Maharashtra BoardEnglish MediumSTD 10MathsP-1 Quadratic Equations3 Marks
Question
Solve the following quadratic equation by completing the square method.
$2 y^2+9 y+10=0$
✓
Answer
$2 y^2+9 y+10=0$
Steps involved in solving quadratic equation by completing the square method are -
1. Making the first variable free of coefficient
Dividing by the coefficient of 2 , we get, $\Rightarrow y^2+\frac{9}{2} y+5=0$
2. The coefficient of linear variable(variable with degree 1) is then squared and then added and subtracted from the equation.
$\Rightarrow y^2+\frac{9}{2} y+\frac{81}{16}-\frac{81}{16}+5=0$
3. Take out the terms following the formula $(a+b)^2=a^2+b^2+2 a b$
$\Rightarrow\left(y^2+\frac{9}{2} y+\frac{81}{16}\right)-\left(\frac{81}{16}-5\right)=0$
$\Rightarrow\left(y+\frac{9}{4}\right)^2=\frac{81}{16}-5$
$\Rightarrow\left(y+\frac{9}{2}\right)^2=\frac{81-80}{16}$
$\Rightarrow\left(y+\frac{9}{2}\right)^2=\frac{1}{16}$
$\Rightarrow y +\frac{9}{2}=\sqrt{\frac{1}{16}}$
$\Rightarrow y +\frac{9}{2}= \pm \frac{1}{4}$
$\Rightarrow y +\frac{9}{2}=\frac{1}{4} \text { or } y +\frac{9}{2}=-\frac{1}{4}$
$\Rightarrow y =\frac{1}{4}-\frac{9}{2} \text { or } y =-\frac{1}{4}-\frac{9}{2}$
$\Rightarrow y =\frac{1-18}{4} \text { or } y =\frac{-1}{4}$
$\Rightarrow y =-\frac{17}{4} \text { or } y =-\frac{19}{4}$
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