Question

2x²-5x+7 please solve step by step by splitting the middle term​

Answer 1

Answer:

$x = -1,\frac{7}{2}$  are two values of $x$.

Step-by-step explanation:

• STEP 1:

    Equate the equation to zero.

$2x^{2} -5x-7 = 0$

• STEP 2:

Factorise by splitting the middle term:

•   Factorising $(2x^{2} -5x)-7$

The first term is  $2x^{2}$  its coefficient is $2.$

The middle term is  $-5x$  its coefficient is $-5.$

The last term, "the constant", is  $-7 .$

• Now multiply the coefficient of the first term by the constant      

                                    $( 2) *(-7 )= -14$

• Find two factors of $-14$  whose sum equals the coefficient of the middle term, which is $-5.$

                                 $-14 + 1 = -13 \\ -7 + 2 = -5$  

• STEP 3 :

Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,$-7$ and $2$.

                           $= 2x^{2}-7x+2x-7$

• STEP 4 :

                          $= 2x^{2}+2x -7x -7$

Now taking $2x$ common from ist two terms :

                    $=2x(x+1)-7(x+1)\\=(2x-7)(x+1)$

$(2x-7)(x-1)$ are two factors we get after splitting the middle term of

$2x^{2} -5x-7.$

• STEP 5 :

Now equating these factors to zero,

$(2x^{2} -5x-7 ) = 0\\(2x-7)(x+1)=0$

• Equating individual factors to zero,

$2x-7 = 0\\2x= 7\\x=\frac{7}{2}$

• Now equating,

                      $x+1 = 0\\x=-1$

we get value of   $x = -1,\frac{7}{2}$

#SPJ3

Answer 2

Answer:

$x=\frac{5}{4}+\frac{\sqrt{31}i}{4}$    and    $x=\frac{5}{4}-\frac{\sqrt{31}i}{4}$  are the two values of x for the given quadratic equation $2x^2-5x+7=0$

Step-by-step explanation:

Given  :

We are given the quadratic Equation

                             $2x^2+5x+7=0$

To Find :

Zeros of the the given quadratic equation by splitting middle term

Solution :

Since we are given the standard form of the quadratic equation

                              $2x^2-5x+7=0$

                              $x^2-\frac{5}{2}x+\frac{7}{2}=0$

Since -5/2 x can be written as ($-\frac{5}{2}x=-(\frac{5}{4}+\frac{\sqrt{31}i}{4}+\frac{5}{4}-\frac{\sqrt{31}i}{4})x$ )

                   $(x^2-(\frac{5}{4}+\frac{\sqrt{31}i}{4}))(x-\frac{5}{4}+\frac{\sqrt{31}i}{4})=0$

Since multiplication of two terms is zero so either first term is 0 or second term is 0

Hence we get

 $(x-\frac{5}{4}+\frac{\sqrt{31}i}{4})=0$            and         $(x-(\frac{5}{4}+\frac{\sqrt{31}i}{4})) = 0$

solving above equations for x we get the following solutions

   $x=\frac{5}{4}-\frac{\sqrt{31}i}{4}$           and       $x=\frac{5}{4}+\frac{\sqrt{31}i}{4}$

Hence we can say that the zeros of the given quadratic equation are complex numbers.

#SPJ2