Question-7360

Question Number 7360 by rohit meena last updated on 24/Aug/16

Answered by Rasheed Soomro last updated on 25/Aug/16

(3) (4−k)x^2 +2(k+2)x+8k+1 The value of k,for which the above expression be perfect square, is required. −−−−−−−−−−−−−−−−−−−−−−−− A quadratic expression will be perfect square if it is of the following form: a^2 ±2ab+b^2 If first term, middle term and last term are f, m and l respectively then m=±2(√f) (√l) Dividing by 4−k to the given expression x^2 +((2(k+2))/(4−k))x+((8k+1)/(4−k)) [To make the first term perfect square] x^2 +((2(k+2))/(4−k))x+((√((8k+1)/(4−k))))^2 [Writing last term in square form] Now , m=±2(√f) (√l) ((2(k+2))/(4−k))x=±2(x)((√((8k+1)/(4−k)))) (((k+2))/(4−k))=±(√((8k+1)/(4−k))) (((k+2)/(4−k)))^2 =(±(√((8k+1)/(4−k))))^2 (((k+2)^2 )/((4−k)^2 ))=((8k+1)/(4−k)) ((k^2 +4k+4)/(4−k))=8k+1 k^2 +4k+4=32k+4−8k^2 −k 9k^2 −27k=0 k(k−3)=0 k=0 ∣ k=3 (4) x−α is a common factor of a_1 x^2 +b_1 x+c and a_2 x^2 +b_2 x+c x−k is a factor of p(x)⇔k is a root of p(x)=0 x−α is a common factor of a_1 x^2 +b_1 x+c and a_2 x^2 +b_2 x+c ∴ α is a root of both equations: a_1 x^2 +b_1 x+c=0 and a_2 x^2 +b_2 x+c=0 Hence, a_1 α^2 +b_1 α+c=0 ∧ a_2 α^2 +b_2 α+c=0 On subtracting: (a_1 −a_2 )α^2 +(b_1 −b_2 )α=0 (a_1 −a_2 )α+(b_1 −b_2 )=0 (a_1 −a_2 )α=−(b_1 −b_2 ) (a_1 −a_2 )α=b_2 −b_1 P r o v e d

(3)

(4 - k) x^{2} + 2 (k + 2) x + 8 k + 1

T h e v a l u e o f k, f o r w h i c h t h e a b o v e

e x p r e s s i o n b e p e r f e c t s q u a r e, i s r e q u i r e d .

- - - - - - - - - - - - - - - - - - - - - - - -

A q u a d r a t i c e x p r e s s i o n w i l l b e p e r f e c t s q u a r e

i f i t i s o f t h e f o l l o w i n g f o r m :

a^{2} \pm 2 a b + b^{2}

I f f i r s t t e r m, m i d d l e t e r m a n d l a s t t e r m a r e

f, m a n d l r e s p e c t i v e l y t h e n

m = \pm 2 \sqrt{f} \sqrt{l}

D i v i d i n g b y 4 - k t o t h e g i v e n e x p r e s s i o n

x^{2} + \frac{2 (k + 2)}{4 - k} x + \frac{8 k + 1}{4 - k} [T o m a k e t h e f i r s t t e r m p e r f e c t s q u a r e]

x^{2} + \frac{2 (k + 2)}{4 - k} x + {(\sqrt{\frac{8 k + 1}{4 - k}})}^{2} [W r i t i n g l a s t t e r m i n s q u a r e f o r m]

N o w, m = \pm 2 \sqrt{f} \sqrt{l}

\frac{2 (k + 2)}{4 - k} x = \pm 2 (x) (\sqrt{\frac{8 k + 1}{4 - k}})

