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2x^(2)+5x-k=0
In the given equation, 
k is a constant. For what value of 
k does the equation have exactly one distinct real solution?
Choose 1 answer:
(A) 
-(25)/(8)
(B) 
-(5)/(4)
(c) 
(5)/(4)
(D) 
(25)/(8)

2x2+5xk=0 2 x^{2}+5 x-k=0 \newlineIn the given equation, k k is a constant. For what value of k k does the equation have exactly one distinct real solution?\newlineChoose 11 answer:\newline(A) 258 -\frac{25}{8} \newline(B) 54 -\frac{5}{4} \newline(C) 54 \frac{5}{4} \newline(D) 258 \frac{25}{8}

Full solution

Q. 2x2+5xk=0 2 x^{2}+5 x-k=0 \newlineIn the given equation, k k is a constant. For what value of k k does the equation have exactly one distinct real solution?\newlineChoose 11 answer:\newline(A) 258 -\frac{25}{8} \newline(B) 54 -\frac{5}{4} \newline(C) 54 \frac{5}{4} \newline(D) 258 \frac{25}{8}
  1. Quadratic Equation Discriminant: A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 has exactly one distinct real solution when its discriminant b24acb^2 - 4ac is equal to zero. This is because the discriminant determines the nature of the roots of the quadratic equation.
  2. Given Equation and Variables: For the given equation 2x2+5xk=02x^2 + 5x - k = 0, a=2a = 2, b=5b = 5, and c=kc = -k. We will set the discriminant equal to zero and solve for kk.
    Discriminant: b24ac=0b^2 - 4ac = 0
  3. Substitute Values: Substitute the values of aa, bb, and cc into the discriminant equation:\newline(5)24(2)(k)=0(5)^2 - 4(2)(-k) = 0
  4. Simplify Equation: Simplify the equation: 25+8k=025 + 8k = 0
  5. Solve for k: Solve for k:\newline8k=258k = -25\newlinek=258k = -\frac{25}{8}
  6. Final Solution: The value of kk for which the equation has exactly one distinct real solution is 258-\frac{25}{8}.

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