Question

solve for x √(3x square -5x +2) - √(3x square -5x-1) =1​

Answer 1

Step-by-step explanation:

Solution

Given :-

\sf \implies \: Equation \: 3x + 4y - 9 = 0⟹Equation3x+4y−9=0

\sf \implies Point \: (1,3) \: and \: (2,7)⟹Point(1,3)and(2,7)

Let

\sf \implies \: Ratio \: = P \ratio \: 1⟹Ratio=P:1

Using section formula

\implies \sf \: p \bigg( \dfrac{x_2m + nx_1}{m + n} , \: \dfrac{y_2m + y_1n}{m + n} \bigg)⟹p(

m+n

x

2

m+nx

1

,

m+n

y

2

m+y

1

n

)

Where

\begin{gathered} \sf \implies \: x_1 = 1, y_1 = 3 \\ \sf \implies \: x_2 = 2,y_2 = 7\end{gathered}

⟹x

1

=1,y

1

=3

⟹x

2

=2,y

2

=7

\sf \implies \: m = p \: \: and \: n = 1⟹m=pandn=1

Now put the value on formula

\sf \implies\: p \bigg( \dfrac{2 \times p + 1 \times1}{p + 1} , \: \dfrac{7 \times p + 3 \times 1}{p + 1} \bigg)⟹p(

p+1

2×p+1×1

,

p+1

7×p+3×1

)

\sf \implies\: p \bigg( \dfrac{2p + 1 }{p + 1} , \: \dfrac{7 p + 3 }{p + 1} \bigg)⟹p(

p+1

2p+1

,

p+1

7p+3

)

Now put the value of x and y on Given equation

\sf \implies \: 3x + 4y - 9 = 0⟹3x+4y−9=0

\sf \implies3 \bigg( \dfrac{2p + 1}{p + 1} \bigg) + 4 \bigg( \dfrac{7p + 3}{p + 1} \bigg) - 9 = 0⟹3(

p+1

2p+1

)+4(

p+1

7p+3

)−9=0

\sf\implies \: \dfrac{6p + 3}{p + 1} + \dfrac{28p + 12}{p + 1} - 9 = 0⟹

p+1

6p+3

+

p+1

28p+12

−9=0

Takin Lcm

\sf \implies \: \dfrac{6p + 3 + 28p + 12 - 9p - 9}{p + 1} = 0⟹

p+1

6p+3+28p+12−9p−9

=0

\sf \implies\: 6p + 3 + 28p + 12 - 9p - 9 = 0⟹6p+3+28p+12−9p−9=0

\sf \implies \: 34p + 15 - 9p - 9 = 0⟹34p+15−9p−9=0

\sf \implies 25p + 6 = 0⟹25p+6=0

\sf \implies \: p = \dfrac{ - 6}{25}⟹p=

25

−6

Answer

The ratio is -6/25 or -6:25

Answer 2

Given:-

• $\sf \sqrt{3 {x}^{2} - 5x + 2 } - \sqrt{3 {x}^{2} - 5x - 1} = 1$

To find:-

• Value of x

Answer:-

$\sf \sqrt{3 {x}^{2} - 5x + 2 } - \sqrt{3 {x}^{2} - 5x - 1} = 1$

Let $\sf 3x^2 - 5x$ be $\sf a.$

Now,

$\sf \longrightarrow \: \sqrt{a + 2} - \sqrt{a - 1} = 1$

Squaring both sides,

$\sf \longrightarrow {( \sqrt{a + 2} - \sqrt{a - 1} ) }^{2} = {1}^{ 2 }$

Using (m - n)² = m² + n² - 2mn,

$\sf \longrightarrow ({ \sqrt{a + 2} })^{2} + ({ \sqrt{a - 1} })^{2} - 2 \sqrt{(a + 2)(a - 1)} = 1$

$\sf \longrightarrow a + 2 + a - 1 - 2 \sqrt{(a + 2)(a - 1)} = 1$

$\sf \longrightarrow 2a + 1 - 2\sqrt{(a + 2)(a - 1)} = 1$

$\sf \longrightarrow 2a - 2 \sqrt{(a + 2)(a - 1)} = 0$

$\sf \longrightarrow 2a = 2 \sqrt{(a + 2)(a - 1)}$

$\sf \longrightarrow a = \sqrt{(a + 2)(a - 1)}$

Squaring both sides,

$\sf \longrightarrow {a}^{2} = (a + 2)(a - 1)$

$\sf \longrightarrow {a}^{2} = {a}^{2} + 2a - a - 2$

$\sf \longrightarrow a = 2$

But as we have assumed $\sf 3x^2 - 5x = a,$

$\sf 3 {x}^{2} - 5x = a$

$\sf \longrightarrow 3 {x}^{2} - 5x = 2$

$\sf \longrightarrow 3 {x}^{2} - 5x - 2 = 0$

Now we will split the middle term in L.H.S.

$\sf \longrightarrow3 {x}^{2} - 6x + x - 2 = 0$

$\sf \longrightarrow 3x(x - 2) + 1(x - 2) = 0$

$\sf\longrightarrow (x - 2)(3x + 1) = 0$

Either,

$\sf x - 2 = 0$

$\sf \longrightarrow x = 2 \:$

Or,

$\sf 3x + 1 = 0$

$\sf \longrightarrow x = - \dfrac{1}{3}$

Hence,

$\longrightarrow \underline{\underline{\sf{\green{ x = 2 \:or\:x = - \dfrac{1}{3} }}}}$