Question

Solve x2 + 10x = 24 by completing the square. Which is the solution set of the equation?

Answer 1

Answer:

The solution set of the equation are (2,0) and (-12,0)

Step-by-step explanation:

Given : Equation $x^2+10x=24$

To find : Solve equation by completing the square ?

Solution :

The general form of quadratic equation is $ax^2+bx+c=0$

So, $x^2+10x-24=0$

To complete the square we add and subtract $(\frac{b}{2})^2$

i.e. $(\frac{b}{2})^2=(\frac{10}{2})^2=5^2$

Applying in equation,

$x^2+10x-24+5^2-5^2=0$

Re-writ equation as,

$x^2+2\times 5\times x+5^2-24-25=0$

Apply $a^2+2ab+b^2=(a+b)^2$

$(x+5)^2-49=0$

$(x+5)^2=49$

Taking root both side,

$x+5=\pm \sqrt{49}$

$x+5=\pm 7$

$x+5=7$ or $x+5=-7$

$x=2$ or $x=-12$

Therefore, the solution set of the equation are (2,0) and (-12,0)

Answer 2

$\large\begin{array}{l} \textsf{So the question gives us this equation:}\\\\ \mathsf{x^2+10x=24}\\\\ \mathsf{x^2+2\cdot 5x=24}\\\\ \textsf{In order to complete the square on the left side, just add }\mathsf{5^2}\\\textsf{to both sides of the equation above:}\\\\ \mathsf{x^2+2\cdot 5x+5^2=24+5^2}\\\\ \mathsf{x^2+5x+5x+5^2=24+25}\\\\ \mathsf{x(x+5)+5(x+5)=49} \end{array}$

$\large\begin{array}{l} \mathsf{(x+5)(x+5)=49}\\\\ \mathsf{(x+5)^2=49}\\\\\\ \textsf{So,}\\\\ \mathsf{x+5=\pm\,\sqrt{49}}\\\\ \mathsf{x+5=\pm\,7}\\\\ \mathsf{x=\pm\,7-5} \end{array}$

$\large\begin{array}{l} \begin{array}{rcl} \mathsf{x=-7-5}&~\textsf{ or }~&\mathsf{x=7-5}\\\\ \mathsf{x=-12}&~\textsf{ or }~&\mathsf{x=2} \end{array}\\\\\\ \textsf{Solution set: }\mathsf{S=\{-12,\,2\}.}\qquad\checkmark \end{array}$


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Tags: solve quadratic equation completing square solution algebra