Math, asked by kucky30, 4 months ago

Find the zeros of quadratic polynomials p(x)= √3x²-10x+7√3 and verify the

relationship between the zeros and the coefficients​

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Answers

Answered by anushaalika
1

Answer:

Given,

q(x) = √3x2 + 10x + 7√3

We put q(x) = 0

⇒ √3x2 + 10x + 7√3 = 0

⇒ √3x2 + 3x + 7x + 7√3x = 0

⇒ √3x(x + √3) + 7 (x + √3) = 0

⇒ (x + √3)(√3x + 7) = 0

This gives us 2 zeros, for

x = -√3 and x = -7/√3

Hence, the zeros of the quadratic equation are -√3 and -7/√3.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

-√3 + (-7/√3) = – (10) /√3

(-3-7)/ √3 = -10/√3

-10/ √3 = -10/√3

Product of roots = constant / coefficient of x2

(-√3) x (-7/√3) = (7√3)/√3

7 = 7

Therefore, the relationship between zeros and their coefficients is verified

Answered by jismonth2006
0

Answer:

Given,

q(x) = √3x2 + 10x + 7√3

We put q(x) = 0

⇒ √3x2 + 10x + 7√3 = 0

⇒ √3x2 + 3x + 7x + 7√3x = 0

⇒ √3x(x + √3) + 7 (x + √3) = 0

⇒ (x + √3)(√3x + 7) = 0

This gives us 2 zeros, for

x = -√3 and x = -7/√3

Hence, the zeros of the quadratic equation are -√3 and -7/√3.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

-√3 + (-7/√3) = – (10) /√3

(-3-7)/ √3 = -10/√3

-10/ √3 = -10/√3

Product of roots = constant / coefficient of x2

(-√3) x (-7/√3) = (7√3)/√3

7 = 7

Step-by-step explanation:

Hence verified

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