Solve the following quadratic equation by factorization:
$(x-5)(x-6)=\frac{25}{(24)^2}$

Given:

Given quadratic equation is $(x-5)(x-6)=\frac{25}{(24)^2}$.

To do:

We have to solve the given quadratic equation.

Solution:

$(x-5)(x-6)=\frac{25}{(24)^2}$

$x^2-5x-6x+30=\frac{25}{(24)^2}$

$x^2-11x+30=\frac{25}{576}$

$x^2-11x+30-\frac{25}{576}=0$

$x^2-11x+\frac{30\times576-25}{576}=0$

$x^2-11x+\frac{17280-25}{576}=0$

$x^2-11x+\frac{17255}{576}=0$

$x^2-\frac{264}{24}x+\frac{17255}{(24)^2}=0$

$x^2-(\frac{145+119}{24})x+\frac{145\times119}{(24)^2}=0$

$x^2-\frac{145}{24}x-\frac{119}{24}x+\frac{145\times119}{(24)^2}=0$

$x(x-\frac{145}{24})-\frac{119}{24}(x-\frac{145}{24})=0$

$(x-\frac{145}{24})(x-\frac{119}{24})=0$

$(x-\frac{145}{24})=0$ or $(x-\frac{119}{24})=0$

$x=\frac{145}{24}$ or $x=\frac{119}{24}$

The values of $x$ are $\frac{145}{24}$ and $\frac{119}{24}$.

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Updated on: 10-Oct-2022

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