Question

The quadratic equation whose roots are twice the roots of 2x²-5x +2=0 is

Answer 1

Answer:

2x2 -5x +2

2x2 -4x -x +2

2x(x-2) -1(x-2)

(2x -1)( x -2)

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Answer 2

$\huge\boxed{\underline{\underline{\green{\tt Solution}}}}$

Let $\alpha \: \: and \: \: \beta$are the roots of the equation $2x²-5x +2=0$

So

$\alpha + \beta = \frac{5}{2}$

$\alpha \beta \: = \frac{2}{2} = 1$

Now we have to form the quadratic equation whose roots are twice the roots of

$2x²-5x +2=0$

So $2\alpha \: \: and \: \: 2\beta$ are the roots of the required equation

Hence the required Quadratic Equation is

${x}^{2} - ( \: sum \: of \: the \: roots \: )x + (product \: of \: the \: roots \: ) = 0$

$\implies \: {x}^{2} - (2\alpha + 2 \beta)x + (2\alpha \times 2 \beta) = 0$

$\implies \: {x}^{2} - 2(\alpha + \beta)x + (4\alpha \beta) = 0$

$\implies \: {x}^{2} - 2 \times ( \frac{5}{2} )x + (4 \times 1) = 0$

$\implies \: {x}^{2} - 5x + 4 = 0$

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