How do you find all zeros of the function $f (x) = 4 x^{2} + 29 x + 30$ $f(x)=4x^2+29x+30$ ?

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Answer 1 · 2016-03-30T22:49:30

$x_{1} = - \frac{5}{4}$ $x_1=-5/4$ and $x_{2} = - 6$ $x_2=-6$

From the given equation $f (x) = 4 x^{2} + 29 x + 30$ $f(x) = 4x^2+29x+30$

Set $f (x) = 0$ $f(x)=0$

$4 x^{2} + 29 x + 30 = 0$ $4x^2+29x+30=0$

Try factoring

$(4 x + 5) (x + 6) = 0$ $(4x+5)(x+6)=0$

Set both factors to zero to solve for the roots

$4 x + 5 = 0$ $4x+5=0$

$4 x = - 5$ $4x=-5$

$x = - \frac{5}{4}$ $x=-5/4$ this is a zero

the other one

$x + 6 = 0$ $x+6=0$

$x = - 6$ $x=-6$
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We can use another method. Using the Quadratic Formula

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

From $4 x^{2} + 29 x + 30 = 0$ $4x^2+29x+30=0$ , let $a = 4$ $a=4$ , $b = 29$ $b=29$ , $c = 30$ $c=30$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- 29 \pm \sqrt{29^{2} - 4 (4) (30)}}{2 (4)}$ $x=(-29+-sqrt(29^2-4(4)(30)))/(2(4))$

$x = \frac{- 29 \pm \sqrt{841 - 480}}{8}$ $x=(-29+-sqrt(841-480))/(8)$

$x = \frac{- 29 \pm \sqrt{361}}{8}$ $x=(-29+-sqrt(361))/(8)$

$x = \frac{- 29 \pm 19}{8}$ $x=(-29+-19)/(8)$

$x_{1} = \frac{- 10}{8} = - \frac{5}{4}$ $x_1=(-10)/(8)=-5/4$

$x_{2} = \frac{- 48}{8} = - 6$ $x_2=(-48)/(8)=-6$

God bless....I hope the explanation is useful

How do you find all zeros of the function f(x)=4x2+29x+30f(x)=4x^2+29x+30?