On the number of solutions of a quadratic equation in a normed space二次算子二次泛函方程向量空间研究方程Q(u)=g,其中Q是映射一个赋范空间到另一个的连续二次算子.显然,若u是该方程的一个解,则-u也是一个解.给出没有其它解的条件,并应用其研究偏微方程u△u=g的Dirichlet边值问题.Alexandrov Victo黑龙江大学自然科学学报
Quadratic equations can have $$0$$, $$1$$, or $$2$$ solutions. The number of real solutions of a quadratic equation depends on the sign of the discriminant $$b^2-4ac$$ of that quadratic equation. $$b^2-4ac>0$$ Two real solutions $$b^2-4ac=0$$ One real solution $$b^2-4a...
Use the discriminant to determine the number of real solutions of the equation. Do not solve the equation. $$4x^2 + 5x + \dfrac{13}{8} = 0 $$ Nature of Solutions of a Quadratic Equation Algebraically, a quadratic equation is...
NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year pap...
Number of solutions of the equation z^3+[3(barz)^2]/|z|=0 where z is a... 03:27 If the roots of the quadratic equation x^2+p x+q=0 are tan30^0a n dtan... 03:03 If xa n dy are complex numbers, then the system of equations (1+i)x+(1... 03:18 If a ,b ,a ...
Consider a general quadratic with the coefficient of x1 being 1 and the roots being r and s. It can be factored as (x−r)(x−s) which is just x2−(r+s)x+rs. Thus, the sum of the roots is the negative of the coefficient of x and the product is the constant term. (In...
Given that for any rational number a, quadratic equation 2 x^2+(a+1) x-(3 a^2-4 a+b)=0 always has two rational roots for x, what is the value of b?相关知识点: 试题来源: 解析 b=1 因为对任何有理数 a, 关于 x 的二次方程 2 x^2+(a+1) x-(3 a^2-4 a+b)=0 都有有...
The type of the solution of a quadratic equation can be determined by a standard value that is called the discriminant value. D=b2−4ac If D>0, then the solutions are real and distinct. If D<0, then the solutions are imaginary. ...
To figure out a fraction of a number, all you need to do is divide that number by the denominator of the fraction and then multiply that result by the numerator.
The number of solutions of the equation "tan" x + "sec"x = 2"cos" x lying in the interval [0, 2 pi] is