Question

Solve the following equations by using the general formula, if the equation has a solution in R :
$\left(1\right)x^2+6x+5=0$
$\left(2\right)x^2-3x-4=0$
$\left(3\right)5x^2+8x+3=0$
$\left(4\right)5x^2-3x-2=0$
$\left(5\right)3x^2+5\sqrt{2}x-2=0$
$\left(6\right)x^2+5x+3=0$
$\left(7\right)9x^2+6x+4=0$
$\left(8\right)48m^2-13m-1=0$

Answer 1

$\left(1\right)$ $x^2+6x+5=0$

$\Rightarrow x=\frac{-6\pm \sqrt{36-20}}{2\left(1\right)}$

$\Rightarrow x=\frac{-6\pm 4}{2}$

$\Rightarrow x=-3\pm 2$

$\Rightarrow x=-1,-5$

$\left(2\right)x^2-3x-4=0$

$x=\frac{3\pm \sqrt{9+16}}{2}$

$x=4,-1$

$\left(3\right)$ $5x^2+8x+3=0$

$x=\frac{-8\pm \sqrt{64-60}}{2\left(5\right)}$

$x=\frac{3}{5},-1$

$\left(4\right)$ $5x^2-3x-2=0$

$x=\frac{3\pm \sqrt{9+40}}{2\left(5\right)}$

$x=1,\frac{-2}{5}$

$\left(5\right)3x^2+5\sqrt{2}x-2=0$

$x=\frac{-5\sqrt{2}\pm \sqrt{50+24}}{2\left(3\right)}$

$x=\frac{-5\sqrt{2}+\sqrt{7}4}{6},\frac{-5\sqrt{2}-\sqrt{7}4}{6}$

$\left(6\right)x^2+5x+3=0$

$x=\frac{-5\pm \sqrt{25-12}}{2}$

$x=\frac{-5+\sqrt{13}}{2},\frac{-5-\sqrt{13}}{2}$

$\left(7\right)9x^2+6x+4=0$

$D\equiv 36-4\left(4\right)\left(9\right)<0$

Since discriminant is negative, roots of the given quadratic equation are not real.

8)$48m^2-13m-1=0$

$x=\frac{13\pm \sqrt{169+192}}{2\left(48\right)}$

$x=\frac{1}{3},\frac{-1}{16}$