Question

Solve the equation $35{{y}^{2}}+13y-12=0$ and the find value of $y$.
(a) $y=\dfrac{3}{7}$ and $y=\dfrac{-4}{5}$
(b) $y=\dfrac{1}{7}$ and $y=\dfrac{-2}{5}$
(c) $y=\dfrac{-5}{7}$ and $y=\dfrac{2}{5}$
(d) $y=\dfrac{2}{9}$ and $y=\dfrac{-11}{5}$

Answer 1

Hint: We will derive a formula to solve the quadratic equation by completing the square using a binomial formula. Then use the derived formula in the given equation to find the value of y.

Complete Step-by-Step solution:
Let us consider a general quadratic equation $a{{y}^{2}}+by+c=0\cdots \cdots \left( i \right)$, where $a,b,c$ are constants and $y$ is a variable.
To find the value of $y$ let us try write this equation in the form of ${{\left( m+n \right)}^{2}}={{m}^{2}}+2mn+{{n}^{2}}\cdots \cdots \left( ii \right)$, such that remaining terms after writing this square may not contain variable $y$, that is, it will be constant. So, then we will be able to take the square root of the equation to form a linear equation, which we can solve easily.
Firstly, we divide equation$\left( i \right)$ by $a$ to make first term perfectly square, so we get,
${{y}^{2}}+\dfrac{b}{a}y+\dfrac{c}{a}=0$.
Now we will write this equation in the form of equation $\left( ii \right)$. For this, we will write the term $\dfrac{b}{a}y$ in the form of $2mn$ and then add and subtract ${{n}^{2}}$ term in the equation considering $m$ to be $y$. So, we get,
${{y}^{2}}+2\times y\times \dfrac{b}{2a}+{{\left( \dfrac{b}{2a} \right)}^{2}}-{{\left( \dfrac{b}{2a} \right)}^{2}}+\dfrac{c}{a}=0$
$\Rightarrow {{y}^{2}}+2\times y\times \dfrac{b}{2a}+{{\left( \dfrac{b}{2a} \right)}^{2}}={{\left( \dfrac{b}{2a} \right)}^{2}}-\dfrac{c}{a}$
Applying equation $\left( ii \right)$ here, we get,
$\Rightarrow {{\left( y+\dfrac{b}{2a} \right)}^{2}}={{\left( \dfrac{b}{2a} \right)}^{2}}-\dfrac{c}{a}$
Taking square root on both sides of this equation, we get,
 $\sqrt{{{\left( y+\dfrac{b}{2a} \right)}^{2}}}=\pm \sqrt{{{\left( \dfrac{b}{2a} \right)}^{2}}-\dfrac{c}{a}}$
$\Rightarrow y+\dfrac{b}{2a}=\pm \sqrt{{{\left( \dfrac{b}{2a} \right)}^{2}}-\dfrac{c}{a}}$
Subtracting $\dfrac{b}{2a}$ from both sides of the equation, we get,
$\Rightarrow y=\pm \sqrt{{{\left( \dfrac{b}{2a} \right)}^{2}}-\dfrac{c}{a}}-\dfrac{b}{2a}$
$\Rightarrow y=\pm \sqrt{\dfrac{{{b}^{2}}}{{{\left( 2a \right)}^{2}}}-\dfrac{c}{a}}-\dfrac{b}{2a}$
Taking LCM of the terms inside the root, we get,
$\Rightarrow y=\pm \sqrt{\dfrac{{{b}^{2}}}{{{(2a)}^{2}}}-\dfrac{4ac}{4{{a}^{2}}}}-\dfrac{b}{2a}$
$\Rightarrow y=\pm \sqrt{\dfrac{{{b}^{2}}-4ac}{{{(2a)}^{2}}}}-\dfrac{b}{2a}$
Separating root of denominator and numerator, we get,
$\Rightarrow y=\pm \dfrac{\sqrt{{{b}^{2}}-4ac}}{\sqrt{{{(2a)}^{2}}}}-\dfrac{b}{2a}$
$\Rightarrow y=\dfrac{\pm \sqrt{{{b}^{2}}-4ac}}{2a}-\dfrac{b}{2a}$
Taking LCM, we get,
$\Rightarrow y=\dfrac{\pm \sqrt{{{b}^{2}}-4ac}-b}{2a}$
$\Rightarrow y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\cdots \cdots \left( iii \right)$
Here, the right hand side of the equation is a constant term as $a,b,c$ are constants. Hence, this will be the value of $y$.
Now, we will use equation $\left( iii \right)$ to solve the given equation $35{{y}^{2}}+13y-12=0$, in which,
$a=35,\text{ }b=13\text{ and }c=-12$
Therefore,
$y=\dfrac{-13\pm \sqrt{{{13}^{2}}-4\times 35\times \left( -12 \right)}}{2\times 35}$

$\Rightarrow y=\dfrac{-13\pm \sqrt{169+4\times 35\times 12}}{2\times 35}$
Solving thus further by multiplying terms, we get,
$\Rightarrow y=\dfrac{-13\pm \sqrt{169+1680}}{70}$
$\Rightarrow y=\dfrac{-13\pm \sqrt{1849}}{70}$
Writing the value of root, we get,
$\Rightarrow y=\dfrac{-13\pm 43}{70}$
Separating the + and – terms, we get,
$\Rightarrow y=\dfrac{-13+43}{70}$ or $y=\dfrac{-13-43}{70}$
$\Rightarrow y=\dfrac{30}{70}$ or $y=\dfrac{-56}{70}$
$\Rightarrow y=\dfrac{3}{7}$ or $y=\dfrac{-4}{5}$
These are the required values of y.
Hence, the correct answer is option (a).

Note: In this question, you can take any approach to solving quadrating equations using factorization. Also, remember the formula we derived, then you directly apply that to solve the equation without first deriving it.