If x - -3 - 1-2- find the value of 4x3 - 2x2 - 8x - 7-

X = √(3) + 1/2⇒ 2x = √(3) + 1 ⇒ 2x 1=√(3) Squaring both sides. (2x 1)^2 = (√(3))^2 ⇒ 4x^2 4x + 1 =3 4x^2 4x + 1 3 =0 ⇒ 4x^2 4x 2 = 0 ∴ 2x^2 2x 1 = ...

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Byju's Answer

Standard IX

Mathematics

Operations on Rational and Irrational Numbers

If x = √3 + 1...

Question

If $x = \frac{\sqrt{3} + 1}{2}$ , find the value of $4 x^{3} + 2 x^{2} - 8 x + 7$ .

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Solution

$x = \frac{\sqrt{3} + 1}{2} \Rightarrow 2 x = \sqrt{3} + 1$

$\Rightarrow 2 x - 1 = \sqrt{3}$

Squaring both sides.

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

$4 x^{2} - 4 x + 1 - 3 = 0 \Rightarrow 4 x^{2} - 4 x - 2 = 0$

$∴ 2 x^{2} - 2 x - 1 = 0$

Now, $4 x^{3} + 2 x^{2} - 8 x + 7$

$= (4 x^{3} - 4 x^{2} - 2 x) + (6 x^{2} - 6 x - 3) + 10$

$= 2 x (2 x^{2} - 2 x - 1) + 3 (2 x^{2} - 2 x - 1) + 10$

$= 2 x \times 0 + 3 \times 0 + 10$

$= 0 + 0 + 10 = 10$

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If x=√3+12, find the value of 4x3+2x2−8x+7.

x=√3+12⇒2x=√3+1 ⇒2x−1=√3 Squaring both sides. (2x−1)2=(√3)2⇒4x2−4x+1=3 4x2−4x+1−3=0⇒4x2−4x−2=0 ∴ 2x2−2x−1=0 Now, 4x3+2x2−8x+7 =(4x3−4x2−2x)+(6x2−6x−3)+10 =2x(2x2−2x−1)+3(2x2−2x−1)+10 =2x×0+3×0+10 =0+0+10=10

If $x = \frac{\sqrt{3} + 1}{2}$ , find the value of $4 x^{3} + 2 x^{2} - 8 x + 7$ .