Question

for the previous problem some of the work was done for you remember when solving a quadratic equation you must start with all terms on one side and put it in standard form for the following equations first put the equation in standard form and them use quadratic formula to solve 9x^2 -25 = -12-10x
4n^2-1=-12n +3n^2+11

Answer 1

Solving the given equations we get,

• Solutions of quadratic equation $9x^2-25=-12-10x$ are x=0.77 and -1.88
• Solutions of quadratic equation $4n^2-1=12n+3n^2+11$ are n=12.93 and -0.93

If the quadratic equation is $ax^2+bx+c=0$ then the solutions of this are given by,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here in the problem given that,

The first quadratic equation is $9x^2-25=-12-10x$

Converting it into standard form of quadratic equation we get,

$9x^2-25=-12-10x\\9x^2-25+12+10x=0\\9x^2+10x-13=0$

So here a=9,b=10,c=-13

The solutions of the equation is given by,

$x=\frac{-10\pm\sqrt{10^2-4\times9\times(-13)}}{2\times9}=\frac{-10\pm\sqrt{100+468}}{18}\approx\frac{-10\pm23.83}{18}$

So either,

$x=\frac{-10+23.83}{18}=\frac{13.83}{18}\approx0.77$

Or,

$x=\frac{-10-23.83}{18}\approx-1.88$

So the solutions are given by x=0.77 and -1.88

Second quadratic equation is $4n^2-1=12n+3n^2+11$

Converting the equation in its standard form we get,

$4n^2-1=12n+3n^2+11\\4n^2-1-12n-3n^2-11=0\\n^2-12n-12=0$

so here a=1,b=-12,c=-12

Solutions are given by,

$n=\frac{-(-12)\pm\sqrt{(-12)^2-4\times1\times(-12)}}{2\times1}=\frac{12\pm\sqrt{144+48}}{2}\approx\frac{12\pm13.86}{2}$

So either

$n=\frac{12+13.86}{2}=12.93$

Or,

$n=\frac{12-13.86}{2}=-0.93$

Solutions are given by n = 12.93 and -0.93

Hence the solutions are x=0.77 and -1.88 for first equation and n=12.93 and -0.93 for second equation.

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