Question

Find the value of $y$, if $x^2-7y=8,x-y=1$, and $y>0$

• A. $1.64$
• B. $4.78$
• C. $5.64$
• D. $6.14$
• E. $7.28$

Answer 1

The correct option is D $6.14$

Given $x-y=1$ and $y>0$

$x^2-7y=8\Rightarrow x^2-7y-8=0$

Put the value $x=y+1$, we get

$(1+y{)}^2-7y-8=0$

$\Rightarrow 1+2y+y^2-7y-8=0$

$\Rightarrow y^2-5y-8=0$

$y=\frac{-(-5)\pm \sqrt{(-5{)}^2+4\times 1\times (-7)}}{2\times 1}=\frac{25\pm \sqrt{25+28}}{2}=\frac{25\pm \sqrt{43}}{2}=\frac{25\pm 7.28}{2}$

Then $y=$ $\frac{5+7.28}{2}=\frac{12.28}{2}=6.14$

and $y=$ $\frac{5-7.28}{2}=\frac{-2.28}{2}=-1.14$

But given $y>0$

Then $y=6.14$