WebIf \(kx^{2}+5x-\frac{5}{4}=0\) has equal roots, then \(b^2-4ac=0\). The discriminant is \({b^2} - 4ac\), which comes from the quadratic formula and we can use this to find the nature of … WebJan 13, 2024 · IVT states that if a continuous function f(x) on the interval [a,b] has values of opposite sign inside an interval, then there must be some value x=c on the interval (a,b) for which f(c)=0. Because f(-2) is negative and f(-1) is positive, and f(x) is continuous on the closed interval [-2,-1], there must be some value x=c on the interval [-2,-1 ...
If one root of x 2 - x - k = 0 be square of the other, then k - Toppr
WebStep 1: Know about the discrimination( b^2-4ac ) of the quadratic equation.The discriminant indicates whether there are two, one, or no solutions. 1) b^2 - 4ac 0 (Positive), there are 2 real solutions 2) b^2-4ac = 0 (Zero), there is one real solution 3) b^2-4ac real roots.Step 2: Understand the given question Quadratic equation: kx^2 + 12x + k = 0 Constant k = … WebThe real part of the other side is a polynomial in cos x and sin x, in which all powers of sin x are even and thus replaceable through the identity cos 2 x + sin 2 x = 1. By the same reasoning, sin nx is the imaginary part of the polynomial, in which all powers of sin x are odd and thus, if one factor of sin x is factored out, the remaining factors can be replaced to … talbot health services mansfield ohio
Geometrical properties of polynomial roots - Wikipedia
WebLet `alpha and beta `be the roots of quadratic equation`2x^2 + kx + 4 = 0` in such a way that `alpha = 2` Here, a = 2, b = k and , c = 4. Then , according to question sum of the roots `alpha + beta = (-b)/a` `2+ beta = (-k)/2` `beta = (-k)/2 - 2` `beta = (-k -4)/2` And the product of the roots `alpha .beta = c /a` `= 4/2` `= 2` Putting the ... WebApproach 1 First suppose that the roots of the equation \[\begin{equation} x^2-bx+c=0 \label{eq:1} \end{equation}\] are real and positive. From the quadratic formula, we see that the roots of \(\eqref{eq:1}\) are of the form \[\frac{b\pm\sqrt{b^2-4c}}{2}.\] For the root or roots to be real, we require that \(b^2-4c \geq 0\), that is, \(b^2 \geq 4c\).In order for them … WebClick here👆to get an answer to your question ️ Find the least positive value of k for which the equation x^2 + kx + 4 = 0 has real roots. Solve Study ... If the equation is supposed to have … twitter lxeagle