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x^(2)-13 x+30=0
What are the solutions to the given equation?
Choose 1 answer:
(A) 
x=-15 and 
x=2
(B) 
x=-10 and 
x=-3
(C) 
x=3 and 
x=10
(D) 
x=-2 and 
x=15

x213x+30=0 x^{2}-13 x+30=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A) x=15 x=-15 and x=2 x=2 \newline(B) x=10 x=-10 and x=3 x=-3 \newline(C) x=3 x=3 and x=10 x=10 \newline(D) x=2 x=-2 and x=15 x=15

Full solution

Q. x213x+30=0 x^{2}-13 x+30=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A) x=15 x=-15 and x=2 x=2 \newline(B) x=10 x=-10 and x=3 x=-3 \newline(C) x=3 x=3 and x=10 x=10 \newline(D) x=2 x=-2 and x=15 x=15
  1. Given equation: The given equation is a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0. We need to find the values of xx that satisfy the equation x213x+30=0x^2 - 13x + 30 = 0. To do this, we can factor the quadratic if possible.
  2. Factoring the quadratic: We look for two numbers that multiply to give the constant term, c=30 c = 30 , and add up to give the coefficient of the x x term, b=13 b = -13 . The numbers that satisfy these conditions are 3 -3 and 10 -10 because (3)×(10)=30 (-3) \times (-10) = 30 and (3)+(10)=13 (-3) + (-10) = -13 .
  3. Setting up the equations: We can now factor the quadratic equation as (x3)(x10)=0(x - 3)(x - 10) = 0.
  4. Solving the first equation: To find the solutions, we set each factor equal to zero and solve for x. This gives us two equations: x3=0x - 3 = 0 and x10=0x - 10 = 0.
  5. Solving the second equation: Solving the first equation, x3=0x - 3 = 0, we get x=3x = 3.
  6. Solution to the quadratic equation: Solving the second equation, x10=0x - 10 = 0, we get x=10x = 10.
  7. Solution to the quadratic equation: Solving the second equation, x10=0x - 10 = 0, we get x=10x = 10. We have found the two solutions to the quadratic equation: x=3x = 3 and x=10x = 10. These are the values of xx that satisfy the equation x213x+30=0x^2 - 13x + 30 = 0.

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