\frac{(k + 2)}{4 - k} = \pm \sqrt{\frac{8 k + 1}{4 - k}}

{(\frac{k + 2}{4 - k})}^{2} = {(\pm \sqrt{\frac{8 k + 1}{4 - k}})}^{2}

\frac{{(k + 2)}^{2}}{{(4 - k)}^{2}} = \frac{8 k + 1}{4 - k}

\frac{k^{2} + 4 k + 4}{4 - k} = 8 k + 1

k^{2} + 4 k + 4 = 32 k + 4 - 8 k^{2} - k

9 k^{2} - 27 k = 0

k (k - 3) = 0

k = 0 ∣ k = 3

(4)

x - α i s a c o m m o n f a c t o r o f a_{1} x^{2} + b_{1} x + c a n d

a_{2} x^{2} + b_{2} x + c

x - k i s a f a c t o r o f p (x) \Leftrightarrow k i s a r o o t o f p (x) = 0

x - α i s a c o m m o n f a c t o r o f a_{1} x^{2} + b_{1} x + c a n d a_{2} x^{2} + b_{2} x + c

∴ α i s a r o o t o f b o t h e q u a t i o n s : a_{1} x^{2} + b_{1} x + c = 0

a n d a_{2} x^{2} + b_{2} x + c = 0

H e n c e, a_{1} α^{2} + b_{1} α + c = 0 \land a_{2} α^{2} + b_{2} α + c = 0

O n s u b t r a c t i n g : (a_{1} - a_{2}) α^{2} + (b_{1} - b_{2}) α = 0

(a_{1} - a_{2}) α + (b_{1} - b_{2}) = 0

(a_{1} - a_{2}) α = - (b_{1} - b_{2})

(a_{1} - a_{2}) α = b_{2} - b_{1}

P r o v e d

Commented by rohit meena last updated on 25/Aug/16

o r Q v e s t i o n k e a n s w e r

Commented by rohit meena last updated on 25/Aug/16

or Qvestion ke answer c kaha gya what value of c

o r Q v e s t i o n k e a n s w e r

c k a h a g y a w h a t v a l u e o f c

Commented by Rasheed Soomro last updated on 26/Aug/16

a_1 α^2 +b_1 α+c=0 _− a_2 α^2 +_(−) b_2 α+_(−) c=0 −−−−−−−−−−−−− (a_1 −a_2 )α^2 +(b_1 −b_2 )α+0=0

a_{1} α^{2} + b_{1} α + c = 0

_{-} a_{2} α^{2} \underline{+} b_{2} α \underline{+} c = 0

- - - - - - - - - - - - -

(a_{1} - a_{2}) α^{2} + (b_{1} - b_{2}) α + 0 = 0

Answered by sandy_suhendra last updated on 25/Aug/16

3) perfect square ⇒ x_1 =x_2 ⇒ D=b^2 −4ac=0 a=4−k ; b=2(k+2) ; c=8k+1 D=[2(k+2)]^2 −4(4−k)(8k+1)=0 4k^2 +16k+16+32k^2 −124k−16=0 36k^2 −108k=0 36k(k−3)=0 k=0 or k=3 4) Mr Soomro has given the great answer continued

3) p e r f e c t s q u a r e \Rightarrow x_{1} = x_{2} \Rightarrow D = b^{2} - 4 a c = 0

a = 4 - k; b = 2 (k + 2); c = 8 k + 1

D = {[2 (k + 2)]}^{2} - 4 (4 - k) (8 k + 1) = 0

4 k^{2} + 16 k + 16 + 32 k^{2} - 124 k - 16 = 0

36 k^{2} - 108 k = 0

36 k (k - 3) = 0

k = 0 o r k = 3

4) M r S o o m r o h a s g i v e n t h e g r e a t a n s w e r

c o n t i n u e d

Answered by sandy_suhendra last updated on 25/Aug/16

6) a=4 ; b=−4(α−2) ; c=α−2 i) real n distinct ⇒ D>0 [−4(α−2)]^2 −4.4(α−2)>0 16α^2 −64α+64−16α+32>0 16α^2 −80α+96>0 α^2 −5α+6>0 (α−2)(α−3)>0 α<2 or α>3 ii)equal ⇒D=0 (α−2)(α−3)=0 α=2 or α=3 iii)imaginary ⇒D<0 (α−2)(α−3)<0 2<α<3 iv) opposite in sign ⇒ x_1 .x_2 <0 (c/a)<0 ((α−2)/4)<0 α−2<0 α<2 v)equal in magnitude but opposite sign ⇒ x_1 +x_2 =0 ((−b)/a)=0 b=0 −4(α−2)=0 α=2 continued

