Question:

Solution of quadratic equation formula, recall if ax^2+bx+c=0 then x=-b±√b^2-4ac/2a

Use the quadratic formula to solve

x+2/4+3/x-1=7

Question:

Solution of quadratic equation formula, recall if ax^2+bx+c=0 then x=-b±√b^2-4ac/2a

Use the quadratic formula to solve

x+2/4+3/x-1=7

Tutors, please sign in to answer this question.

Multiply both sides by 4(x-1),

(x+2)(x-1) + 3*4 = 7*4(x-1)

Collect all terms in one side, and put each term in decreasing order,

x^2 -27x +38 = 0

a = 1, b = -27, and c = 38.

Now you can use quadratic formula. Can you finish it?

Given equation: x+2/4+3/x-1=7 LHS: Left hand side RHS: Right hand side Let us begin by rewriting the given equation more clearly: (x+2)/4 + 3/(x-1) =7 Remember: Equation does not change in value if the same operation is performed on the LHS and RHS of the equality sign. Step 1: Multiply LHS and RHS by 4 to get (x+2) +12/(x-1) = 28 Step 2 Multiply LHS and RHS by (x-1) to get (x-1)(x+2) + 12 = 28(x-1) Step 3 Expand the LHS and RHS of the equation: x^2-x+2x-2+12 = 28x-28 Step 4 Simplifying yields x^2+x+10 =28x-28 Step 5 Add -28x to LHS and RHS to get: x^2+x+10-28x = 28x-28-28x Step 6 Simplifying yields: x^2-27x+10 = -28 Step 7 Add 28 to LHS and RHS of the equation to get: x^2-27x+38 =0 This is a quadratic equation of the form: ax^2+bx+c=0 where a=1 b=-27 c=38 Remember: The roots of a quadratic equation ax^2+bx+c=0 are given by the quadratic formula: x=-b±vb^2-4ac/2a where the symbol "±" indicates that both x= -b+vb^2-4ac/2a and -b-vb^2-4ac/2a are solutions of the quadratic equation Therefore, the roots to this equation are given by: x= -(-27) + v[(27)2-4(1*38)] /2(1) and x = -(-27) - v[(27)2-4(1*8)] /2(1) solving for x yields: x= 27 + v(729-152) /2(1) = 27 + v577 /2(1) and x= 27 - v577 /2(1) v577 = 24.02082429……… In other words, v577 is not an exact number. As a first approximation, let us round this off to 24. Solving for gives: x= 27 + 24 /2(1) = 25.5 and x= 27 - 24 /2(1) = 1.5 Let us now verify if the solutions obtained are correct by substituting for x in the original equation rewritten below: (x+2)/4 + 3/(x-1) =7 Substituting x = 25.5, the equation LHS reduces to: (25.5+2)/4 + 3/(25.5-1) = (27.5/4) + 3/(24.5) = (6.875 + 0.1224489….) =6.9974489….. Substituting x = 1.5, the equation LHS reduces to: (1.5+2)/4 + 3/(1.5-1) = (3.5/4) + 3/(0.5) = (0.875 + 6) =6.875 Note that both solutions closely approximate the RHS value of 7 in the original equation. This is because we approximated v577, which is not an exact number, to 24. Let us redo the solution by approximating v577 to the fourth decimal value, that is, 24.0208 Solving for gives: x= 27 + 24.0208 /2(1) = 25.5104 and x= 27 – 24.0208 /2(1) = 1.4896 Let us now verify if the solutions obtained are correct by substituting for x in the original equation rewritten below: (x+2)/4 + 3/(x-1) =7 Substituting x = 25.5104, the equation LHS reduces to: (25.5104+2)/4 + 3/(25.5104-1) = 6.8776 + 0.1223970….. =6.99997023… Substituting x = 1.4896, the equation LHS reduces to: (1.4896+2)/4 + 3/(1.4896-1) = 0.8724 + 6.1274509….. = 6.9998509…. As you can notice, the match with the RHS of the original equation gets closer as we take a better approximation of v577 that is not an exact number, For most cases, approximation to the secdn decimal should be enough. Final answer: x=25.5104 and x= 1.4896