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6x^(2)-7x-5=0 Let x=j and x=k be solutions to the given equation, with j > k. What is the value of j-k ?

6x2−7x−5=0 6 x^{2}-7 x-5=0 \newlineLet x=j x=j and x=k x=k be solutions to the given equation, with j>k . What is the value of j−k j-k ?

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Q. 6x2−7x−5=0 6 x^{2}-7 x-5=0 \newlineLet x=j x=j and x=k x=k be solutions to the given equation, with j>k j>k . What is the value of j−k j-k ?
  1. Use Quadratic Formula: Use the quadratic formula to find the solutions to the equation 6x2−7x−5=06x^2-7x-5=0. The quadratic formula is given by x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}, where aa, bb, and cc are the coefficients of the quadratic equation ax2+bx+c=0ax^2+bx+c=0. For our equation, a=6a=6, b=−7b=-7, and c=−5c=-5.
  2. Calculate Discriminant: Calculate the discriminant, which is the part under the square root in the quadratic formula: b2−4acb^2-4ac. Discriminant = (−7)2−4⋅6⋅(−5)=49+120=169(-7)^2 - 4\cdot6\cdot(-5) = 49 + 120 = 169.
  3. Apply Quadratic Formula: Since the discriminant is positive, there are two real and distinct solutions. Now, calculate the solutions using the quadratic formula.\newlinex=−(−7)±1692⋅6x = \frac{-(-7) \pm \sqrt{169}}{2\cdot 6}\newlinex=7±1312x = \frac{7 \pm 13}{12}
  4. Find Solutions: Find the two solutions, which are jj and kk.j=(7+13)/12=20/12=5/3j = (7 + 13) / 12 = 20 / 12 = 5/3k=(7−13)/12=−6/12=−1/2k = (7 - 13) / 12 = -6 / 12 = -1/2Since j > k, we have correctly assigned the values to jj and kk.
  5. Calculate Difference: Calculate the difference j−kj - k. \newlinej−k=(53)−(−12)=(53)+(12)=(106)+(36)=136j - k = \left(\frac{5}{3}\right) - \left(-\frac{1}{2}\right) = \left(\frac{5}{3}\right) + \left(\frac{1}{2}\right) = \left(\frac{10}{6}\right) + \left(\frac{3}{6}\right) = \frac{13}{6}

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