Advertisements
Advertisements
प्रश्न
Solve` x^2-4ax+4a^2-b^2=0`
उत्तर
`x^2-4ax+4a^2-b^2=0`
⇒`x^2-4ax+(2a+b) (2a-b)=0`
⇒`x^2-[(2a+b)+(2a-b)]x+(2a+b) (2a-b)=0`
⇒`x^2-(2a+b)x-(2a-b)x+(2a+b)(2a-b)=0`
⇒`x[x-(2a+b)]-(2a-b)[x-(2a+b)=0]`
⇒`[x-(2a+b)][x-(2a-b)]=0`
⇒`x-(2a+b)=0 or x-(2a-b)=0`
⇒`x=(2a+b) or x=(2a-b)`
Hence, `(2a+b) and (2a-b)` are the roots of the given equation.
APPEARS IN
संबंधित प्रश्न
Show the x= -3 is a solution of `x^2+6x+9=0`
If one root of the quadratic equation `3x^2-10x+k=0` is reciprocal of the other , find the value of k.
Write the following equation in the form ax2 + bx + c= 0, then write the values of a, b, c for the equation.
x2 + 5x = –(3 – x)
Suyash scored 10 marks more in second test than that in the first. 5 times the score of the second test is the same as square of the score in the first test. Find his score in the first test.
Choose the correct answer for the following question.
The roots of x2 + kx + k = 0 are real and equal, find k.
Solve the following quadratic equation.
\[x^2 - \frac{3x}{10} - \frac{1}{10} = 0\]
Solve the following quadratic equation.
x2 - 4x - 3 = 0
Convert 15 : 20 into percentage.
If P(y) = y² - 2y + 5, find P(2) .
In a right-angled triangle, altitude is 2 cm longer than its base. Find the dimensions of the right-angled triangle given that the length of its hypotenuse is 10 cm.
