Question

the number of points of intersection of the quadratic polynomial X square + 4 x + 4 with the x-axis is​

Answer 1

Answer:

x=--2-2

Step-by-step explanation:

x^2+4x+4=0

x^2+2x+2x+4=0

x(x+2)+2(x+2)=0

x+2=0;x+2=0

x=-2,-2

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Answer 2

Answer:

One point .

At x = -2

Note:

• The possible values of variable for which the polynomial becomes zero are called its zeros.

• No. of zeros of a polynomial = No. of points of intersection on x-axis.

• For a quadratic polynomial ax² + bx + c , the discriminant is given by ; D = b² - 4ac

• If D = 0 , then the zeros are real and equal and hence its curve (graph) will intersect at a single point on x-axis.

• If D > 9 , then the zeros are real and distinct and hence its curve (graph) will intersect at two distinct points on x-axis.

• If D < 0 , then the zeros are imaginary and hence its curve (graph) would not intersect at x-axis.

Solution:

The given quadratic polynomial is :

x² + 4x + 4 .

Clearly,

a = 1

b = 4

c = 4

Thus,

The discriminant will be ;

=> D = b² - 4ac

=> D = 4² - 4•1•4

=> D = 16 - 16

=> D = 0

Since,

The discriminant of the given quadratic polynomial is zero , hence it has real and equal .

Thus,

The curve (graph) of the given quadratic polynomial will touch the x - axis at a single point.

Now,

Let's find the point at with the graph of the given quadratic polynomial intersect at x - axis.

Let the given quadratic polynomial be y .

( Since, the graph is drawn x v/s y )

Thus,

y = x² + 4x + 4

We know that ,

At x-axis , the y-coordinate is zero.

Thus,

=> y = 0

=> x² + 4x + 4 = 0

=> x² + 2•x•2 + 2² = 0

=> (x + 2)² = 0

=> (x + 2)(x + 2) = 0

=> x = - 2 , - 2 .

Clearly,

We got a single value of x for which the polynomial becomes zero.

Hence,

The graph of the given quadratic polynomial will touch the x-axis at a single point , ie ; x = –2 .