Math, asked by aakarshj23, 2 months ago

Find the quadratic polynomial whose zeros are -3 and 4

Answers

Answered by amansharma264
5

EXPLANATION.

Quadratic equation.

Zeroes are = -3 and 4.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = - 3 + 4 = 1.

Product of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = (-3)(4) = -12.

As we know that,

Formula of the quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Put the values in the equation, we get.

⇒ x² - (1)x + (-12) = 0.

⇒ x² - x - 12 = 0.

                                                                                                                         

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answered by piyushnehra2006
36

Answer:

Hey mate your answer is as follows

Quadratic equation.

Quadratic equation Zeroes are = -3 and 4.

We know that

_____________________________

Sum of the zeroes of the quadratic equation.

Sum of the zeroes of the quadratic equation.⇒ α + β = -b/a.

Sum of the zeroes of the quadratic equation.⇒ α + β = -b/a.⇒ α + β = - 3 + 4 = 1.

______________________________

Product of the zeroes of the quadratic equation.

Product of the zeroes of the quadratic equation.⇒ αβ = c/a.

Product of the zeroes of the quadratic equation.⇒ αβ = c/a.⇒ αβ = (-3)(4) = -12.

______________________________

Formula of the quadratic polynomial.

Formula of the quadratic polynomial.⇒ x² - (α + β)x + αβ.

_______________________________

By Putting the values in the equation, we get.

the values in the equation, we get.⇒ x² - (1)x + (-12) = 0.

the values in the equation, we get.⇒ x² - (1)x + (-12) = 0.⇒ x² - x - 12 = 0.

_______________________________         

         hope it helps you.

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