Question

If x=√3+1 / 2
find the value of 4xcube + 2x square – 8x + 7.
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Answer 1

Solution!!

$\sf x=\dfrac{\sqrt{3}+1}{2}$

$\sf 4x^{3}+2x^{2}-8x+7$

Substitute the value of x.

$\sf =4\left(\dfrac{\sqrt{3}+1}{2}\right)^{3}+2\left(\dfrac{\sqrt{3}+1}{2}\right)-8\left(\dfrac{\sqrt{3}+1}{2}\right)+7$

Reduce the numbers with the greatest common factor 2.

$\sf =4\left(\dfrac{\sqrt{3}+1}{2}\right)^{3}+2\left(\dfrac{\sqrt{3}+1}{2}\right)-4(\sqrt{3}+1)+7$

To raise a fraction to a power, raise the numerator and denominator to that power.

$\sf =4\left(\dfrac{(\sqrt{3}+1)^{3}}{8}+2\left(\dfrac{\sqrt{3}+1}{2}\right)-4(\sqrt{3}+1)+7$

To raise a fraction to a power, raise the numerator and denominator to that power.

$\sf =4\left(\dfrac{(\sqrt{3}+1)^{3}}{8}+2\left(\dfrac{(\sqrt{3}+1)^{2}}{4}-4(\sqrt{3}+1)+7$

Reduce the numbers with the greatest common factor 4 and 2.

$\sf =\dfrac{(\sqrt{3}+1)^{3}}{2}+\dfrac{(\sqrt{3}+1)^{2}}{2}-4(\sqrt{3}+1)+7$

Distribute -4 through the parentheses.

$\sf =\dfrac{(\sqrt{3}+1)^{3}}{2}+\dfrac{(\sqrt{3}+1)^{2}}{2}-4\sqrt{3}-4+7$

Use (a + b)³ = a³ + 3a²b + 3ab² + b³ to expand the expression.

$\sf =\dfrac{\sqrt{3}+9+\sqrt{3}+1}{2}+\dfrac{(\sqrt{3}+1)^{2}}{2}-4\sqrt{3}-4+7$

Use (a + b)² = a² + 2ab + b² to expand the expression.

$\sf =\dfrac{3\sqrt{3}+9+3\sqrt{3}+1}{2}+\dfrac{3+2\sqrt{3}+1}{2}-4\sqrt{3}-4+7$

Collect the like terms and add and subtract the numbers.

$\sf =\dfrac{6\sqrt{3}+10}{2}+\dfrac{4+2\sqrt{3}}{2}-4\sqrt{3}+3$

Factor out 2 from the expressions.

$\sf =\dfrac{2(3\sqrt{3}+5)}{2}+\dfrac{2(2+\sqrt{3})}{2}-4\sqrt{3}+3$

Reduce the fractions with 2.

$\sf =3\sqrt{3}+5+2+\sqrt{3}-4\sqrt{3}+3$

Collect the like terms.

$\sf =3\sqrt{3}+\sqrt{3}-4\sqrt{3}+5+2+3$

Add and subtract the terms.

$\sf =10$