Answer:

(x+ 8/3 ) ^2 =− 64/71

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Step-by-step explanation:

4x ^2 +3x+5=0

x ^2 +3/ 4x+ 5/4=0

x ^2 +2( 3/8x)x=− 5/4

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x ^2 +2( 3/8x)+( 3/8 )^2 =( 3/8 ) ^2 −5/ 4

(x+ 3/8 ) ^2 =−71/ 64

I might've done it wrong I just tired to do it the way my professor taught me. There's a lot of numbers sorry! (  the "/" are fraction symbols)  ​

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Verified answer

Stepwise Explanation

The method complete the square is used in quadratic equation solving.

[tex]\text{$\rm 4{x}^{2}+3x+5=0$}[/tex]

[tex]\;[/tex]

It is an equal manipulation on both sides that allows us to solve equations.

[tex]\begin{aligned}&\bullet \textrm{Addition}\\&\bullet \textrm{Subtraction}\\&\bullet \textrm{Multiplication}\\&\bullet \textrm{Nonzero division}\end{aligned}[/tex]

[tex]\;[/tex]

Polynomial Identity

[tex]\boxed{\rm (a+b)^{2}={a}^{2}+2ab+{b}^{2}}[/tex]

[tex]\;[/tex]

First, we divide by four to eliminate the leading coefficient.

[tex]\text{$\rm \dfrac{1}{4} \cdot (4{x}^{2}+3x+5)=0$}[/tex]

[tex]\text{$\rm {x}^{2}+\dfrac{3}{4}x+\dfrac{5}{4}=0$}[/tex]

It is better to transpose the constant term.

[tex]\;[/tex]

To complete the square of

[tex]\begin{aligned}&\text{$\rm (a+b)^{2}={a}^{2}+2ab+{b}^{2}$}\end{aligned}[/tex]

-we need a constant term.

[tex]\;[/tex]

What we have here is [tex]\text{$\rm {a}^{2}+2ab$}[/tex]. To find [tex]\text{$\rm {b}^{2}$}[/tex], we can divide the linear term coefficient by two and then square it.

[tex]\;[/tex]

[tex]\text{$\rm {x}^{2}+\dfrac{3}{4}x=-\dfrac{5}{4}$}[/tex]

[tex]\text{$\rm {x}^{2}+\dfrac{3}{4}x+\bigg(\dfrac{3}{8}\bigg)^{2}=\bigg(\dfrac{3}{8}\bigg)^{2}-\dfrac{5}{4}$}[/tex]

[tex]\text{$\rm {\bigg(x+\dfrac{3}{8}\bigg)}^{2}=-\dfrac{71}{64}$}[/tex]

[tex]\;[/tex]

Trichotomy

Real numbers are positive, zero, or negative. Their square is non-negative.

[tex]\;[/tex]

Hence, such real roots don't exist.

[tex]\;[/tex][tex]\;[/tex]

The More, The Better

Derivation

The generalized method for [tex]\text{$\rm a{x}^{2}+bx+c=0$, $\rm a\neq 0$}[/tex] is called the quadratic formula.

[tex]\;[/tex]

Proof

[tex]\text{$\rm a{x}^{2}+bx+c=0$, $\rm a\neq 0$}[/tex]

[tex]\text{$\rm {x}^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=0$}[/tex]

[tex]\text{$\rm {x}^{2}+\dfrac{b}{a}x=-\dfrac{c}{a}$}[/tex]

[tex]\text{$\rm {x}^{2}+\dfrac{b}{a}x+{\bigg(\dfrac{b}{2a}\bigg)}^{2}={\bigg(\dfrac{b}{2a}\bigg)}^{2}-\dfrac{c}{a}$}[/tex]

[tex]\text{$\rm {\bigg(x+\dfrac{b}{2a}\bigg)}^{2}=\dfrac{b^{2}}{4{a}^{2}}-\dfrac{c}{a}$}[/tex]

[tex]\text{$\rm {\bigg(x+\dfrac{b}{2a}\bigg)}^{2}=\dfrac{b^{2}-4ac}{4{a}^{2}}$}[/tex]

[tex]\;[/tex]

Square Root

To solve the equation, we can use the calculation opposite of squaring a number. It is the square root.

[tex]\text{$\rm x+\dfrac{b}{2a}=\pm \dfrac{\sqrt{b^{2}-4ac}}{2a}$}[/tex]

[tex]\text{$\rm \therefore x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$}[/tex]

[tex]\;[/tex]

Hence it is proven.

[tex]\;[/tex]

Even Coefficient Formula

Suppose we have a quadratic equation [tex]\text{$\rm a{x}^{2}+2b'x+c=0$}[/tex]. Substituting the coefficients into the formula, we have-

[tex]\text{$\rm \therefore x=\dfrac{-b'\pm\sqrt{b'^{2}-ac}}{a}$}[/tex]

[tex]\;[/tex]

Discriminant

In the derivation of the quadratic formula, we found [tex]\text{$\rm b^{2}-4ac$}[/tex]. The nature of roots differs from the square root.

[tex]\begin{aligned}&\bullet D > 0: \text{Real and distinct roots}\\&\bullet D=0: \text{Real and equal roots}\\&\bullet D < 0: \text{Imaginary and distinct roots}\\\end{aligned}[/tex]