Chapter 10, Problem 4CA

(a)

To determine

To find the value of a and b.

Answer to Problem 4CA

  $a=2$ and $b=-1$

Explanation of Solution

Given information: $x=\sqrt{ax+b}$ and 1 real solution

Formula Used: $D=b^2-4ac$

Calculation:

  $x=\sqrt{ax+b}$

Now, squaring both sides

  ${\left(x\right)}^2={\left(\sqrt{ax+b}\right)}^2$

So,

  $x^2=ax+b$

Now,

  $x^2-ax-b=0$

Discriminant be used to confirm the number of real solutions of a quadratic equation because discriminant method is much easier and shorter than completing the square method.

We can easily calculate discriminant by $D=b^2-4ac$ and can know how many real numbers equation has.

• If $D$ equal to negative then $x$ has no real solution.
• If $D$ equal to 0 then $x$ has 1 real solution.
• If $D$ equal to positive then $x$ has 2 real solution.

This equation has 1 real solution.

So,

  $D=0$

As,

  $D=b^2-4ac$

So,

  $b^2-4ac=0$

Now,

  ${\left(a\right)}^2-4\times 1\times \left(-b\right)=0$

Hence,

  $a^2+4b=0$

If $a=2$ and $b=-1$

Then,

  $a^2+4b=0$

(b)

To determine

To find the value of a and b.

Answer to Problem 4CA

  $a=-1$ and $b=1$

Explanation of Solution

Given information: $x=\sqrt{ax+b}$ and 2 real solutions

Formula Used: $D=b^2-4ac$

Calculation:

  $x=\sqrt{ax+b}$

Now, squaring both sides

  ${\left(x\right)}^2={\left(\sqrt{ax+b}\right)}^2$

So,

  $x^2=ax+b$

Now,

  $x^2-ax-b=0$

Discriminant be used to confirm the number of real solutions of a quadratic equation because discriminant method is much easier and shorter than completing the square method.

We can easily calculate discriminant by $D=b^2-4ac$ and can know how many real numbers equation has.

• If $D$ equal to negative then $x$ has no real solution.
• If $D$ equal to 0 then $x$ has 1 real solution.
• If $D$ equal to positive then $x$ has 2 real solution.

This equation has 2 real solution.

So,

  $D>0$

As,

  $D=b^2-4ac$

So,

  $b^2-4ac>0$

Now,

  ${\left(a\right)}^2-4\times 1\times \left(-b\right)>0$

Hence,

  $a^2+4b>0$

If $a=-1$ and $b=1$

Then,

  $a^2+4b>0$

(c)

To determine

To find the value of a and b.

Answer to Problem 4CA

  $a=1$ and $b=-1$

Explanation of Solution

Given information: $x=\sqrt{ax+b}$ and No real solutions

Formula Used: $D=b^2-4ac$

Calculation:

  $x=\sqrt{ax+b}$

Now, squaring both sides

  ${\left(x\right)}^2={\left(\sqrt{ax+b}\right)}^2$

So,

  $x^2=ax+b$

Now,

  $x^2-ax-b=0$

Discriminant be used to confirm the number of real solutions of a quadratic equation because discriminant method is much easier and shorter than completing the square method.

We can easily calculate discriminant by $D=b^2-4ac$ and can know how many real numbers equation has.

• If $D$ equal to negative then $x$ has no real solution.
• If $D$ equal to 0 then $x$ has 1 real solution.
• If $D$ equal to positive then $x$ has 2 real solution.

This equation has no real solution.

So,

  $D<0$

As,

  $D=b^2-4ac$

So,

  $b^2-4ac<0$

Now,

  ${\left(a\right)}^2-4\times 1\times \left(-b\right)>0$

Hence,

  $a^2+4b<0$

If $a=1$ and $b=-1$

Then,

  $a^2+4b<0$