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