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
algebra
quadratic equation
Roots
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 \((d-1)(e-1)\)=\((1-1)(\frac{-5}{2} -1)\)=\(0\)
Other useful links
- https://www.cheenta.com/ratio-of-lcm-gcf-algebra-amc-8-2013-problem-10/
- https://www.youtube.com/watch?v=V01neV8qmh4