How to solve x2−10x=−24?

Question

remairaskf · Accepted Answer

Given,(displaystyle{color{white}{times{x}}}{x}^{{2}}-{10}{x}=-{24})(displaystyleRightarrow{x}^{{2}}-{10}{x}+{24}=cancel{{-{24}}}+cancel{{24}}) [Add 24 to both sides](displaystyleRightarrow{x}^{{2}}-{10}{x}+{24}={0})(displaystyleRightarrow{x}^{{2}}-{left({6}+{4}right)}{x}+{24}={0}) [Well, you can write 10=6+4.](displaystyleRightarrow{x}^{{2}}-{6}{x}-{4}{x}+{24}={0}) [Use Distributive Property to destroy it.](displaystyleRightarrow{x}{left({x}-{6}right)}-{4}{left({x}-{6}right)}={0}) [Group the like terms](displaystyleRightarrow{left({x}-{6}right)}{left({x}-{4}right)}={0}) [Group again]Now, Either (displaystyle{x}-{6}={0}Rightarrow{x}={6})Or, (displaystyle{x}-{4}={0}Rightarrow{x}={4})Then we have two options (displaystyle{x}={4},{6}).We need use Quadratic Formula too.According to the Quadratic Formula,If there is a Quadratic Equation in the form of (displaystyle{a}{x}^{{2}}+{b}{x}+{c}={0}),The roots of the equation are (displaystyle{x}=frac{{-{b}pmsqrt{{{D}}}}}{{{2}{a}}}), where (displaystyle{D}={b}^{{2}}-{4}{a}{c}), which is called the Discriminant.The Equation in General Form is now presented: ((displaystyle{a}{x}^{{2}}+{b}{x}+{c}={0})) (displaystyle{x}^{{2}}-{10}{x}+{24}={0}).Therefore, (displaystyle{D}={left(-{10}right)}^{{2}}-{4}cdot{24}cdot{1}={100}-{96}={4}{&amp;gt;}{0})As (displaystyle{D}{&amp;gt;}{0}), we will have two real and distinct roots for the equation.Now, (displaystyle{x}=frac{{-{b}pmsqrt{{{D}}}}}{{{2}{a}}}=frac{{-{left(-{10}right)}pmsqrt{{{4}}}}}{{{2}cdot{1}}}=frac{{{10}+{2}}}{{2}},frac{{{10}-{2}}}{{2}}=frac{{12}}{{2}},frac{{8}}{{2}}={6},{4})So we get the same solution, x = 4,6.

Sage Farley · Accepted Answer

So solution:rearrange in  standard form  ;ax2+bx+c=0add 24 to both sidesx2−10x+24=0the factors of + 24 which sum to - 10 are - 6 and - 4(x−6)(x−4)=0equate each factor to zero and solve for xx−4=0⇒x=4x−6=0⇒x=6

How to solve x^2-10x=-24?

Answered question

Answer & Explanation

New Questions in Algebra I