MATHEMATICS HIGH SCHOOL

1. x=1/(root3-root2). find rootx-(1/rootx) 2. if x=[root(a+2b)+root(a-2b)]/[root(a+2b)-root(a-2b]. show that bx^2-ax+b=0

Answers

Answer 1
Answer:

Answer with explanation:

Ques 1)

x=(1)/(√(3)-√(2))

Now we are asked to find the value of:

√(x)-(1)/(√(x))

We know that:

(√(x)-(1)/(√(x)))^2=x+(1)/(x)-2

Also:

x=(1)/(√(3)-√(2)) could be written as:

x=(1)/(√(3)-√(2))* (√(3)+√(2))/(√(3)+√(2))\n\n\nx=(√(3)+√(2))/((√(3))^2-(√(2))^2)

since, we know that:

(a+b)(a-b)=a^2-b^2

Hence,

x=(√(3)+√(2))/(3-2)\n\n\nx=√(3)+√(2)

Also,

(1)/(x)=√(3)-√(2)

Hence, we get:

(√(x)-(1)/(√(x)))^2=√(3)+√(2)+√(3)-√(2)-2\n\n\n(√(x)-(1)/(√(x)))^2=2√(3)-2\n\n\n√(x)-(1)/(√(x))=\sqrt{2√(3)-2}

Hence,

       √(x)-(1)/(√(x))=\sqrt{2√(3)-2}

Ques 2)

x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))

on multiplying and dividing by conjugate of denominator we get:

x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))* (√(a+2b)+√(a-2b))/(√(a+2b)+√(a-2b))\n\n\nx=((√(a+2b)+√(a-2b))^2)/((√(a+2b))^2-(√(a-2b))^2)\n\n\nx=((√(a+2b))^2+(√(a-2b))^2+2√(a+2b)√(a-2b))/(a+2b-a+2b)\n\n\nx=(a+2b+a-2b+2√(a+2b)√(a-2b))/(4b)\n\n\nx=(2a+2√(a^2-4b^2))/(4b)\n\n\nx^2=((2a+2√(a^2-4b^2))/(4b))^2\n\n\nx^2=((2a+2√(a^2-4b^2))^2)/(16b^2)

Hence, we have:

x^2=(4a^2+4(a^2-4b^2)+8a√(a^2-4b^2))/(16b^2)\n\n\nx^2=(4a^2+4a^2-16b^2+8a√(a^2-4b^2))/(16b^2)\n\n\n\nx^2=(8a^2-16b^2+8a√(a^2-4b^2))/(16b^2)\n\n\nbx^2=(8a^2-16b^2+8a√(a^2-4b^2))/(16b)\n\n\nbx^2=(8a(a+√(a^2-4b^2))-16b^2)/(16b)\n\n\nbx^2=(8a(a+√(a^2-4b^2)))/(16b)-(16b^2)/(16b)\n\n\nbx^2=(a(a+√(a^2-4b^2)))/(2b)-b\n\n\nbx^2=ax-b\n\n\ni.e.\n\n\nbx^2-ax+b=0
Answer 2
Answer: 1. x = 1/ ( √(3) - √(2)) = √(3)+ √(2);
 ( √(x) -1/ √(x) )^(2) = x + 1/x - 2 = = 

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