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What is the equation of a line that passes through the point 0 2 and is perpendicular to a line with a slope of 3?
Slope of line: 3 Perpendicular slope: -1/3 Equation: y-2 = -1/3(x-0) => y = -1/3x+2
Y equals -5x plus 5 and 3x plus 2y equals 3 solve this equation using substitution?
y = -5x+5 (Equation 1)3x+2y = 3 (Equation 2)you see the the first equation is equal to y. just put that into the second equation.3x + 2(-5x+5) = 3 (now do algebra and find x)3x - 10x + 10 = 3-7x = -7-x = -1x = 1Now put x into one of the equations to find yy = -5(1) + 5 (Equation 1)y = -5 + 5y = 0Now put both the x and y into both equations to see if it is truee.0 = -5(1) + 5 (Equation 1)0 = -5 + 50 = 0 CORRECT3(1) + 2(0) = 3 (Equation 2)3 + 0 = 33 = 3 CORRECTTherefore your answers arex = 1y = 0
What equation gives 25 using 8 6 1 2 3 0?
It is: 6*(2+3-1)+80 = 25
How do you solve the Relationship between the coefficient and roots of quadratic equations?
TO FIND THE RELATION BETWEEN ROOTS AND COEFFICIENTS OF A QUADRATIC EQUATION:Let us take the general form of a quadratic equation:ax2 + bx + c = 0 (1)where a(≠0) is the coefficient of x2 , b is the coefficient of x and c is a constant term. If and ß be the roots of the equation, then we have to find the relations of and ß with a, b and c.Since a ≠0, hence multiplying both sides of (1) by 4a we get,4a2x2 + 4abx + 4ac = 0 or (2ax)2 + 2.2ax.b + b2 - b2 + 4ac = 0Or, (2ax + b)2 = b2 - 4ac2ax + b = b2 - 4acx =Hence, the roots of (1) areLet, = and ß =Hence, + ß = +Or + ß = = - b/a = - (2)Again ß = xOr ß = =Or ß = = = (3)Equations (2) and (3) represent the required relations between roots (that is, and ß) and coefficients (that is, a, b and c) of equation (1).Example 1:If the roots of the equation 2x2 - 9x - 3 = 0 be and ß, then find + ß and ß.Solution:We know that + ß = - = - =And ß = = (Answer)Example 2:If one root of the quadratic equation x2 - x - 1 = 0 is a, prove that its other root is 3 - 3.Solution:x2 - x - 1 = 0 (1)Let ß be the other root of the equation (1). Then,+ ß = = 1 or ß = 1 -Since is a root of the equation (1) hence, 2 - - 1 = 0 or 2 = + 1Now, 3 - 3 = . 2 - 3 = ( + 1) - 3 [Since 2 = + 1]= 2 + - 3 = + 1 - 2 = 1 - = ß [Since ß = 1 - ]Hence, the other root of equation (1) is 3 - 3. (Proved)Example 3:If a2 = 5a - 3 and b2 = 5b - 3, (a ≠b), find the quadratic equation whose roots are and .Solution:Given (a ≠b) and a2 = 5a - 3 and b2 = 5b - 3, hence it is clear that a and b are the roots of the equation x2 = 5x - 3 or x2 - 5x + 3 = 0.Hence, a + b = - = 5 and ab = = 3.Now, the sum of the roots of the required equation= + = = = = =And the product of the roots of the required equation = . = 1.Hence, the required equation is x2 - x + 1 = 0 or 3x2 - 19x + 3 = 0. (Answer)
What is the equation of the line that passes through 2 2 and 6 3 in standard form?
Points: (2, 2) and (6, 3) Slope: 1/4 Equation: y = 1/4x+3/2 In standard form: x-4y+6 = 0
