Find the quadractic equation whose roots are twice the
roots of 2x - 5x+2=0
Answers
Answer:
Let \alpha \: \: and \: \: \betaαandβ are the roots of the equation 2x²-5x +2=02x²−5x+2=0
So
\alpha + \beta = \frac{5}{2}α+β=
2
5
\alpha \beta \: = \frac{2}{2} = 1αβ=
2
2
=1
Now we have to form the quadratic equation whose roots are twice the roots of
2x²-5x +2=02x²−5x+2=0
So 2\alpha \: \: and \: \: 2\beta2αand2β are the roots of the required equation
Hence the required Quadratic Equation is
{x}^{2} - ( \: sum \: of \: the \: roots \: )x + (product \: of \: the \: roots \: ) = 0x
2
−(sumoftheroots)x+(productoftheroots)=0
\implies \: {x}^{2} - (2\alpha + 2 \beta)x + (2\alpha \times 2 \beta) = 0⟹x
2
−(2α+2β)x+(2α×2β)=0
\implies \: {x}^{2} - 2(\alpha + \beta)x + (4\alpha \beta) = 0⟹x
2
−2(α+β)x+(4αβ)=0
\implies \: {x}^{2} - 2 \times ( \frac{5}{2} )x + (4 \times 1) = 0⟹x
2
−2×(
2
5
)x+(4×1)=0
\implies \: {x}^{2} - 5x + 4 = 0⟹x
2
−5x+4=0
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