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16x^(2)-8x-3=0 Let x=q and x=r be solutions to the equation shown, with q > r. What is the value of q-r ?

16x2−8x−3=0 16 x^{2}-8 x-3=0 \newlineLet x=q x=q and x=r x=r be solutions to the equation shown, with q>r . What is the value of q−r q-r ?

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Q. 16x2−8x−3=0 16 x^{2}-8 x-3=0 \newlineLet x=q x=q and x=r x=r be solutions to the equation shown, with q>r q>r . What is the value of q−r q-r ?
  1. Identify quadratic equation: Identify the quadratic equation and its standard form.\newlineThe given quadratic equation is 16x2−8x−3=016x^2 - 8x - 3 = 0. The standard form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0.
  2. Use quadratic formula: Use the quadratic formula to find the solutions for xx. The quadratic formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=16a = 16, b=−8b = -8, and c=−3c = -3.
  3. Calculate discriminant: Calculate the discriminant b2−4acb^2 - 4ac. Discriminant, D=(−8)2−4×16×(−3)=64+192=256D = (-8)^2 - 4 \times 16 \times (-3) = 64 + 192 = 256.
  4. Calculate two solutions: Calculate the two solutions using the quadratic formula.\newlinex=−(−8)±2562×16x = \frac{-(-8) \pm \sqrt{256}}{2 \times 16}\newlinex=8±25632x = \frac{8 \pm \sqrt{256}}{32}\newlinex=8±1632x = \frac{8 \pm 16}{32}
  5. Find two solutions: Find the two solutions qq and rr. Since q > r, we take the positive value for qq and the negative value for rr. q=(8+16)/32=24/32=3/4q = (8 + 16) / 32 = 24 / 32 = 3/4 r=(8−16)/32=−8/32=−1/4r = (8 - 16) / 32 = -8 / 32 = -1/4
  6. Calculate difference: Calculate the difference q−rq - r. \newlineq−r=(34)−(−14)=34+14=44=1q - r = \left(\frac{3}{4}\right) - \left(-\frac{1}{4}\right) = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1

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