solve-simultaneously-2-k-h-9-i-k-2-h-3-ii-

Question Number 41984 by Tawa1 last updated on 16/Aug/18

solve simultaneously : 2 \sqrt{k} + h = 9 \dots \dots . (i)

k + 2 \sqrt{h} = 3 \dots \dots . (ii)

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18

x + 2 \sqrt{y} - 9 = 0

2 \sqrt{x} + y - 3 = 0

2 \sqrt{y} = (9 - x)

4 y = {(9 - x)}^{2}

{(9 - x)}^{2} = 4 \times 1 \times y

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18

t h i s i s a p a r a b o l a e q n

2 \sqrt{x} = (y - 3)

4 x = {(y - 3)}^{2}

{(y - 3)}^{2} = 4 x

t h i s i s a l s o a p a r a b o l a

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18

n o s o l u t i o n \dots

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18

i f c o n v e r t e d t o p a r a b o l a b y s q u a r i n g \dots t h e n s o l u t i o n

x = 3.8 (a p p r o x) y = 7.5 a p p r o x

h = 3.8 k = 7.5

Commented by MJS last updated on 16/Aug/18

in this case we have 2 real and 2 complex

points

A = (\begin{matrix} 3.76 \\ 6.88 \end{matrix}) B = (\begin{matrix} 15.60 \\ 10.90 \end{matrix})

C = (\begin{matrix} 8.32 + 3 . 47 i \\ - 2.89 - 1 . 18 i \end{matrix}) D = (\begin{matrix} 8.32 - 3 . 47 i \\ - 2.89 + 1 . 18 i \end{matrix})

Answered by MJS last updated on 16/Aug/18

(i)

\sqrt{k} = \frac{9 - h}{2} \Rightarrow 9 - h ⩾ 0 \Leftrightarrow h ⩽ 9

k = \frac{{(9 - h)}^{2}}{4}

(i i)

k = 3 - 2 \sqrt{h} \Rightarrow h ⩾ 0

k = \frac{{(9 - h)}^{2}}{4} \land 0 ⩽ h ⩽ 9 \Rightarrow 0 ⩽ k ⩽ \frac{81}{4}

k = 3 - 2 \sqrt{h} \land 0 ⩽ h ⩽ 9 \Rightarrow - 3 ⩽ h ⩽ 3

\Rightarrow 0 ⩽ k ⩽ 3

(i)

h = 9 - 2 \sqrt{k} \land 0 ⩽ k ⩽ 3 \Rightarrow (9 - 2 \sqrt{3}) ⩽ h ⩽ 9

(i i)

\sqrt{h} = \frac{3 - k}{2} \Rightarrow 3 - k ⩾ 0 \Leftrightarrow k ⩽ 3

h = \frac{{(3 - k)}^{2}}{4} \land 0 ⩽ k ⩽ 3 \Rightarrow 0 ⩽ h ⩽ \frac{9}{4}

so we have 2 musts :

(9 - 2 \sqrt{3}) ⩽ h ⩽ 9 and 0 ⩽ h ⩽ \frac{9}{4}

9 - 2 \sqrt{3} \approx 5.54

\frac{9}{4} = 2.25

h ⩾ 5.54 and h ⩽ 2.25 is impossible \Rightarrow no real

solution

Commented by Tawa1 last updated on 16/Aug/18

God bless you sir . am still expecting the polynomial equation sir .

any time

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18

\sqrt{k} = \frac{9 - h}{2}

k = \frac{81 - 18 h + h^{2}}{4}

\sqrt{h} = \frac{3 - k}{2}

h = \frac{9 - 6 k + k^{2}}{4}

s o k = \frac{81 - 18 (\frac{9 - 6 k + k^{2}}{4}) + \frac{81 + 36 k^{2} + k^{4} - 108 k - 12 k^{3} + 18 k^{2}}{16}}{4}

4 k = 81 \times 16 - 18 \times 4 (9 - 6 k + k^{2}) + 81 - 108 k + 54 k^{2} - 12 k^{3} + k^{4}

4 k = 1296 - 18 \times 4 \times 9 + 18 \times 4 \times 6 k - 18 \times 4 k^{2} - 108 k + 54 k^{2} - 12 k^{3} + k^{4}

k^{4} - 12 k^{3} + k^{2} (54 - 72) + k (- 108 + 18 \times 4 \times 6 - 4) + 1296 - 18 \times 4 \times 9 = 0

k^{4} - 12 k^{3} - 18 k^{2} + k (- 108 + 432 - 4) + 1296 - 648 = 0

k^{4} - 12 k^{3} - 18 k^{2} + 320 k + 648 = 0

Commented by MJS last updated on 10/Sep/18

typo somewhere . it must be

k^{4} - 12 k^{3} - 18 k^{2} + 260 k + 729 = 0

k_{1} = 6.87588 \dots

k_{2} = 10.9001 \dots

but we squared 2 times to get here and both

resulting pairs of h, k {don}^{'} t solve the given

equation system

Facebook Tweet Pin