Question

Please solve this question

If you know the answer then answer if you don't know the answer then don't answer.

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Answer 1

Answer:

Step-by-step explanation:

See the ans in attachment,

Answer 2

Answer:

The value of x is $\frac{1}{2},\:x=\frac{9+\sqrt{57}}{4},\:x=\frac{9-\sqrt{57}}{4}$

Step-by-step explanation:

The provided expression is $\sqrt{2x^2-9x+4}+3\sqrt{2x-1}=\sqrt{2x^2+21x-11}$

squaring both the sides in above expression

$(\sqrt{2x^{2}-9x+4} +3\sqrt{2x-1})^{2}=(\sqrt{2x^2+21x-11})^{2}$

since (a + b)² = a² +b² +2ab

$(\sqrt{2x^{2}-9x+4})^{2} +(3\sqrt{2x-1})^{2} + 6\sqrt{2x-1}\sqrt{2x^{2}-9x+4} = (\sqrt{2x^2+21x-11})^{2}$

$2x^{2}-9x+4+9(2x-1) +6\sqrt{2x-1}\sqrt{2x^{2}-9x+4} =2x^{2}+21x-11$

$2x^{2}-9x+4+18x-9 + 6\sqrt{2x-1}\sqrt{2x^{2}-9x+4}=2x^{2}+21x-11$

$2x^{2}+9x-5 +6\sqrt{2x-1}\sqrt{2x^{2}-9x+4}=2x^{2}+21x-11$

Subtract both the sides by $2x^{2}$

$2x^{2}-2x^{2}+9x-5 + 6\sqrt{2x-1}\sqrt{2x^{2}-9x+4}=2x^{2}-2x^{2}+21x-11$

$9x-5 + 6\sqrt{2x-1}\sqrt{2x^{2}-9x+4}=21x-11$

Subtract both the sides by 9x,

$9x-5-9x +6\sqrt{2x-1}\sqrt{2x^{2}-9x+4}=21x-11-9x$

$-5 +6\sqrt{2x-1}\sqrt{2x^{2}-9x+4}=12x-11$

Add both the sides by 11,

$6 +6\sqrt{2x-1}\sqrt{2x^{2}-9x+4}=12x$

divide both the sides by 6,

$1 +\sqrt{2x-1}\sqrt{2x^{2}-9x+4}=2x$

subtract both the sides by 1,

$(2x-1)(2x^{2}-9x+4)=2x-1$

$4x^{3}-20x^{2}+17x-4=2x-1$

subtract both the sides by 2x, in above expression

$4x^{3}-20x^{2}+15x-4=-1$

add 1 both the sides,

$4x^{3}-20x^{2}+15x-3=0$

Solve by factoring

Factors: $\left(2x-1\right)\left(2x^2-9x+3\right)$

$\left(2x-1\right)\left(2x^2-9x+3\right)=0$

$\frac{1}{2}=x$

$\mathrm{Solve\:}\:2x-1=0:\quad x=\frac{1}{2}$

$\mathrm{Solve\:}\:2x^2-9x+3=0:\quad x=\frac{9+\sqrt{57}}{4},\:x=\frac{9-\sqrt{57}}{4}$

$x=\frac{1}{2},\:x=\frac{9+\sqrt{57}}{4},\:x=\frac{9-\sqrt{57}}{4}$

The value of x is $\frac{1}{2},\:x=\frac{9+\sqrt{57}}{4},\:x=\frac{9-\sqrt{57}}{4}$