6) a = 4; b = - 4 (α - 2); c = α - 2

i) r e a l n d i s t i n c t \Rightarrow D > 0

{[- 4 (α - 2)]}^{2} - 4.4 (α - 2) > 0

16 α^{2} - 64 α + 64 - 16 α + 32 > 0

16 α^{2} - 80 α + 96 > 0

α^{2} - 5 α + 6 > 0

(α - 2) (α - 3) > 0

α < 2 o r α > 3

i i) e q u a l \Rightarrow D = 0

(α - 2) (α - 3) = 0

α = 2 o r α = 3

i i i) i m a g i n a r y \Rightarrow D < 0

(α - 2) (α - 3) < 0

2 < α < 3

i v) o p p o s i t e i n s i g n \Rightarrow x_{1} . x_{2} < 0

\frac{c}{a} < 0

\frac{α - 2}{4} < 0

α - 2 < 0

α < 2

v) e q u a l i n m a g n i t u d e b u t o p p o s i t e s i g n \Rightarrow x_{1} + x_{2} = 0

\frac{- b}{a} = 0

b = 0

- 4 (α - 2) = 0

α = 2

c o n t i n u e d

Commented by sandy_suhendra last updated on 25/Aug/16

α<2 it means −∞<α<2 ⇒(−∞,2) α>3 it means 3<α<∞ ⇒(3,∞)

α < 2 i t m e a n s - \infty < α < 2 \Rightarrow (- \infty, 2)

α > 3 i t m e a n s 3 < α < \infty \Rightarrow (3, \infty)

Commented by rohit meena last updated on 25/Aug/16

a n s w e r i s (i) (- \infty, 2) (3, \infty)

Answered by sandy_suhendra last updated on 25/Aug/16

7) P(x).Q(x)=0 (ax^2 +bx+c)(−ax^2 +dx+c)=0 ax^2 +bx+c=0 ⇒ the roots x_1 and x_2 or −ax^2 +dx+c=0 ⇒ the roots x_3 and x_4 x_(1,2) = ((−b±(√(b^2 −4ac)))/(2a)) x_(3,4) = ((−d±(√(d^2 −4(−a)c)))/(−2a)) = ((−d±(√(d^2 +4ac)))/(−2a)) if ac>0 ∧ b^2 −4ac<0 ⇒ x_(1,2) are imaginary and x_(3,4) are real (2 real roots) if ac>0 ∧ b^2 −4ac>0 ⇒ x_(1,2) and x_(3,4) are real (4 real roots) if ac<0 ∧ d^2 +4ac>0 ⇒ x_(1,2) and x_(3,4) are real (4 real roots) if ac<0 ∧ d^2 +4ac<0 ⇒ x_(1,2) are real and x_(3,4) are imaginary (2 real roots)

