Write a quadratic equation of one variable with 3, - 1 as root and quadratic coefficient of 1

Write a quadratic equation of one variable with 3, - 1 as root and quadratic coefficient of 1

For the quadratic equation with 3, - 1 as roots and the coefficient of quadratic term being 1, the following equation is obtained
(x-3)(x+1)=0
x²-2x-3=0

It is necessary to write a quadratic equation with 4, - 5 as roots and the coefficient of quadratic term is 1______ ．

∵ 4 + (- 5) = - 1, 4 × (- 5) = - 20, ∵ the univariate quadratic equation with roots of 4, - 5 and coefficient of quadratic term of 1 is x2 + x-20 = 0. So the answer is x2 + x-20 = 0

The quadratic equation with roots 2 and 5 and quadratic coefficient 1 is?

(x-2)(x-5)=x^2-7x+10=0

Write a quadratic equation of one variable, its quadratic coefficient is 1, its two roots are 2 and 3 respectively

x²+bx+c=0
x1=2,x2=3
Then X1 + x2 = - B
x1x2=c
So B = - 5, C = 6
x²-5x+6=0

Write out the quadratic equation of one variable with quadratic coefficient of 1 and roots of 2 and 3

(x-2)(x-3)=0

Please write out a quadratic equation with one variable so that one root is 1 and the coefficient of quadratic term is - 2

The general solution is: - 2 (x-1) (x-a) = 0, it may be substituted into a = 0, then a satisfactory equation is: - 2x ^ 2 + 2x = 0

The coefficient of quadratic term is 2, and two quadratic equations with roots of 1,3 / 2 are

When the coefficients of two terms and quadratic terms are known, the following results are obtained by means of Weida's theorem
x1+x2=-b/a ----->-b/2=(1+3/2)
b=-5.
x1*x2=c/a -----> c/2=1*3/2.
c=3,
The quadratic equation of one variable is: 2x ^ 2-5x + 3 = 0

Please write a quadratic term whose coefficient is - 3, one root is 3, and the other root satisfies - 1

-3x^2+9x=0

Univariate quadratic equation 3 (x-3) = (x-3) ^ 2

3(x-3)=(x-3)(x-3)
[3-(x-3)](x-3)=0
(6-x)(x-3)=0
6-x = 0 or x-3 = 0
∴x1=6 x2=3

Univariate quadratic equation (2 + √ 3) x ^ 2 - (2 - √ 3) x = 0
The method of factorization

(2+√3)X^2-(2-√3)x=0
X^2-(2-√3)/(2+√3)*x=0
X[X-(2-√3)/(2+√3)]=0
Then X1 = 0, X2 = (2 - √ 3) / (2 + √ 3) = (2 - √ 3) ^ 2 = 4-4 √ 3 + 3 = 7-4 √ 3