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The sum of a number and its square is $\frac{63}{4}$, find the numbers.
Given:
The sum of a number and its square is $\frac{63}{4}$.
To do:
We have to find the numbers.
Solution:
Let the required number be $x$.
This implies,
The square of the number is $x^2$.
According to the question,
$x+x^2=\frac{63}{4}$
$4(x+x^2)=63$
$4x^2+4x-63=0$
Solving for $x$ by factorization method, we get,
$4x^2+18x-14x-63=0$
$2x(2x+9)-7(2x+9)=0$
$(2x+9)(2x-7)=0$
$2x+9=0$ or $2x-7=0$
$2x=-9$ or $2x=7$
$x=\frac{-9}{2}$ or $x=\frac{7}{2}$
The required numbers are $\frac{-9}{2}$ and $\frac{7}{2}$.
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