Advertisements
Advertisements
प्रश्न
Solve the quadratic equation: `x^2 + 2sqrt(2)x - 6` = 0 for x.
Advertisements
उत्तर
We have, `x^2 + 2sqrt(2)x - 6` = 0
Now, by using the formula x = `(-b +- sqrt(b^2 - 4ac))/(2a)`, we get
x = `(-2sqrt(2) +- sqrt((2sqrt(2))^2 - 4 xx 1 xx (-6)))/(2 xx 1)`
⇒ x = `(-2sqrt(2) +- sqrt(8 + 24))/2`
⇒ x = `(-2sqrt(2) +- 4sqrt(2))/2`
= `- sqrt(2) +- 2sqrt(2)`
⇒ x = `-sqrt(2) + 2sqrt(2)`, x = `-sqrt(2) - 2sqrt(2)`
⇒ x = `sqrt(2)`, x = `-3sqrt(2)`
APPEARS IN
संबंधित प्रश्न
Solve for x : ` 2x^2+6sqrt3x-60=0`
Solve for x: `1/(x+1)+2/(x+2)=4/(x+4), `x ≠ -1, -2, -3
Without solving, examine the nature of roots of the equation 2x2 + 2x + 3 = 0
Find the values of k for which the roots are real and equal in each of the following equation:
k2x2 - 2(2k - 1)x + 4 = 0
Determine the nature of the roots of the following quadratic equation :
x2 -4x + 4=0
Find the nature of the roots of the following quadratic equations: `x^2 - 2sqrt(3)x - 1` = 0 If real roots exist, find them.
Find the least positive value of k for which the equation x2 + kx + 4 = 0 has real roots.
Find the values of k so that the quadratic equation (4 – k) x2 + 2 (k + 2) x + (8k + 1) = 0 has equal roots.
State whether the following quadratic equation have two distinct real roots. Justify your answer.
3x2 – 4x + 1 = 0
Find whether the following equation have real roots. If real roots exist, find them.
8x2 + 2x – 3 = 0
