Let - and - are the roots of the equation ax2 -bx-c - 0- If 1- and 1- are the roots of px2 -qx-r - 0 and q p - k- b c- then the value of k is

Let f(x) = ax^2 +bx+c = 0 has roots α, β y =1 xx, where x =α, β y = 1 x 1 ⇒ y +1 = 1 x ⇒ x nbsp; = 1y+1 The equation whose nbsp;1 αα and 1 ββ is, f ( ...

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Byju's Answer

Standard XIII

Mathematics

Relation between Roots and Coefficients for Quadratic

Let α and β a...

Question

Let $α$ and $β$ are the roots of the equation $a x^{2} + b x + c = 0.$ If $\frac{1 - α}{α}$ and $\frac{1 - β}{β}$ are the roots of $p x^{2} + q x + r = 0$ and $\frac{q}{p} = k + \frac{b}{c}$ , then the value of $k$ is

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Solution

Let $f (x) = a x^{2} + b x + c = 0$ has roots $α, β$
$y = \frac{1 - x}{x}$ , where $x = α, β$
$y = \frac{1}{x} - 1 \Rightarrow y + 1 = \frac{1}{x} \Rightarrow x = \frac{1}{y + 1}$

The equation whose $\frac{1 - α}{α}$ and $\frac{1 - β}{β}$ is,
$f (\frac{1}{x + 1}) = 0 \Rightarrow a + b (x + 1) + c (x + 1)^{2} = 0 \Rightarrow c x^{2} + (2 c + b) x + (a + b + c) = 0$
Comparing with
$p x^{2} + q x + r = 0$
$\frac{c}{p} = \frac{(2 c + b)}{q} = \frac{(a + b + c)}{r} = k$
Now,
$\frac{c}{p} = \frac{(2 c + b)}{q}$
$\Rightarrow \frac{q}{p} = \frac{(2 c + b)}{c} \Rightarrow \frac{q}{p} = 2 + \frac{b}{c} ∴ k = 2$

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Let α and β are the roots of the equation ax2+bx+c=0. If 1−αα and 1−ββ are the roots of px2+qx+r=0 and qp=k+bc, then the value of k is

Let f(x)=ax2+bx+c=0 has roots α,β y=1−xx, where x=α,β y=1x−1⇒y+1=1x⇒x=1y+1 The equation whose 1−αα and 1−ββ is, f(1x+1)=0⇒a+b(x+1)+c(x+1)2=0⇒cx2+(2c+b)x+(a+b+c)=0 Comparing with px2+qx+r=0 cp=(2c+b)q=(a+b+c)r=k Now, cp=(2c+b)q ⇒qp=(2c+b)c⇒qp=2+bc∴k=2

Let $α$ and $β$ are the roots of the equation $a x^{2} + b x + c = 0.$ If $\frac{1 - α}{α}$ and $\frac{1 - β}{β}$ are the roots of $p x^{2} + q x + r = 0$ and $\frac{q}{p} = k + \frac{b}{c}$ , then the value of $k$ is