Find the discriminant of the quadratic equation: 4x^2-2/3x-1/16=0 00:55 Determine the discriminant of quadratic equation (i)3x^2+2x-1=0 (ii) x... 02:42 Find the discriminant of the quadratic equation: 4x^(2)-(2)
Bx = - 3 Cx = - 1 Dx = - 4Submit Solve the following equation for x and give your answer correct to 2 decimal places : 3x2+5x−9=0 View Solution Solve the x using the quadratic formula. Write your answer correct to two significant figures. (x−1)2−3x+4=0. View Solution...
Solve the quadratic equation: -8x^2 - 24x + 8 = 0 Solve the quadratic equation. -0.3x^{2}+12x+10=0 Solve the following quadratic equation: 3x(x - 4) = x - 4 Solve the quadratic equation : 0 = 54x - 6x^2 + 60 Solve the quadratic equation 5x^{2} - 80 = 0. Solve the ...
The quadratic formula is a method used to solve quadratic equations. It consists of substituting the coefficients of the equation and solving to obtain the two values of the variable if it has a solution for the real numbers. The quadratic equation in its general form is ax...
1.1 has many ingredients and is based on the next five lemmas. lemma 3.3 let \(k-2\) be a multiple of 4, and \(g^*\) be a \(k\times k\) grid. if \((g^*,\varphi )\) is p -separated, then there exists an affine transformation \(\nu \) such that \((g^*[-1],\...
7 A curve has equation y=(ln(3x^2-5))/(2x+1)for3x^25.(a) Find the equation of the no
Quadratic EquationThe quadratic equation is the one that has as principal exponent a two; we can also call it a second-degree equation. We can solve this by factoring in or applying the quadratic formula. We must identify the ...
Write a quadratic equation in the variable x having the given numbers as solutions. Give the equation in standard form: ax^2 + bx + c = 0. -sqrt 3, 3*sqrt 3 Write 5x^{10}-6x^5-3 in quadratic form. What is the standard form of qu...
Critical points Critical point x 1c c xx c −1 Figure 4.3: A quadratic: One critical point Figure 4.4: f (x) = x3 + x + 1: No critical points Figure 4.5: Many critical points 178 Chapter Four USING THE DERIVATIVE Testing For Local Maxima and Minima If f has different signs on ...
5x - 1 = 2x2 ⇒ 2x2 - 5x + 1 = 0 This has the form ax2 + bx + c = 0 a = 2 b = -5 c = 1 The quadratic formula gives solutions: x = [1/(2a)] [-b ±√(b2 - 4ac)] There will be two solutions, one with the positive square root, and one with the negative sq...