Question

plz solve this question ,i have posted this question more than 10 times but no one is answering properly so if you the answer then only post it , so that you will get marked as brainlist
thank you ​

Answer 1

Question 1

What are the roots of quadratic equation (x-2)(x+3)=0

Given

• equation- (x-2)(x+3)=0

To Find

• value of x or roots of the equation

Solution

First we have to made this equation in the form of quadratic

$\rm \: (x - 2)(x + 3) = 0 \\ \\ \implies \rm \: {x}^{2} + 3x - 2x - 6 = 0 \\ \\ \implies \boxed{ \bf \: {x}^{2} + x - 6 = 0 } \: \: \: \: \: \: \: \:$

Now we have two methods to do this question quadratic formula and Middle term split

Easiest method to do this question is Middle term split

$\rm \therefore \: {x}^{2} + x - 6 = 0 \\ \\ \implies \rm \: {x}^{2} + 3x - 2x - 6 = 0 \\ \\ \implies \rm \: x(x + 3) - 2(x + 3) = 0 \\ \\ \implies \rm \: (x - 2)(x + 3) = 0 \: \: \: \: \: \: \: \: \: \\ \\ \implies \boxed{ \huge \bf \: x = 2 \: or \: - 3}$

Question 2

If the discriminant of the quadratic equation kx^2-6x+1=0 is 0

Given

• Discriminant of the above quadratic equation is 0

To Find

• value of k

Concept used

$\boxed{\bf \: d = {b}^{2} - 4ac }$

Solution

We have the discriminant and we know that

$\rm \: d = {b}^{2} - 4ac$

Here

• d=0
• a=k
• b=-6
• c=1

$\bf \: putting \: values \\ \\ \rm \: 0 = { (- 6)}^{2} - 4 \times k \times 1 \\ \\ \implies \rm \: 0 = 36 - 4k \: \: \: \: \: \\ \\ \implies \rm \: 4k = 36 \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \rm \: k = \frac{36}{4} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \underline{ \boxed{ \huge \bf \: k = 9}}$

Hence the value of k is 9