Solve अभ्यास 7.1 1 . दिइएका मध्ये कुन कुन वर्ग समीकरण हुन् ? कार ( क ) (x - 2) ^ 2 + 1 = 2x - 3 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$( x - 2 ) ^ { 2 } + 1 = 2 x - 3$$

Solve for x

$x=2$
$x=4$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-2\right)^{2}$.

$$x^{2}-4x+4+1=2x-3$$

Add $4$ and $1$ to get $5$.

$$x^{2}-4x+5=2x-3$$

Subtract $2x$ from both sides.

$$x^{2}-4x+5-2x=-3$$

Combine $-4x$ and $-2x$ to get $-6x$.

$$x^{2}-6x+5=-3$$

Add $3$ to both sides.

$$x^{2}-6x+5+3=0$$

Add $5$ and $3$ to get $8$.

$$x^{2}-6x+8=0$$

To solve the equation, factor $x^{2}-6x+8$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-6$$ $$ab=8$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $8$.

$$-1,-8$$ $$-2,-4$$

Calculate the sum for each pair.

$$-1-8=-9$$ $$-2-4=-6$$

The solution is the pair that gives sum $-6$.

$$a=-4$$ $$b=-2$$

Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.

$$\left(x-4\right)\left(x-2\right)$$

To find equation solutions, solve $x-4=0$ and $x-2=0$.

$$x=4$$ $$x=2$$

Steps Using Factoring By Grouping

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-2\right)^{2}$.

$$x^{2}-4x+4+1=2x-3$$

Add $4$ and $1$ to get $5$.

$$x^{2}-4x+5=2x-3$$

Subtract $2x$ from both sides.

$$x^{2}-4x+5-2x=-3$$

Combine $-4x$ and $-2x$ to get $-6x$.

$$x^{2}-6x+5=-3$$

Add $3$ to both sides.

$$x^{2}-6x+5+3=0$$

Add $5$ and $3$ to get $8$.

$$x^{2}-6x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+8$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-6$$ $$ab=1\times 8=8$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $8$.

$$-1,-8$$ $$-2,-4$$

Calculate the sum for each pair.

$$-1-8=-9$$ $$-2-4=-6$$

The solution is the pair that gives sum $-6$.

$$a=-4$$ $$b=-2$$

Rewrite $x^{2}-6x+8$ as $\left(x^{2}-4x\right)+\left(-2x+8\right)$.

$$\left(x^{2}-4x\right)+\left(-2x+8\right)$$

Factor out $x$ in the first and $-2$ in the second group.

$$x\left(x-4\right)-2\left(x-4\right)$$

Factor out common term $x-4$ by using distributive property.

$$\left(x-4\right)\left(x-2\right)$$

To find equation solutions, solve $x-4=0$ and $x-2=0$.

$$x=4$$ $$x=2$$

Steps Using the Quadratic Formula

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-2\right)^{2}$.

$$x^{2}-4x+4+1=2x-3$$

Add $4$ and $1$ to get $5$.

$$x^{2}-4x+5=2x-3$$

Subtract $2x$ from both sides.

$$x^{2}-4x+5-2x=-3$$

Combine $-4x$ and $-2x$ to get $-6x$.

$$x^{2}-6x+5=-3$$

Add $3$ to both sides.

$$x^{2}-6x+5+3=0$$

Add $5$ and $3$ to get $8$.

$$x^{2}-6x+8=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-6$ for $b$, and $8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8}}{2}$$

Square $-6$.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 8}}{2}$$

Multiply $-4$ times $8$.

$$x=\frac{-\left(-6\right)±\sqrt{36-32}}{2}$$

Add $36$ to $-32$.

$$x=\frac{-\left(-6\right)±\sqrt{4}}{2}$$

Take the square root of $4$.

$$x=\frac{-\left(-6\right)±2}{2}$$

The opposite of $-6$ is $6$.

$$x=\frac{6±2}{2}$$

Now solve the equation $x=\frac{6±2}{2}$ when $±$ is plus. Add $6$ to $2$.

$$x=\frac{8}{2}$$

Divide $8$ by $2$.

$$x=4$$

Now solve the equation $x=\frac{6±2}{2}$ when $±$ is minus. Subtract $2$ from $6$.

$$x=\frac{4}{2}$$

Divide $4$ by $2$.

$$x=2$$

The equation is now solved.

$$x=4$$ $$x=2$$

Steps for Completing the Square

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-2\right)^{2}$.

$$x^{2}-4x+4+1=2x-3$$

Add $4$ and $1$ to get $5$.

$$x^{2}-4x+5=2x-3$$

Subtract $2x$ from both sides.

$$x^{2}-4x+5-2x=-3$$

Combine $-4x$ and $-2x$ to get $-6x$.

$$x^{2}-6x+5=-3$$

Subtract $5$ from both sides.

$$x^{2}-6x=-3-5$$

Subtract $5$ from $-3$ to get $-8$.

$$x^{2}-6x=-8$$

Divide $-6$, the coefficient of the $x$ term, by $2$ to get $-3$. Then add the square of $-3$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square $-3$.

$$x^{2}-6x+9=-8+9$$

Add $-8$ to $9$.

$$x^{2}-6x+9=1$$

Factor $x^{2}-6x+9$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-3\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-3=1$$ $$x-3=-1$$

Add $3$ to both sides of the equation.

$$x=4$$ $$x=2$$