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二次方程・零点与韦达定理进阶

\( x^2+27x+1=0 \)
What is one of the solutions to the given equation?

二次方程・零点与韦达定理挑战

The function h is defined by \( h\left(x\right)=-\sqrt{\left(x^2+bx+c\right)} \) , where b and c are constants. In the xy-plane, the graph of \( y=h(x) \) contains the points \( \left(2,\ 0\right) \) and \( (0,-\sqrt{314}) \) . If \( h(m)=0 \) , what is the greatest possible value of m?

二次方程・零点与韦达定理基础

\( (z+2)(z-4)=0 \)
What are all possible solutions to the given equation?

二次方程・零点与韦达定理挑战

\( x^2+y+5=5 \)
\( 4x+4-y=0 \)
The solution to the given system of equations is (x, y). What is the value of x?

二次方程・零点与韦达定理基础

\( (x-44)^2=0 \)
What value of x is the solution to the given equation?

二次方程・零点与韦达定理进阶

\( x(x-14)(x+6)(x-5)^2=0 \)
What is the greatest solution to the given equation?

二次方程・零点与韦达定理挑战

\( (x-k)^2=(k-5a)(x-k) \)
In the given equation, a and k are constants, where \( k>5a \). The sum of the solutions to the equation is \( 3k+37 \). What is the value of a?

二次方程・零点与韦达定理进阶

\( 7x(2x-34)=0 \)
What is the sum of the solutions to the given equation?

二次方程・零点与韦达定理进阶

\( x^2-2x=30 \)
What is one of the solutions to the given equation?

二次方程・零点与韦达定理进阶

\( \left(3x-8\right)^2-11\left(3x-8\right)+28=0 \)
What is a solution to the given equation?

原库参考解析

二次方程・零点与韦达定理挑战

\( w^2+8w-28=0 \)
Which of the following is a solution to the given equation?

原库参考解析

二次方程・零点与韦达定理进阶

\( (x+9)^{2}(x−49)^{2}=0 \)
What is the positive solution to the given equation?

原库参考解析