7) P (x) . Q (x) = 0

({a x}^{2} + b x + c) (- {a x}^{2} + d x + c) = 0

{a x}^{2} + b x + c = 0 \Rightarrow t h e r o o t s x_{1} a n d x_{2}

o r - {a x}^{2} + d x + c = 0 \Rightarrow t h e r o o t s x_{3} a n d x_{4}

x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

x_{3, 4} = \frac{- d \pm \sqrt{d^{2} - 4 (- a) c}}{- 2 a} = \frac{- d \pm \sqrt{d^{2} + 4 a c}}{- 2 a}

i f a c > 0 \land b^{2} - 4 a c < 0 \Rightarrow x_{1, 2} a r e i m a g i n a r y a n d x_{3, 4} a r e r e a l (2 r e a l r o o t s)

i f a c > 0 \land b^{2} - 4 a c > 0 \Rightarrow x_{1, 2} a n d x_{3, 4} a r e r e a l (4 r e a l r o o t s)

i f a c < 0 \land d^{2} + 4 a c > 0 \Rightarrow x_{1, 2} a n d x_{3, 4} a r e r e a l (4 r e a l r o o t s)

i f a c < 0 \land d^{2} + 4 a c < 0 \Rightarrow x_{1, 2} a r e r e a l a n d x_{3, 4} a r e i m a g i n a r y (2 r e a l r o o t s)

Commented by Rasheed Soomro last updated on 25/Aug/16

x_{3, 4} = \frac{- d \pm \sqrt{d^{2} - 4 (- a) c}}{- 2 a}

Commented by sandy_suhendra last updated on 25/Aug/16

Thank′s Rasheed for your correction. I′ve corrected it

{T h a n k}^{'} s R a s h e e d f o r y o u r c o r r e c t i o n . I^{'} v e c o r r e c t e d i t

Commented by Rasheed Soomro last updated on 25/Aug/16

Why d^2 +4ac is always positive? Is there a guarantee that a>0 ∧ c>0 or a<0 ∧ c<0 However we can say that at least one of b^2 −4ac and d^2 +4ac is positive.Because there are two possibilities ac>0 or ac<0. If ac>0 then d^2 +4ac>0 If ac<0 then b^2 −4ac>0 (Third possibility ac=0 is excluded in the question.)

W h y d^{2} + 4 a c i s a l w a y s p o s i t i v e ? I s t h e r e a g u a r a n t e e

t h a t a > 0 \land c > 0 o r a < 0 \land c < 0

H o w e v e r w e c a n s a y t h a t a t l e a s t o n e o f

b^{2} - 4 a c a n d d^{2} + 4 a c i s p o s i t i v e . B e c a u s e

t h e r e a r e t w o p o s s i b i l i t i e s a c > 0 o r a c < 0 .

I f a c > 0 t h e n d^{2} + 4 a c > 0

I f a c < 0 t h e n b^{2} - 4 a c > 0

(T h i r d p o s s i b i l i t y a c = 0 i s e x c l u d e d i n t h e q u e s t i o n .)

Commented by sandy_suhendra last updated on 25/Aug/16

That′s great, you′re right! Thank′s Rasheed for reminding me. I have made a mistake. It has been corrected.

{T h a t}^{'} s g r e a t, {y o u}^{'} r e r i g h t! {T h a n k}^{'} s R a s h e e d f o r r e m i n d i n g m e . I h a v e m a d e a m i s t a k e . I t h a s b e e n c o r r e c t e d .

Commented by Rasheed Soomro last updated on 25/Aug/16

Y o u r l o g i c i s s i m p l e a n d a d m i r a b l e .

