Solve Quadraticequations 97=215t-16t^2 Tiger Algebra Solver

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 97-(215*t-16*t^2)=0

Step by step solution :

Step 1 :

Equation at the end of step 1 :

Step 2 :

Trying to factor by splitting the middle term

2.1 Factoring 16t2-215t+97 The first term is, 16t2 its coefficient is 16 .The middle term is, -215t its coefficient is -215 .The last term, "the constant", is +97 Step-1 : Multiply the coefficient of the first term by the constant 16 • 97 = 1552 Step-2 : Find two factors of 1552 whose sum equals the coefficient of the middle term, which is -215 .

For tidiness, printing of 14 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 2 :

Step 3 :

Parabola, Finding the Vertex :

3.1 Find the Vertex of y = 16t2-215t+97Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 16 , is positive (greater 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,At2+Bt+C,the t -coordinate of the vertex is given by -B/(2A) . In our case the t coordinate is 6.7188 Plugging into the parabola formula 6.7188 for t we can calculate the y -coordinate : y = 16.0 * 6.72 * 6.72 - 215.0 * 6.72 + 97.0 or y = -625.266

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = 16t2-215t+97 Axis of Symmetry (dashed) {t}={ 6.72} Vertex at {t,y} = { 6.72,-625.27} t -Intercepts (Roots) : Root 1 at {t,y} = { 0.47, 0.00} Root 2 at {t,y} = {12.97, 0.00}

Solve Quadratic Equation by Completing The Square

3.2 Solving 16t2-215t+97 = 0 by Completing The Square . Divide both sides of the equation by 16 to have 1 as the coefficient of the first term : t2-(215/16)t+(97/16) = 0Subtract 97/16 from both side of the equation : t2-(215/16)t = -97/16Now the clever bit: Take the coefficient of t , which is 215/16 , divide by two, giving 215/32 , and finally square it giving 46225/1024 Add 46225/1024 to both sides of the equation : On the right hand side we have : -97/16 + 46225/1024 The common denominator of the two fractions is 1024 Adding (-6208/1024)+(46225/1024) gives 40017/1024 So adding to both sides we finally get : t2-(215/16)t+(46225/1024) = 40017/1024Adding 46225/1024 has completed the left hand side into a perfect square : t2-(215/16)t+(46225/1024) = (t-(215/32)) • (t-(215/32)) = (t-(215/32))2 Things which are equal to the same thing are also equal to one another. Since t2-(215/16)t+(46225/1024) = 40017/1024 and t2-(215/16)t+(46225/1024) = (t-(215/32))2 then, according to the law of transitivity, (t-(215/32))2 = 40017/1024We'll refer to this Equation as Eq. #3.2.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of (t-(215/32))2 is (t-(215/32))2/2 = (t-(215/32))1 = t-(215/32)Now, applying the Square Root Principle to Eq. #3.2.1 we get: t-(215/32) = √ 40017/1024 Add 215/32 to both sides to obtain: t = 215/32 + √ 40017/1024 Since a square root has two values, one positive and the other negative t2 - (215/16)t + (97/16) = 0 has two solutions: t = 215/32 + √ 40017/1024 or t = 215/32 - √ 40017/1024 Note that √ 40017/1024 can be written as √ 40017 / √ 1024 which is √ 40017 / 32

Solve Quadratic Equation using the Quadratic Formula

3.3 Solving 16t2-215t+97 = 0 by the Quadratic Formula . According to the Quadratic Formula, t , the solution for At2+Bt+C = 0 , where A, B and C are numbers, often called coefficients, is given by : - B ± √ B2-4AC t = ———————— 2A In our case, A = 16 B = -215 C = 97 Accordingly, B2 - 4AC = 46225 - 6208 = 40017Applying the quadratic formula : 215 ± √ 40017 t = ———————— 32 √ 40017 , rounded to 4 decimal digits, is 200.0425 So now we are looking at: t = ( 215 ± 200.042 ) / 32Two real solutions: t =(215+√40017)/32=12.970 or: t =(215-√40017)/32= 0.467

Two solutions were found :

1. t =(215-√40017)/32= 0.467
2. t =(215+√40017)/32=12.970