Math, asked by mdnasar041976, 3 months ago

Factorize the following

x2+4x-7​

Answers

Answered by anusitharatr
4

Step-by-step explanation:

x²+4x-7

Using the quadratic formula to solve,

b²-4ac=4²-4*1*-7

=16+28=44 = (2*√11)²

x=(-b±√b²-4ac)/2a

=(-4±2√11)2*1

=(-4-2√11)2 , (-4+2√11)/2

=2(-2-√11)/2 , 2(-2+√11)/2

=-2-√11 ,-2+√11

Hope it helps:)

Answered by Anonymous
24

 \huge \colorbox{red}{given}

 \huge \tt x {}^{2} + 4x - 7

 \huge \tt \colorbox{gold}{solution}

 \huge \tt \colorbox{r}{step - 1}

==》Trying to factor by splitting the middle term

Factoring x2+4x-7

▪︎The first term is, x2 its coefficient is 1 .

▪︎The middle term is, +4x its coefficient is 4 .

▪︎The last term, "the constant", is -7

Step-1 : Multiply the coefficient of the first term by the constant 1 • -7 = -7

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

which is 4 .

 \huge \tt \: 	-7	   +   	1	   =   	-6	 \\  \\  \huge 	-1	   +   	7	   =   	6

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

 \tt \fcolorbox{green}{r}{Equation \:  at  \: the  \: end  \: of step  - 1}

 \bf x2 + 4x - 7  = 0

 \huge \tt \colorbox{r}{step - \: 2}

Parabola, Finding the Vertex:

==》 Find the Vertex of y=x2+4x-7

▪︎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 -2.0000

▪︎ Plugging into the parabola formula -2.0000 for x we can calculate the y -coordinate :

==》 y = 1.0 × -2.00 ×-2.00 + 4.0× -2.00 - 7.0

==》 y = -11.000

▪︎Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2+4x-7

Axis of Symmetry (dashed) {x}={-2.00}

Vertex at {x,y} = {-2.00,-11.00}

x -Intercepts (Roots) :

Root 1 at {x,y} = {-5.32, 0.00}

Root 2 at {x,y} = { 1.32, 0.00}

Solve Quadratic Equation by Completing The Square

==》 Solving x2+4x-7 = 0 by Completing The Square .

==》 Add 7 to both side of the equation : x2+4x = 7

==》Now the clever bit: Take the coefficient of x , which is 4 , divide by two, giving 2 , and finally square it giving 4

Add 4 to both sides of the equation :

On the right hand side we have :

==》7 + 4 or, (7/1)+(4/1)

==》The common denominator of the two fractions is 1

==》 Adding (7/1)+(4/1) gives 11/1

So adding to both sides we finally get : x2+4x+4 = 11

Adding 4 has completed the left hand side into a perfect square :

 \tt \: x {}^{2}  + 4x + 4 \\  \\  =  \bf \: (x + 2)(x + 2) \\  \\  =  \tt \:  {(x + 2)}^{2}

Things which are equal to the same thing are also equal to one another. Since

 \tt \: x {}^{2}  + 4x + 4 = 11 \\  \\  \bf \: and \\  \bf \: x {}^{2}  + 4x + 4 = (x + 2) {}^{2}

then, according to the law of transitivity,

 \tt =》   (x+2)² = 11

We'll refer to this Equation as Eq. #2.2.1

==》The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(x + 2) {}^{2}  \\  \\  =  {(x + 2)}^{ \frac{2}{2} }  \\  \\  =  {(x + 2)}^{1}  \\  \\  =  \tt \: x + 2

Now, applying the Square Root Principle to Eq. #2.2.1 we get:

 \huge \tt \: x+2 = √ 11

Subtract 2 from both sides to obtain:

 \huge \tt \:   x = -2 + √ 11

Since a square root has two values, one positive and the other negative

  \huge\bf \: x2 + 4x - 7 = 0

  \tt\fcolorbox{red}{gr}{has \: two \: solutions}

  \huge\tt \: x = -2 + √ 11

 \huge \fcolorbox{green}{red}{or}

 \huge \tt \: x = -2 - √ 11

 \huge \tt \fcolorbox{red}{u}{be \: brainly}

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