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Zorluk: Çok zorQuadratic Equations

The quadratic equation x24x+2=0x^2 - 4x + 2 = 0 has roots r1r_1 and r2r_2. A second quadratic equation, x2px+q=0x^2 - px + q = 0, has roots (r1+1r2)\left(r_1 + \frac{1}{r_2}\right) and (r2+1r1)\left(r_2 + \frac{1}{r_1}\right), where pp and qq are constants. What is the value of pp?

Cevap: 6

Cevap

The value of pp is 6.
By Vieta's formulas, the sum of the roots of x24x+2=0x^2 - 4x + 2 = 0 is r1+r2=4r_1 + r_2 = 4 and the product of the roots is r1r2=2r_1 r_2 = 2. The sum of the roots of the second equation x2px+q=0x^2 - px + q = 0 is pp. Therefore, p=(r1+1r2)+(r2+1r1)=(r1+r2)+r1+r2r1r2p = \left(r_1 + \frac{1}{r_2}\right) + \left(r_2 + \frac{1}{r_1}\right) = (r_1 + r_2) + \frac{r_1 + r_2}{r_1 r_2}. Substituting the known values yields p=4+42=6p = 4 + \frac{4}{2} = 6.

Adım Adım Çözüm

1
Determine the sum and product of the roots of the first equation.
r1+r2=4r_1 + r_2 = 4 and r1r2=2r_1 r_2 = 2
Vieta's formulas state that for x2Bx+C=0x^2 - Bx + C = 0, the sum of the roots is BB and the product is CC.
2
Express the sum of the roots of the second equation, which is pp.
p=(r1+r2)+(1r1+1r2)p = (r_1 + r_2) + \left(\frac{1}{r_1} + \frac{1}{r_2}\right)
The sum of the roots of x2px+q=0x^2 - px + q = 0 is pp.
3
Simplify the fractional part of the equation.
1r1+1r2=r1+r2r1r2\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}
Finding a common denominator allows us to write the sum of reciprocals in terms of the sum and product of the roots.
4
Substitute the values of the sum and product into the expression for pp.
p=4+42=6p = 4 + \frac{4}{2} = 6
This evaluates the expression to find the final value of pp.

Anahtar Kavram

Relating the roots and coefficients of quadratic equations using Vieta's formulas.
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