To solve the problem, we need to analyze the quadratic equation given by ax2+bx+c=0 under the condition that both roots are zero. Step 1: Understanding the RootsIf both roots of the equation are zero, it implies that the equation can be expressed in the form:a(x−0)(x−0)=0Th...
BDivide both sides of the equation 3ax2+3bx+3c=0 by 3 to get the equivalent equation ax2+bx+c=0 with the same roots.如果ax2+bx+c=0的根是6和9,那么3ax2+3bx+3c=0的根是( ).A.2和3 B.6和9 C.18和27 D.−2和−3将等式3ax2+3bx+3c=0的两边除以3,得到与ax2+b...
Dax2+b(x+1)+c(1+x)2=0 Ifα&βare the roots of equationax2+bx+c=0then find the Quadratic equation whose roots are1α&1β View Solution The equation formed by multiplying each root ofax2+bx+c=0by2isx2=36x+24=0 If the roots of a quadratic equationax2+bx+c=0areαandβ,the...
If a and B are the roots of the quadrati c ax2 + bx + c = 0, ( a # 0),determine the values of the following expressions in terms of a, b and c.① 1/α+(1)/β 四 1/(a^2)+1/(β^2) 相关知识点: 试题来源: 解析 m9= (0)1(—)= ()—()= /+= (18)...
Quadratic equation consists of at most two roots.The roots of the quadratic equation {eq}ax^2 + bx + c = 0 {/eq} are {eq}\frac{ - b \pm \sqrt {b^2 - 4ac}}{2a} {/eq}. This {eq}{b^2 - 4ac} {/eq} i...
Pay attention to y=ax^2+bx+c(a≠0), and answer the question.If the two roots of this equation are m&n, which situation can be m0 orm>0, nNote: y=0 (I typed wrong.)
百度试题 结果1 题目If the roots of a quadratic equation are A + 1 and A - 1, write the equation in ax^2+bx+c=0 form. 相关知识点: 试题来源: 解析 x^2 − 2Ax+A^2 − 1 = 0x^2 − 2Ax+A^2 − 1 = 0 反馈 收藏 ...
The general solution for second order differential equations vary depending on the roots of the characteristic equation. If we have real and distinct roots, say r1 and r2, then the general solution is y(t)=c1er1t+c2er2t. If we have complex...
百度试题 结果1 题目Pay attention to y=ax^2+bx+c(a≠0), and answer the question.If the two roots of this equation are m&n, which situation can be m0 orm>0, nNote: y=0 (I typed wrong.) 相关知识点: 试题来源: 解析 c/a<0 反馈 收藏 ...
so the graph will touch the x-axis at only one points.(b2 - 4ac) < 0 : The square root will be imaginary, and the roots of the equation will be two complex numbers, so the graph will not touch the x-axis.So by looking at the graph, you ...