Question

from the quadratic equation root. 5 and -4​

Answer 1

EXPLANATION.

Quadratic equation whose roots are = 5 and -4.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ 5 + (-4).

⇒ 5 - 4 = 1.

⇒ α + β = 1.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ (5)(-4) = -20.

As we know that,

Equation of the quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Put the values in the equation, we get.

⇒ x² - (1)x + (-20) = 0.

⇒ x² - x - 20 = 0.

                                                                                                                     

MORE NFORMATION.

Quadratic expression.

A polynomial of degree two of the form ax² + bx + c (a ≠ 0) is called a quadratic expression in x.

The quadratic equation.

ax² + bx + c = 0 (a ≠ 0) has two roots, given by

α = -b + √D/2a.

β = -b - √D/2a.

D = Discriminant  Or b² - 4ac.

Answer 2

Given :-

Zeroes of quadratic equation are 5 and -4

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What to do?

Formation of quadratic equation

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Solution :-

We know that a quadratic equation can be expressed as:

⇒ x²-(sum of zeroes)x+(Product)

So firstly we have to find the value of sum and product of zeroes to substitute in this equation.

Let's assume the given roots be α=5 and β=-4

~Sum

Sum of zeroes=(α+β)

Sum of zeroes=[5+(-4)]

Sum of zeroes=(5-4)

Sum of zeroes=1

~Product

Product of zeroes=(αβ)

Product of zeroes=[(5)×(-4)]

Product of zeroes=[-20]

Product of zeroes=-20

Now substitute these values in formula.

⇒ x²-(sum of zeroes)x+(Product of zeores)

⇒ x²-(1)x+(-20)

⇒ x²-x-20

So the required quadratic equation is x²-x-20.

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Learn More!!

General form of equation:

ax²+bx+c=0

Methods to find zeroes of quadratic equation:

First is middle term splitting where we have to split the middle term or coefficient of x in such a way that product of both the terms is equal to the product of coefficient of x² and constant term and also Algebric sum of splitted terms should be equal to the coefficient of x.

Second method is quadratic formula . We have Formula to find zeroes of any quadratic equation:

$\boxed{\star\;\;\sf X=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}$

$\sf Here\begin{cases}\sf a\:is\: Coefficient\:of\:x^2\\\sf b \:is\: coefficient \:of\; x\\\sf c \:is\: constant\:term\\\sf X\;is\:zeroes\: of\: equation\end{cases}$

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