Question

X^2-5x-24=0 solve by complete square

Answer 1

Popular Problems Algebra Solve by Completing the Square x^2+5x-24=0

x

2

+

5

x

−

24

=

0

Add

24

to both sides of the equation.

x

2

+

5

x

=

24

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of

b

.

(

b

2

)

2

=

(

5

2

)

2

Add the term to each side of the equation.

x

2

+

5

x

+

(

5

2

)

2

=

24

+

(

5

2

)

2

Answer 2

Answer: x=-3,8

Steps to complete the square method

Assume that ax2 + bx + c = 0 is the given quadratic equation. Then follow the steps given to solve it by completing the square method.

Step 1: Write the equation in the form so that c is on the right-hand side.

Step 2: If a is not equal to 1, divide the entire equation by a so that the coefficient of x2 is 1.

Step 3: Now add the square of half the coefficient of the x-term, (b/2a)2, to both sides.

Step 4: Factor the left side of the equation into the square of the binomial term.

Step 5: Take the square root of both sides

Step 6: Solve for the variable x and find the roots

Now the given equation is $x^{2}-5x-24=0$

Place the constant part on the right-hand side of the equation, now the equation becomes $x^{2}-5x=24$

Here coefficient of b is -5

So according to the method add and subtract half of the square of the coefficient of b that is $(-5 /2)^{2}$

Hence the equation becomes  $x^{2}-5x+(-5/2)^{2}-(-5/2)^{2} =24$

Put the negative part of the squaring section on RHS so that we can make LHS a perfect square

now the equation becomes  $x^{2}-5x+(-5/2)^{2}=(-5/2)^{2}+24$$x^{2}-5x+25/4=25/4+24$

On solving it becomes $(x-5/2)^{2}=121/4$

Removing square from LHS and shifting under root on RHS the equation becomes,

(x-5/2)=±11/2

taking one time as positive and one time as negative we get two values of x which is -3 and 8 respectively.

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