Math, asked by arpita7920, 11 months ago

+C =0 with rational coefficients is p + 19, then
13. If one root of the quadratic equation ax + bx + c = 0 with rational coefficient p+√q
other root is​

Answers

Answered by abhi178
2

The other root of given quadratic equation is (p - √q)

one root of the quadratic equation,ax² + bx + c = 0 with rational coefficient is p + √q.

as we know, p + √q is an irrational number.

so, other root must be an irrational number because we know sum and product of two conjugate irrational numbers is a rational number.

as (p + √q) + other root = -b/a = rational number

and (p + √q) other = c/a = rational number

so it is clear that the other root is conjugate of (p + √q) .i.e., (p - √q)

also read similar questions : prove that the roots of equations ax²+bx+c=0 and cx²+bx+a=0 are reciprocals of each other

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Answered by stefangonzalez246
2

The other root is (p+√q) and (p-√q).

Given

To find the other root.

If one root of the quadratic equation ax² + bx + c = 0 with rational coefficient p + √q.

Quadratic equation is ax²+bx+c=0.

Where, a, b, c are rational numbers.    

It has two roots α and β,

α+β = - b/a

αβ = c/a

They are rational numbers.

Here α = p+√q and β = p-√q. (By conjugate method)

Where, α = p+√q and β = p-√q are irrational numbers.

(α+β) = [(p+√q)+(p-√q)]

        = [p+√q+p-√q]

        = 2p

(α+β) = 2p

(αβ) = [(p+√q)(p-√q)]

       = [p²-p√q+p√q-(√q×√q)]

       = [p²-(√q)²]

       = p²-q

αβ = p²-q

Therefore, the other root is (p+√q) and (p-√q).

To learn more...

1. brainly.in/question/1347322

2. brainly.in/question/11828765

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