Question

Find the solutions of each quadratic equation factoring.

Answer 1

Step-by-step explanation:

$10x^2=6x\qquad\text{subtract}\ 6x\ \text{from both isdeS}\\\\10x^2-6x=0\qquad\text{distribute}\\\\2x(5x-3)=0\iff2x=0\ \vee\ 5x-3=0\\\\2x=0\qquad\text{divide both sides by 2}\\\boxed{x=0}\\\\5x-3=0\qquad\text{add 3 to both sides}\\5x=3\qquad\text{divide both sides by 5}\\\boxed{x=\dfrac{3}{5}}\\\\\large\boxed{\bold{ANSWER}:\ \boxed{x=0\ or\ x=\dfrac{3}{5}}}$

$16x^2=81\qquad\text{subtract 81 from both sides}\\16x^2-81=0\\4^2x^2-9^2=0\\(4x)^2-9^2=0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\(4x-9)(4x+9)=0\iff4x-9=0\ \vee\ 4x+9=0\\4x-9=0\qquad\text{add 9 to both sides}\\4x=9\qquad\text{divide both sides by 4}\\\boxed{x=\dfrac{9}{4}}\\4x+9=0\qquad\text{subtract 9 from both sides}\\4x=-9\qquad\text{divide both sides by 4}\\\boxed{x=-\dfrac{9}{4}}\\\large\boxed{\bold{ANSWER:}\ \boxed{x=-\dfrac{9}{4}\ or\ x=\dfrac{9}{4}}}$

$5x^2+11x+2=0\\5x^2+10x+x+2=0\qquad\text{distribute}\\5x(x+2)+1(x+2)=0\\(x+2)(5x+1)=0\iff x+2=0\ \vee\ 5x+1=0\\\\x+2=0\qquad\text{subtract 2 from both sides}\\\boxed{x=-2}\\\\5x+1=0\qquad\text{subtract 1 from both sides}\\5x=-1\qquad\text{divide both sides by 5}\\\boxed{x=-\dfrac{1}{5}}\\\large\boxed{\bold{ANSWER:}\ \boxed{\ x=-2\ or\ x=-\dfrac{1}{5}}}$