Question

what is the nature of zeros of quadratic polynomial x2+88x+125

Answer 1

Given:-

→p(x)=0

→x²+88x+125=0

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To Find:-

→The nature of Zeros of The P(x)?

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AnsWer:-

★Using Discriminant Formula★

→x²+88x+125

๛a=1,b=88,c=125

→d=b²-4ac

•Putting The Values•

→d=(88)²-4×1×125

→d=7744-500

→d=7244

♦ Here The Nature of Roots, Is Real and Un-equal, As d>0 ♦

[•°•d>0,Nature of Roots are Real and Unequal]

[•°•d=0,Nature of Roots are Real and Equal]

[•°•d<0,Nature of Roots are not real and un-equal]

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Answer 2

Answer:

Real and distinct

Note:

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

• 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 + 88x + 125 .

Clearly ,

a = 1

b = 88

c = 125

Thus,

The discriminant of the given quadratic polynomial will be given as ;

=> D = b² - 4ac

=> D = 88² - 4•1•125

=> D = 7744 - 500

=> D = 7244 (D > 0)

Clearly,

The discriminant of the given quadratic polynomial is greater than zero .

Hence,

The given quadratic polynomial would have real and distinct zeros .