Question

75.
which of the following is a quadratic equation


y(2y+15)=2(y2+y+8)


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


x(2x+3)=x+2


(x+2)3=2x(x2-1)​

Answer 1

Answer:

$x(2x + 3) = x + 2 \: \: is \: \: a \: \: quadractic \: \: equation$

Explanation:

Given:

Four equations:

$1) \: \: y(2y + 15) = 2( {y}^{2} + y + 8)$

$2) \: \: (x - 2)(x + 1) = (x - 1)(x + 3)$

$3) \: \: x(2x + 3) = (x + 2)$

$4) \: \: {(x + 2)}^{3} = 2x( {x}^{2} - 1)$

A quadratic equation is a equation of order ax²+bx+c where a, b and c are constants and x is the variable. In place of x, the variable can change.

In ax²+bx+c , the value of a, b and c can be negative as well as positive.

Example:

Equation 1) 2y²+8y+4=0

Here , a is +2 , b is +8 and c is +4.

Equation 2) -4a²+47a-13=0

Here , a is -4 , b is +47 and c is -13.

To Find:

Which of the above four equations are quadratic?

Answer:

Equation 1:

$y(2y + 15) = 2( {y}^{2} + y + 8)$

If we multiply and open the bracket, we get:

$2 {y}^{2} + 15y = 2 {y}^{2} + 2y + 16$

Bringing RH to LHS, we get:

$2 {y}^{2} - 2 {y}^{2} - 2y + 15 - 16$

If we perform the calculation, we get:

$- 2y - 1 = 0$

So, finally the equation has come to be a linear equation.

Therefore, equation 1 is linear and not a quadratic equation.

Equation 2:

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

If we multiply and open the bracket, we get:

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

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

Bringing RHS to LHS, we get:

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

If we perform the calculation, we get:

$- 3x + 1 = 0$

So, this equation has also become a linear equation.

Therefore, equation 2 is also a linear equation and not a quadratic equation.

Equation 4:

${(x + 2)}^{3} = 2x( {x}^{2} - 1)$

If we multiply and open the bracket, we get:

$(a + b)^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)$

$( {x}^{3} + 8 + 6x(x + 2)) = 2 {x}^{3} - 2x$

${x}^{3} + 8 + 6 {x}^{2} + 12x = 2 {x}^{3} - 2x$

Performing calculation, we get:

${x}^{3} - 6 {x}^{2} - 14x - 8 = 0$

So, this equation is neither linear nor quadratic. It is a cubic equation.

Therefore, equation 4 is cubic and not quadratic.

Equation 3:

$x(2x + 3) = x + 2$

If we multiply and open the bracket, we get:

$2 {x}^{2} + 3x = x + 2$

Bringing RHS to LHS, we get:

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

Performing calculation, we get:

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

So, equation 3 is a quadratic equation.

Therefore, answer is option 3.