Question

Solve: Factorisation of x²+7x+5=0​

Answer 1

Answer:

x =(-7-√29)/2=-6.193

x =(-7+√29)/2=-0.807

Step-by-step explanation:

The first term is,  x2  its coefficient is  1 .

The middle term is,  +7x  its coefficient is  7 .

The last term, "the constant", is  +5  

Step-1 : Multiply the coefficient of the first term by the constant   1 • 5 = 5  

Step-2 : Find two factors of  5  whose sum equals the coefficient of the middle term, which is   7 .

     -5    +    -1    =    -6  

     -1    +    -5    =    -6  

     1    +    5    =    6  

     5    +    1    =    6  

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step  1  :

 x2 + 7x + 5  = 0  

Step  2  :

Parabola, Finding the Vertex :

2.1      Find the Vertex of   y = x2+7x+5

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero).  

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.  

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.  

For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -3.5000  

Plugging into the parabola formula  -3.5000  for  x  we can calculate the  y -coordinate :  

 y = 1.0 * -3.50 * -3.50 + 7.0 * -3.50 + 5.0

or   y = -7.250

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Answer 2

Answer:

X^2+7X+5=0

=>FINDING 2 SOLUTION OF THIS EQUATIONS, we have to apply this b^2-4ac,

HERE a=1 ,b=7 ,c=5

=>(7)^2-(4*1*5)

=>49-20=29

=>therefore,B^2-4AC >0

THEREFORE,ROOTS OF THIS EQUATION EQUALS TO

$\alpha$=  -b+$\sqrt{b^2-4ac}$/2a

=>   -7+$\sqrt(29\\}$)/2.........................(i)

$\beta$=  -b-$\sqrt{b^2-4ac}$/2a

=>     -7-$\sqrt{29}$/2..........................(ii)

HENCE, (i) and (ii) are required roots of this equation