Question
Solve the following quadratic equations by factorization:
$(\text{x}-5)(\text{x}-6)=\frac{25}{(24)^2}$
$(\text{x}-5)(\text{x}-6)=\frac{25}{(24)^2}$
✓
Answer
We have been given that,
$(\text{x}-5)(\text{x}-6)=\frac{25}{(24)^2}$
$\text{x}^2-11\text{x}+30-\frac{25}{576}=0$
$\text{x}^2-11\text{x}+\frac{17255}{576}=0$
$\text{x}^2-\frac{145}{24}\text{x}-\frac{119}{24}\text{x}+\frac{17255}{576}=0$
$\text{x}\Big(\text{x}-\frac{145}{24}\Big)-\frac{119}{24}\Big(\text{x}-\frac{145}{24}\Big)=0$
$\Big(\text{x}-\frac{119}{24}\Big)\Big(\text{x}-\frac{145}{24}\Big)=0$
Therefore,
$\text{x}-\frac{119}{24}=0$
$\text{x}=\frac{119}{24}$
or, $\text{x}-\frac{145}{24}=0$
$\text{x}=\frac{145}{24}$
Hence, $\text{x}=\frac{119}{24}=4\frac{23}{24}$ or $\text{x}=\frac{145}{24}=6\frac{1}{24}$
$(\text{x}-5)(\text{x}-6)=\frac{25}{(24)^2}$
$\text{x}^2-11\text{x}+30-\frac{25}{576}=0$
$\text{x}^2-11\text{x}+\frac{17255}{576}=0$
$\text{x}^2-\frac{145}{24}\text{x}-\frac{119}{24}\text{x}+\frac{17255}{576}=0$
$\text{x}\Big(\text{x}-\frac{145}{24}\Big)-\frac{119}{24}\Big(\text{x}-\frac{145}{24}\Big)=0$
$\Big(\text{x}-\frac{119}{24}\Big)\Big(\text{x}-\frac{145}{24}\Big)=0$
Therefore,
$\text{x}-\frac{119}{24}=0$
$\text{x}=\frac{119}{24}$
or, $\text{x}-\frac{145}{24}=0$
$\text{x}=\frac{145}{24}$
Hence, $\text{x}=\frac{119}{24}=4\frac{23}{24}$ or $\text{x}=\frac{145}{24}=6\frac{1}{24}$
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