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# 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

algebra

Roots

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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