Question

Solve the Quadratic equation 5x^2-7x-6=0 by completing the square method​

Answer 1

Given,

$\displaystyle\longrightarrow\sf{5x^2-7x-6=0}$

Dividing each term by 5, we get,

$\displaystyle\longrightarrow\sf{x^2-\dfrac{7}{5}x-\dfrac{6}{5}=0}$

Add $\displaystyle\sf{\dfrac{6}{5}}$ to both sides, then,

$\displaystyle\longrightarrow\sf{x^2-\dfrac{7}{5}x=\dfrac{6}{5}}$

Now, we have to add $\displaystyle\sf{\left(-\dfrac{7}{5}\times\dfrac{1}{2}\right)^2=\dfrac{49}{100}}$ to both sides, to complete the square. Then,

$\displaystyle\longrightarrow\sf{x^2-\dfrac{7}{5}x+\dfrac{49}{100}=\dfrac{6}{5}+\dfrac{49}{100}}$

$\displaystyle\longrightarrow\sf{x^2-\dfrac{7}{5}x+\dfrac{49}{100}=\dfrac{169}{100}}$

Factorising the LHS,

$\displaystyle\longrightarrow\sf{\left(x-\dfrac{7}{10}\right)^2=\left(\dfrac{13}{10}\right)^2}$

Taking the square roots,

$\displaystyle\longrightarrow\sf{x-\dfrac{7}{10}=\pm\dfrac{13}{10}}$

Adding $\displaystyle\sf{\dfrac{7}{10}}$ to both sides,

$\displaystyle\longrightarrow\sf{x=\dfrac{7}{10}\pm\dfrac{13}{10}}$

$\displaystyle\longrightarrow\sf{x=\dfrac{7+13}{10}\quad\quad OR\quad\quad x=\dfrac{7-13}{10}}$

$\displaystyle\longrightarrow\sf{x=\dfrac{20}{10}\quad\quad OR\quad\quad x=\dfrac{-6}{10}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{x=2\quad\quad OR\quad\quad x=-\dfrac{3}{5}}}}$

Answer 2

two zeroes of eqn will be

2 and -3/5

REFER TO THE ATTACHMENT for solution