Answered by Yozzia last updated on 25/Aug/16

5) 3x^2 +2αxy+2y^2 +2ax−4y+1 can be resolved into two linear factors. ∴ 3x^2 +2αxy+2y^2 +2ax−4y+1≡(px+by+c)(dx+ey+f) −−−−−−−−−−−−−−−−−−−−−−−−−− y=0⇒3x^2 +2ax+1≡(px+c)(dx+f) 3x^2 +2(a/( (√3)))(√3)x+(a^2 /3)−(a^2 /3)+1≡(px+c)(dx+f) ((√3)x+(a/( (√3)))+(√((a^2 −3)/3)))((√3)x+(a/( (√3)))−(√((a^2 −3)/3)))≡(px+c)(dx+f) ⇒p=d=(√3), c=((a+(√(a^2 −3)))/( (√3))), f=((a−(√(a^2 −3)))/( (√3))) −−−−−−−−−−−−−−−−−−−−−−−−−−− x=0⇒2y^2 −4y+1≡(by+c)(ey+f) 2y^2 −2×(√2)(√2)y+2−2+1≡(by+c)(ey+f) ((√2)y−(√2))^2 −1≡(by+c)(ey+f) ((√2)y−(√2)−1)((√2)y−(√2)+1)≡(by+c)(ey+f) ⇒b=e=(√2) and possibly c=−(√2)−1 and f=1−(√2) or c=1−(√2) and f=−1−(√2). −−−−−−−−−−−−−−−−−−−−−−−− coefficient of xy=2α=pe+bd 2α=(√2)(p+d)=(√2)(2(√3))⇒α=(√6). −−−−−−−−−−−−−−−−−−−−−−− We have that x^2 +4ax+2a^2 +6=0. Try x=α=(√6). ∴ 6+4a(√6)+2a^2 +6=0 2a^2 +4a(√6)+12=0. Now, if c=−1−(√2)⇒a^2 −3=((−1−(√2))(√3)−a)^2 a^2 −3=a^2 +2a(√3)(1+(√2))+3(1+(√2))^2 2a(1+(√2))(√3)+3+3(3+2(√2))=0 2a(1+(√2))(√3)+12+6(√2)=0 a(1+(√2))×3+(6+3(√2))(√3)=0 a(1+(√2))+(2+(√2))(√3)=0 a=((−(2+(√2))(√3))/(1+(√2)))=(2+(√2))(1−(√2))(√3)=−(√6) Also, if f=1−(√2)⇒a=−(√6). [((a−(√(a^2 −3)))/( (√3)))=1−(√2)⇒(a−(1−(√2))(√3))^2 =a^2 −3 2a((√2)−1)(√3)+3(3−2(√2))=−3 a(−1+(√2))+(2−(√2))(√3)=0 a=(((√2)−2)/(−1+(√2)))(√3)=((((√2)−2)(−1−(√2)))/(1−2))(√3) a=(√3)((√2)−2)(1+(√2))=(√3)((√2)−2(√2)−2+2)=−(√6)] ∴2(−(√6))^2 +4(√6)(−(√6))+12=0 12+12−24=0 Hence α satisfies x^2 +4ax+2a^2 +6=0 for some constant a, if we can resolve 3x^2 +2αxy+2y^2 +2ax−4y+1 into two linear factors.

5) 3 x^{2} + 2 α x y + 2 y^{2} + 2 a x - 4 y + 1

c a n b e r e s o l v e d i n t o t w o l i n e a r f a c t o r s .

∴ 3 x^{2} + 2 α x y + 2 y^{2} + 2 a x - 4 y + 1 \equiv (p x + b y + c) (d x + e y + f)

- - - - - - - - - - - - - - - - - - - - - - - - - -

y = 0 \Rightarrow 3 x^{2} + 2 a x + 1 \equiv (p x + c) (d x + f)

3 x^{2} + 2 \frac{a}{\sqrt{3}} \sqrt{3} x + \frac{a^{2}}{3} - \frac{a^{2}}{3} + 1 \equiv (p x + c) (d x + f)

(\sqrt{3} x + \frac{a}{\sqrt{3}} + \sqrt{\frac{a^{2} - 3}{3}}) (\sqrt{3} x + \frac{a}{\sqrt{3}} - \sqrt{\frac{a^{2} - 3}{3}}) \equiv (p x + c) (d x + f)

\Rightarrow p = d = \sqrt{3}, c = \frac{a + \sqrt{a^{2} - 3}}{\sqrt{3}}, f = \frac{a - \sqrt{a^{2} - 3}}{\sqrt{3}}

- - - - - - - - - - - - - - - - - - - - - - - - - - -

x = 0 \Rightarrow 2 y^{2} - 4 y + 1 \equiv (b y + c) (e y + f)

