1
BITSAT 2023
+3
-1

Let $$\alpha, \beta$$ be the roots of the equation $$x^2-p x+r=0$$ and $$\frac{\alpha}{2}, 2 \beta$$ be the roots of the equation $$x^2-q x+r=0$$. Then, the value of $$r$$ is equal to

A
$$\frac{2}{9}(p-q)(2 q-p)$$
B
$$\frac{2}{9}(q-2 p)(2 q-p)$$
C
$$\frac{2}{9}(q-p)(2 p-q)$$
D
$$\frac{2}{9}(2 p-q)(2 q-p)$$
2
BITSAT 2022
+3
-1

If $$\alpha$$ be a root of the equation $$4{x^2} + 2x - 1 = 0$$, then the other root of the equation is

A
4$$\alpha$$2 + 2$$\alpha$$
B
4$$\alpha$$2 $$-$$ 2$$\alpha$$
C
4$$\alpha$$3 $$-$$ 3$$\alpha$$
D
4$$\alpha$$3 + 3$$\alpha$$
3
BITSAT 2022
+3
-1

Let a, b be the solutions of x2 + px + 1 = 0 and c, d be the solution of x2 + qx + 1 = 0. If (a $$-$$ c) (b $$-$$ c) and (a + d)(b + d) are the solution of x2 + ax + $$\beta$$ = 0, then $$\beta$$ is equal to

A
p + q
B
p $$-$$ q
C
p2 + q2
D
q2 $$-$$ p2
4
BITSAT 2021
+3
-1

If a$$\in$$R, b$$\in$$R, then the equation x2 $$-$$ abx $$-$$ a2 = 0 has

A
one positive root and one negative root
B
Both positive roots
C
Both negative roots
D
Non-real roots
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