Solve Quadraticequations -3x^2-5x-30=0 Tiger Algebra Solver

Step by step solution :

Step 1 :

Equation at the end of step 1 :

Step 2 :

Step 3 :

Pulling out like terms :

3.1 Pull out like factors : -3x2 - 5x - 30 = -1 • (3x2 + 5x + 30)

Trying to factor by splitting the middle term

3.2 Factoring 3x2 + 5x + 30 The first term is, 3x2 its coefficient is 3 .The middle term is, +5x its coefficient is 5 .The last term, "the constant", is +30 Step-1 : Multiply the coefficient of the first term by the constant 3 • 30 = 90 Step-2 : Find two factors of 90 whose sum equals the coefficient of the middle term, which is 5 .

For tidiness, printing of 18 lines which failed to find two such factors, was suppressedObservation : No two such factors can be found !! Conclusion : Trinomial can not be factored

Equation at the end of step 3 :

Step 4 :

Parabola, Finding the Vertex :

4.1 Find the Vertex of y = -3x2-5x-30Parabolas have a highest or a lowest point called the Vertex . Our parabola opens down and accordingly has a highest point (AKA absolute maximum) . We know this even before plotting "y" because the coefficient of the first term, -3 , is negative (smaller than zero).Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is -0.8333 Plugging into the parabola formula -0.8333 for x we can calculate the y -coordinate : y = -3.0 * -0.83 * -0.83 - 5.0 * -0.83 - 30.0 or y = -27.917

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = -3x2-5x-30 Axis of Symmetry (dashed) {x}={-0.83} Vertex at {x,y} = {-0.83,-27.92} Function has no real roots

Solve Quadratic Equation by Completing The Square

4.2 Solving -3x2-5x-30 = 0 by Completing The Square . Multiply both sides of the equation by (-1) to obtain positive coefficient for the first term: 3x2+5x+30 = 0 Divide both sides of the equation by 3 to have 1 as the coefficient of the first term : x2+(5/3)x+10 = 0Subtract 10 from both side of the equation : x2+(5/3)x = -10Now the clever bit: Take the coefficient of x , which is 5/3 , divide by two, giving 5/6 , and finally square it giving 25/36 Add 25/36 to both sides of the equation : On the right hand side we have : -10 + 25/36 or, (-10/1)+(25/36) The common denominator of the two fractions is 36 Adding (-360/36)+(25/36) gives -335/36 So adding to both sides we finally get : x2+(5/3)x+(25/36) = -335/36Adding 25/36 has completed the left hand side into a perfect square : x2+(5/3)x+(25/36) = (x+(5/6)) • (x+(5/6)) = (x+(5/6))2 Things which are equal to the same thing are also equal to one another. Since x2+(5/3)x+(25/36) = -335/36 and x2+(5/3)x+(25/36) = (x+(5/6))2 then, according to the law of transitivity, (x+(5/6))2 = -335/36We'll refer to this Equation as Eq. #4.2.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of (x+(5/6))2 is (x+(5/6))2/2 = (x+(5/6))1 = x+(5/6)Now, applying the Square Root Principle to Eq. #4.2.1 we get: x+(5/6) = √ -335/36 Subtract 5/6 from both sides to obtain: x = -5/6 + √ -335/36 In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1 Since a square root has two values, one positive and the other negative x2 + (5/3)x + 10 = 0 has two solutions: x = -5/6 + √ 335/36 • i or x = -5/6 - √ 335/36 • i Note that √ 335/36 can be written as √ 335 / √ 36 which is √ 335 / 6

Solve Quadratic Equation using the Quadratic Formula

4.3 Solving -3x2-5x-30 = 0 by the Quadratic Formula . According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by : - B ± √ B2-4AC x = ———————— 2A In our case, A = -3 B = -5 C = -30 Accordingly, B2 - 4AC = 25 - 360 = -335Applying the quadratic formula : 5 ± √ -335 x = —————— -6In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i) Both i and -i are the square roots of minus 1Accordingly,√ -335 = √ 335 • (-1) = √ 335 • √ -1 = ± √ 335 • i √ 335 , rounded to 4 decimal digits, is 18.3030 So now we are looking at: x = ( 5 ± 18.303 i ) / -6Two imaginary solutions :

Two solutions were found :

1. x =(5-√-335)/-6=5/-6+i/6√ 335 = -0.8333-3.0505i
2. x =(5+√-335)/-6=5/-6-i/6√ 335 = -0.8333+3.0505i