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June 15, 2020

Quadratic equation Problem | AMC-10A, 2003 | Problem 5

Try this beautiful problem from Algebra based on quadratic equation.

Quadratic equation - AMC-10A, 2003- Problem 5

Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$. What is the value of $(d-1)(e-1)$?

  • \(2\)
  • \(0\)
  • \(\frac{7}{2}\)

Key Concepts


quadratic equation


Check the Answer

Answer: \(0\)

AMC-10A (2003) Problem 5

Pre College Mathematics

Try with Hints

To find out the value of \((d-1)(e-1)\),at first we have to find out the value of \(d\) and \(e\).Given that \(d\) and \(e\) are the solutions of the equations $2x^{2}+3x-5=0$ that means \(d\) and \(e\) are the roots of the given equation.so if we find out the values of roots from the given equation then we will get \(d\) and \(e\).Can you find out the roots?

Can you now finish the problem ..........

To find out the roots :

The given equation is \(2x^{2}+3x-5=0\) \(\Rightarrow (2x+5)(x-1)=0\) \(\Rightarrow x=1 or \frac{-5}{2}\)

Therefore the values of \(d\) and \(e\) are \(1\) and \(\frac{-5}{2}\) respectively

can you finish the problem........

Therefore \((d-1)(e-1)\)=\((1-1)(\frac{-5}{2} -1)\)=\(0\)

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