Question

Each of the equations $3x^2-2=25,(2x-1{)}^2=(x-1{)}^2,\sqrt{x^2-7}=\sqrt{x-1}$ has

• A. two integral roots
• B. no root greater than $3$
• C. no root zero
• D. only one root
• E. only negative root and one positive root

Answer 1

Answer: (B)

no root greater than $3$

$⟹3x^2-2=25⟹3x^2=27\therefore x=\pm 3=-3,3⟹{(2x-1)}^2={(x-1)}^2⟹4x^2-4x+1=x^2-2x+1⟹3x^2-2x=0\therefore x=0,\frac{2}{3}\sqrt{x^2-7}=\sqrt{x-1}⟹x^2-7=x-7⟹x^2-x-6=0\therefore x=-2,3$

But x = -2 will not satisfy the given equation.

$\therefore$ x = 3 is the only solution.

$\therefore$ No value of x is greater than 3.

3 integral roots and one root is zero.

4 roots [one negative, 2 positive roots, one zero root]