Number of real roots of $(- 3) x^{4} + 13 x^{2} - 36$

Open in App

Solution

$f' (x) = - 12 x^{3} + 26 x f' (X) = 0 26 x = 12 x^{3} (x \neq 0) x^{2} = \frac{26}{12} x = \pm \sqrt{\frac{13}{6}}, α = \sqrt{\frac{13}{6}}, β = - \sqrt{\frac{13}{6}} f (α) = f (β) < 0$
Curve has zero roots.

Suggest Corrections

Similar questions

Find the roots of the following equation

3 x^{4} - 13 x^{2} + 10 = 0

, then

Q. The number of distinct real roots of equation

3 x^{4} + 4 x^{3} - 12 x^{2} + 4 = 0

Q. 3x⁴ – 13x² + 10 = 0

Q. Solve the quadratic equation

3 x^{4} - 13 x^{2} + 10 = 0

Find the number of positive real roots for equation 3 $x^{5}$ - 4 $x^{4}$ - 42 $x^{3}$ + 56 $x^{2}$ + 27x - 36 = 0. If all the roots of above given equation are real.

Join BYJU'S Learning Program

Number of real roots of (−3)x4+13x2−36

f′(x)=−12x3+26xf′(X)=026x=12x3(x≠0)x2=2612x=±√136,α=√136,β=−√136f(α)=f(β)<0Curve has zero roots.

Number of real roots of $(- 3) x^{4} + 13 x^{2} - 36$

$f' (x) = - 12 x^{3} + 26 x f' (X) = 0 26 x = 12 x^{3} (x \neq 0) x^{2} = \frac{26}{12} x = \pm \sqrt{\frac{13}{6}}, α = \sqrt{\frac{13}{6}}, β = - \sqrt{\frac{13}{6}} f (α) = f (β) < 0$
Curve has zero roots.