If p 1 p -2 q 1- q then which of following statement about the quadratic equations x 2- px - q -0 - x 2- p 1 x - q 1-0 is always true Here-p - p 1- q - q 1- - R A- Both the equations have real rootsB- at least one of the equations has real rootsC- Both the equations have imaginary rootsD- exactly one of them has equal roots

The correct option is B at least one of the equations has real roots Δ_1=p^2 4q; Δ_2=p^2_1 4q_1 Δ_1+Δ_2=p^2+p^2_1 4(q+q_1)=p^2+p^2_1 2p_1p=(p p_1)^2 ≥ 0 Both Δ ...

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Byju's Answer

Standard XII

Mathematics

Common Roots

If p 1 p =2 q...

Question

If $p_{1} p = 2 (q_{1} + q)$ then which of following statement about the quadratic equations $x^{2} + p x + q = 0; x^{2} + p_{1} x + q_{1} = 0$ is always true (Here $p, p_{1}, q, q_{1}, \in R$ )

Both the equations have real roots

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Both the equations have imaginary roots

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at least one of the equations has real roots

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exactly one of them has equal roots

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Solution

The correct option is C
at least one of the equations has real roots

$Δ_{1} = p^{2} - 4 q; Δ_{2} = p_{1}^{2} - 4 q_{1}$

$Δ_{1} + Δ_{2} = p^{2} + p_{1}^{2} - 4 (q + q_{1}) = p^{2} + p_{1}^{2} - 2 p_{1} p = (p - p_{1})^{2} \geq 0$

Both $Δ_{1}, Δ_{2}$ cannot be negative.

Hence C.

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Similar questions

Q. If the equations

x^{2} + p x + q = 0

and

x^{2} + p_{1} x + q_{1} = 0

have a common roots, show that it must be either

\frac{p q_{1} - p_{1} q}{q - q_{1}}

\frac{q - q_{1}}{p_{1} - p}

Q. The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has

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. The equation has real roots which are of opposite signs, then the equation

α (x - β)^{2} + β (x - α)^{2} = 0

Q. If

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, then show that at least one of the equations

x^{2} + p x + q = 0

and

x^{2} + r x + s = 0

has real roots.

Q. If

a, b, c, d \in R,

then the equation

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If p1p=2(q1+q) then which of following statement about the quadratic equations x2+px+q=0;x2+p1x+q1=0 is always true (Here p,p1,q,q1,∈R)

The correct option is C at least one of the equations has real roots Δ1=p2−4q;Δ2=p21−4q1 Δ1+Δ2=p2+p21−4(q+q1)=p2+p21−2p1p=(p−p1)2≥0 Both Δ1,Δ2 cannot be negative. Hence C.

If $p_{1} p = 2 (q_{1} + q)$ then which of following statement about the quadratic equations $x^{2} + p x + q = 0; x^{2} + p_{1} x + q_{1} = 0$ is always true (Here $p, p_{1}, q, q_{1}, \in R$ )

The correct option is C
at least one of the equations has real roots

$Δ_{1} = p^{2} - 4 q; Δ_{2} = p_{1}^{2} - 4 q_{1}$

$Δ_{1} + Δ_{2} = p^{2} + p_{1}^{2} - 4 (q + q_{1}) = p^{2} + p_{1}^{2} - 2 p_{1} p = (p - p_{1})^{2} \geq 0$

Both $Δ_{1}, Δ_{2}$ cannot be negative.

Hence C.