Question

What is length and breadth of the rectangle whose area is 4a^2+ 4x – 3.


please give little short answer​

Answer 1

GivenArea =4a2+4a−3.We know thatArea of rectangle = length × breadth 

GivenArea =4a2+4a−3.We know thatArea of rectangle = length × breadth So, to find the possible expressions for the length and breadth we have to factorise the given expression.

4x2+4x−3=0

4x2+4x−3=04 {x}^{2} + 6x - 2x - 3 = 04x2+6x−2x−3=0

4x2+4x−3=04 {x}^{2} + 6x - 2x - 3 = 04x2+6x−2x−3=02x(2x + 3) - 1(2x + 3) = 02x(2x+3)−1(2x+3)=0

4x2+4x−3=04 {x}^{2} + 6x - 2x - 3 = 04x2+6x−2x−3=02x(2x + 3) - 1(2x + 3) = 02x(2x+3)−1(2x+3)=0(2x + 3)(2x - 1) = 0(2x+3)(2x−1)=0

4x2+4x−3=04 {x}^{2} + 6x - 2x - 3 = 04x2+6x−2x−3=02x(2x + 3) - 1(2x + 3) = 02x(2x+3)−1(2x+3)=0(2x + 3)(2x - 1) = 0(2x+3)(2x−1)=0so , either

4x2+4x−3=04 {x}^{2} + 6x - 2x - 3 = 04x2+6x−2x−3=02x(2x + 3) - 1(2x + 3) = 02x(2x+3)−1(2x+3)=0(2x + 3)(2x - 1) = 0(2x+3)(2x−1)=0so , either\begin{gathered}2x - 1 = 0 \\ \implies \: x = \frac{1}{2} \end{gathered}2x−1=0⟹x=21

4x2+4x−3=04 {x}^{2} + 6x - 2x - 3 = 04x2+6x−2x−3=02x(2x + 3) - 1(2x + 3) = 02x(2x+3)−1(2x+3)=0(2x + 3)(2x - 1) = 0(2x+3)(2x−1)=0so , either\begin{gathered}2x - 1 = 0 \\ \implies \: x = \frac{1}{2} \end{gathered}2x−1=0⟹x=21or

4x2+4x−3=04 {x}^{2} + 6x - 2x - 3 = 04x2+6x−2x−3=02x(2x + 3) - 1(2x + 3) = 02x(2x+3)−1(2x+3)=0(2x + 3)(2x - 1) = 0(2x+3)(2x−1)=0so , either\begin{gathered}2x - 1 = 0 \\ \implies \: x = \frac{1}{2} \end{gathered}2x−1=0⟹x=21or\begin{gathered}2x + 3 = 0 \\ \implies \: x = \frac{ - 3}{2} \: \: \: \: rejected\end{gathered}2x+3=0⟹x=2−3rejected

4x2+4x−3=04 {x}^{2} + 6x - 2x - 3 = 04x2+6x−2x−3=02x(2x + 3) - 1(2x + 3) = 02x(2x+3)−1(2x+3)=0(2x + 3)(2x - 1) = 0(2x+3)(2x−1)=0so , either\begin{gathered}2x - 1 = 0 \\ \implies \: x = \frac{1}{2} \end{gathered}2x−1=0⟹x=21or\begin{gathered}2x + 3 = 0 \\ \implies \: x = \frac{ - 3}{2} \: \: \: \: rejected\end{gathered}2x+3=0⟹x=2−3rejectedSo ,

4x2+4x−3=04 {x}^{2} + 6x - 2x - 3 = 04x2+6x−2x−3=02x(2x + 3) - 1(2x + 3) = 02x(2x+3)−1(2x+3)=0(2x + 3)(2x - 1) = 0(2x+3)(2x−1)=0so , either\begin{gathered}2x - 1 = 0 \\ \implies \: x = \frac{1}{2} \end{gathered}2x−1=0⟹x=21or\begin{gathered}2x + 3 = 0 \\ \implies \: x = \frac{ - 3}{2} \: \: \: \: rejected\end{gathered}2x+3=0⟹x=2−3rejectedSo ,Length = 1/2

Answer 2

$4 {x}^{2} + 4x - 3 = 0$

$4 {x}^{2} + 6x - 2x - 3 = 0$

$2x(2x + 3) - 1(2x + 3) = 0$

$(2x + 3)(2x - 1) = 0$

so , either

$2x - 1 = 0 \\ \implies \: x = \frac{1}{2}$

or

$2x + 3 = 0 \\ \implies \: x = \frac{ - 3}{2} \: \: \: \: rejected$

So ,

• Length = 1/2