If 4x 2 -4ax- a 2 - b 2 -0 and a-b- R - then vaue of x is equal to

The correct option is A a+b / 2 , a b / 2 Given equation is: 4x ^ 2 4ax+( a ^ 2 b ^ 2 ) =0 ⇒ #160; 4x ^ 2 { 2( a+b ) x+2( a b ) x } + a ^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Combination

If 4x 2 -4a...

Question

If ${4 x}^{2} - 4 a x + (a^{2} - b^{2}) = 0$ and $(a, b \in R)$ , then vaue of $x$ is equal to

\frac{a + b}{2}, \frac{a - b}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(a + b)}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

2 (b c o s^{2} \frac{C}{2} + c c o s^{2} \frac{B}{2})

A : \int \frac{1}{4 + 5 sin x} d x = \frac{1}{3} log ∣ ∣ ∣ \frac{2 tan (x / 2) + 1}{2 tan (x / 2) + 4} ∣ ∣ ∣ + c

R

0 < a < b

\int \frac{d x}{a + b sin x} = \frac{1}{\sqrt{b^{2} - a^{2}}} log ∣ ∣ ∣ ∣ \frac{a tan (x / 2) + b - \sqrt{b^{2} - a^{2}}}{a tan (x / 2) + b + \sqrt{b^{2} - a^{2}}} ∣ ∣ ∣ ∣ + c

a^{2} + b^{2} + c^{2} - a b - b c - c a \leq 0 \forall a, b, c \in R

∣ ∣ ∣ ∣ ∣ \begin{matrix} (a - b + 2)^{2} & a^{2} - b^{2} & 1 1 & (b - c + 2)^{2} & b^{2} - c^{2} c^{2} - a^{2} & 1 & (c - a + 2)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

\frac{a + i b}{a - i b}

\frac{2 a}{a^{2} + b^{2}}

\frac{2 a b}{a^{2} - b^{2}}

\frac{a^{2} - b^{2}}{a^{2} + b^{2}}

(1 + i) (1 + 2 i) (1 + 3 i) . . . (1 + n i) = a + i b, then 2 \times 5 \times 10 \times . . . . \times (1 + n^{2})

\sqrt{a^{2} + b^{2}}

\sqrt{a^{2} - b^{2}}

a^{2} + b^{2}

a^{2} - b^{2}

a + b

Join BYJU'S Learning Program