Advertisements
Advertisements
प्रश्न
Solve using formula.
x2 – 3x – 2 = 0
उत्तर
x2 – 3x – 2 = 0
On comparing with the equation \[a x^2 + bx + c = 0\]
a = 1, b = \[-3\] and c = \[-2\]
Now
\[b^2 - 4ac = \left( - 3 \right)^2 - 4 \times 1 \times \left( - 2 \right) = 9 + 8 = 17\]
\[x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[x = \frac{3 \pm \sqrt{17}}{2 \times 1} = \frac{3 \pm \sqrt{17}}{2}\]
\[\Rightarrow x = \frac{3 + \sqrt{17}}{2} \text{ or } x = \frac{3 - \sqrt{17}}{2}\]
APPEARS IN
संबंधित प्रश्न
Compare the given quadratic equation to the general form and write values of a,b, c.
2m2 = 5m – 5
Solve using formula.
3m2 + 2m – 7 = 0
Solve using formula.
y2 + `1/3`y = 2.
Find the value of discriminant of the following equation.
5m2 - m = 0
Find the value of discriminant of the following equation.
\[\sqrt{5} x^2 - x - \sqrt{5} = 0\]
Two roots of quadratic equation is given ; frame the equation.
10 and –10
Two roots of quadratic equation is given ; frame the equation.
\[1 - 3\sqrt{5} \text{ and } 1 + 3\sqrt{5}\]
Two roots of quadratic equation is given ; frame the equation.
0 and 7
Determine the nature of root of the quadratic equation.
\[3 x^2 - 5x + 7 = 0\]
Determine the nature of root of the quadratic equation.
\[\sqrt{3} x^2 + \sqrt{2}x - 2\sqrt{3} = 0\]
Determine the nature of root of the quadratic equation.
m2 - 2m + 1 = 0
Find m if (m – 12) x2 + 2(m – 12) x + 2 = 0 has real and equal roots.
The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.
The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.
If 2 and 5 are the roots of the quadratic equation, then complete the following activity to form quadratic equation:
Activity:
Let α = 2 and β = 5 are the roots of the quadratic equation.
Then quadratic equation is:
x2 − (α + β)x + αβ = 0
∴ `x^2 - (2 + square)x + square xx 5 = 0`
∴ `x^2 - square x + square = 0`