2 y^{2} - 2 \times \sqrt{2} \sqrt{2} y + 2 - 2 + 1 \equiv (b y + c) (e y + f)

{(\sqrt{2} y - \sqrt{2})}^{2} - 1 \equiv (b y + c) (e y + f)

(\sqrt{2} y - \sqrt{2} - 1) (\sqrt{2} y - \sqrt{2} + 1) \equiv (b y + c) (e y + f)

\Rightarrow b = e = \sqrt{2} a n d p o s s i b l y c = - \sqrt{2} - 1 a n d f = 1 - \sqrt{2}

o r c = 1 - \sqrt{2} a n d f = - 1 - \sqrt{2} .

- - - - - - - - - - - - - - - - - - - - - - - -

c o e f f i c i e n t o f x y = 2 α = p e + b d

2 α = \sqrt{2} (p + d) = \sqrt{2} (2 \sqrt{3}) \Rightarrow α = \sqrt{6} .

- - - - - - - - - - - - - - - - - - - - - - -

W e h a v e t h a t x^{2} + 4 a x + 2 a^{2} + 6 = 0 .

T r y x = α = \sqrt{6} .

∴ 6 + 4 a \sqrt{6} + 2 a^{2} + 6 = 0

2 a^{2} + 4 a \sqrt{6} + 12 = 0 .

N o w, i f c = - 1 - \sqrt{2} \Rightarrow a^{2} - 3 = {((- 1 - \sqrt{2}) \sqrt{3} - a)}^{2}

a^{2} - 3 = a^{2} + 2 a \sqrt{3} (1 + \sqrt{2}) + 3 {(1 + \sqrt{2})}^{2}

2 a (1 + \sqrt{2}) \sqrt{3} + 3 + 3 (3 + 2 \sqrt{2}) = 0

2 a (1 + \sqrt{2}) \sqrt{3} + 12 + 6 \sqrt{2} = 0

a (1 + \sqrt{2}) \times 3 + (6 + 3 \sqrt{2}) \sqrt{3} = 0

a (1 + \sqrt{2}) + (2 + \sqrt{2}) \sqrt{3} = 0

a = \frac{- (2 + \sqrt{2}) \sqrt{3}}{1 + \sqrt{2}} = (2 + \sqrt{2}) (1 - \sqrt{2}) \sqrt{3} = - \sqrt{6}

A l s o, i f f = 1 - \sqrt{2} \Rightarrow a = - \sqrt{6} .

[\frac{a - \sqrt{a^{2} - 3}}{\sqrt{3}} = 1 - \sqrt{2} \Rightarrow {(a - (1 - \sqrt{2}) \sqrt{3})}^{2} = a^{2} - 3

2 a (\sqrt{2} - 1) \sqrt{3} + 3 (3 - 2 \sqrt{2}) = - 3

a (- 1 + \sqrt{2}) + (2 - \sqrt{2}) \sqrt{3} = 0

a = \frac{\sqrt{2} - 2}{- 1 + \sqrt{2}} \sqrt{3} = \frac{(\sqrt{2} - 2) (- 1 - \sqrt{2})}{1 - 2} \sqrt{3}

a = \sqrt{3} (\sqrt{2} - 2) (1 + \sqrt{2}) = \sqrt{3} (\sqrt{2} - 2 \sqrt{2} - 2 + 2) = - \sqrt{6}]

∴ 2 {(- \sqrt{6})}^{2} + 4 \sqrt{6} (- \sqrt{6}) + 12 = 0

12 + 12 - 24 = 0

H e n c e α s a t i s f i e s x^{2} + 4 a x + 2 a^{2} + 6 = 0

f o r s o m e c o n s t a n t a, i f w e c a n r e s o l v e

3 x^{2} + 2 α x y + 2 y^{2} + 2 a x - 4 y + 1 i n t o t w o

l i n e a r f a c t o r s .

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