n, such as a<0, a(n+b 2a) has a maximum value of 0, (if and only if n=-b 2a, etc.), so that the n value can be determined, and then the whole can be determined.
1: Let's use the equation to solve it.
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