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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Updated on: 10-Oct-2022

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