Let - - be the roots of the quadratic equation 3 x 2-10 x -2-0- then the quadratic equation whose roots are -5- -5- isA- 27 x 2-2 x -48-0B- 27 x2-48 x-2-0- 46 x 2-27 x -2-0D- 27 x 2-46 x -2-0

The correct option is D 27x^2 +46x +2 = 0Let y = xx+5, where x = α, β xy +5y = x ⇒ x = 5y1 y Putting x = 5y1 y in the quadratic equation nbsp; 3x^2 +10x+2 = ...

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Standard XII

Mathematics

Relation between Roots and Coefficients for Quadratic

Let α, β be t...

Question

Let $α, β$ be the roots of the quadratic equation $3 x^{2} + 10 x + 2 = 0$ , then the quadratic equation whose roots are $\frac{α}{α + 5}, \frac{β}{β + 5}$ , is

46 x^{2} + 27 x + 2 = 0

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27 x^{2} + 2 x + 48 = 0

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27 x^{2} + 48 x + 2 = 0

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27 x^{2} + 46 x + 2 = 0

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Solution

The correct option is D $27 x^{2} + 46 x + 2 = 0$
Let $y = \frac{x}{x + 5}$ , where $x = α, β$
$x y + 5 y = x \Rightarrow x = \frac{5 y}{1 - y}$
Putting $x = \frac{5 y}{1 - y}$ in the quadratic equation
$3 x^{2} + 10 x + 2 = 0 \Rightarrow 3 {(\frac{5 y}{1 - y})}^{2} + 10 (\frac{5 y}{1 - y}) + 2 = 0 \Rightarrow 75 y^{2} + 50 y (1 - y) + 2 (1 - y)^{2} = 0 \Rightarrow 27 y^{2} + 46 y + 2 = 0$

Hence, the quadratic equation whose roots are $\frac{α}{α + 5}, \frac{β}{β + 5}$ is
$27 x^{2} + 46 x + 2 = 0$

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Let α,β be the roots of the quadratic equation 3x2+10x+2=0, then the quadratic equation whose roots are αα+5,ββ+5, is

The correct option is D 27x2+46x+2=0Let y=xx+5, where x=α,β xy+5y=x⇒x=5y1−y Putting x=5y1−y in the quadratic equation 3x2+10x+2=0⇒3(5y1−y)2+10(5y1−y)+2=0⇒75y2+50y(1−y)+2(1−y)2=0⇒27y2+46y+2=0 Hence, the quadratic equation whose roots are αα+5,ββ+5 is 27x2+46x+2=0

Let $α, β$ be the roots of the quadratic equation $3 x^{2} + 10 x + 2 = 0$ , then the quadratic equation whose roots are $\frac{α}{α + 5}, \frac{β}{β + 5}$ , is