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Determine how many REAL solutions the quadratic equation has. 2x^(2)-xsqrt5+2=0 No real solutions One real solution Two real solutions

44. Determine how many REAL solutions the quadratic equation has.\newline2x2−x5+2=0 2 x^{2}-x \sqrt{5}+2=0 \newlineNo real solutions\newlineOne real solution\newlineTwo real solutions

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Q. 44. Determine how many REAL solutions the quadratic equation has.\newline2x2−x5+2=0 2 x^{2}-x \sqrt{5}+2=0 \newlineNo real solutions\newlineOne real solution\newlineTwo real solutions
  1. Calculate Discriminant: To determine the number of real solutions, we can use the discriminant, which is b2−4acb^2 - 4ac from the quadratic formula.
  2. Identify Coefficients: For the equation 2x2−x5+2=02x^2 - x\sqrt{5} + 2 = 0, a=2a = 2, b=−5b = -\sqrt{5}, and c=2c = 2.
  3. Simplify Discriminant: Now calculate the discriminant: D=b2−4ac=(−5)2−4(2)(2)D = b^2 - 4ac = (-\sqrt{5})^2 - 4(2)(2).
  4. Check Discriminant Sign: Simplify the discriminant: D=5−16=−11D = 5 - 16 = -11.
  5. Conclusion: Since the discriminant is negative D=−11D = -11, there are no real solutions to the equation.

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