Question

3. If one of the zeroes of the
quadratic polynomial (k-1) x2+kx
+1 is -3 then the value of k is *
a) 4/3
b) - 4/3
c) 2/3
d) - 2/3
​

Answer 1

Step-by-step explanation:

$\green{\bold{\underline{ ✪ UPSC-ASPIRANT✪ }}}$

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-solve and verify the equation

$\frac{1}{3} x - 4 = x - ( \frac{1}{2} + \frac{x}{ 3} )$

$\huge\tt\underline\blue{Answer }$

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$⟹</p><p> \frac{1}{3} x - 4 = x - ( \frac{1}{2} + \frac{x}{3} )$

$⟹</p><p> \frac{x}{3} - 4 = x - ( \frac{3 + 2x}{6} )$

$⟹</p><p> \frac{x - 12}{3} = x - ( \frac{2x + 3}{6} )$

$⟹</p><p> \frac{x - 12}{3} = x - \frac{2x - 3}{6}$

$⟹</p><p> \frac{x - 12}{3} = \frac{6x - 2x - 3}{6}$

$⟹</p><p> \frac{x - 12}{3} = \frac{4x - 3}{6}$

cancelling 6( R.H.S) By 3 From L.H.S

$⟹ \frac{x - 12}{1} = \frac{4x - 3}{2} </p><p>$

$⟹</p><p>2(x - 12) = 4x - 3$

$⟹</p><p>2x - 24 = 4x - 3$

$⟹</p><p> - 24 + 3 = 4x - 2x$

$⟹</p><p> - 21 = 2x$

$⟹</p><p>x = - \frac{21}{2}$

CHECK:-

$⟹ \frac{ - \frac{21}{2} }{3} - 4 = - \frac{21}{2} - ( \frac{1}{2} + ( - ) \frac{ \frac{21}{2} }{3} )</p><p>$

$⟹</p><p> - \frac{21}{6} - 4 = - \frac{21}{2} - ( \frac{1}{2} - \frac{21}{6} )$

$⟹</p><p> - \frac{7}{2} - 4 = - \frac{21}{2} - ( \frac{1}{2} - \frac{7}{2} )$

$⟹</p><p> \frac{ - 7 - 8}{2} = - \frac{21}{2} - ( - \frac{6}{2} )$

$⟹ - \frac{15}{2} = - \frac{21}{2} - ( - 3) </p><p>$

$⟹</p><p> - \frac{15}{2} = - \frac{21}{2} + 3$

$⟹</p><p> - \frac{15}{2} = \frac{ - 21 + 6}{2} = - \frac{15}{2}$

THEREFORE,L.H.S=R.H.S

VERIFIED✔️

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HOPE IT HELPS YOU..

_____________________

Thankyou:)

Answer 2

Answer:

(a) k= 4/3

Step-by-step explanation:

(k-1)x^2+kx+1 = 0

(k-1)(-3)^2+k(-3)+1 =0

(k-1)9-3k+1 = 0

9k-9-3k+1 = 0

6k-8 = 0

6k = 8

k = 8/6

k = 4/3